Electric circuits: Physics unit 17 notes (FBISE class 10)

17.1Circuit elements and diagrams

An electric circuit is an interconnection of electrical elements (components) joined in a specific way. Electric circuit theory is the foundation of power, electric machines, control, electronics, communications and instrumentation, and circuits work inside radios, TVs, cell phones, computers and even the nervous systems of animals.

Electric circuit: An interconnection of electrical elements, formed by connecting electrical components in a specific way. It is a closed pathway through which electricity flows, typically made of a power source, wires and components such as resistors.

A basic circuit, like the one in Figure 17.1, typically includes conductors (often wires), a power source (such as a battery), a load (for instance a lamp) and a switch to turn the circuit on and off. A circuit of this kind can be used independently for purposes such as a torch light or a search light. The key thing to remember is that a circuit creates a closed loop, starting and ending at the same point.

Figure 17.1: A simple circuit. The battery (power source), wires, switch and bulb (load) form one closed loop.

Bare copper wire can be used as a conductor for demonstration, but insulated wires are generally used for safety reasons.

Active and passive elements

Electric circuits consist of two types of elements.

Active and passive elements.
Type	What it does	Examples
Active element	Can supply or generate energy	Batteries, generators
Passive element	Cannot generate energy; it absorbs, stores or dissipates energy	Resistors, capacitors, inductors

Circuit diagrams

Circuit diagram: A simplified map showing how the various components are interconnected. It helps in understanding the flow of current.

In a physical circuit (Figure 17.2(a) of the book) the components are connected with electrical wires. In a circuit diagram (Figure 17.2(b)) the wires are drawn as straight lines and each component is drawn as a standard symbol. Using symbols helps us recognise the electric components and the flow of electricity. Such diagrams give a fundamental understanding of electrical systems and of the relationships between voltage, current and resistance.

Figure 17.2(b): The circuit diagram of a physical circuit made of a cell (

ε

), a resistor

R

and a capacitor

C

. Wires become straight lines and every component becomes its symbol.

Figure 17.3 shows the symbols used for the electric and electronic components in a circuit diagram. They are redrawn below in three groups.

Figure 17.3 (a): Symbols for wires and connectors, and for power supplies and earth. Wires that cross are joined only where there is a dot.

Figure 17.3 (b): Symbols for resistors (fixed, variable, thermistor, LDR), capacitor, lamp, heating element and electric motor.

Figure 17.3 (c): Symbols for measuring devices, switches, a safety fuse, diodes (including LED and photodiode), transformer and relay.

17.2Resistors and the colour codeOften examined

Tiny components in electronics are essential for managing the flow of electricity. Among them the resistor stands out as a key player, controlling the current in a circuit.

Resistor: The component of an electric circuit whose function is to provide a specified value of resistance.

Unlike some components that generate power, resistors are passive. They consume a small amount of electrical energy and dissipate it as heat, thereby regulating the overall power flow within the circuit. The unit of resistance is the ohm ( $Ω$ ).

Carbon resistors

The most commonly used resistors in electrical and electronic circuits are carbon resistors. A carbon resistor consists of a ceramic core on which a thin layer of crystalline carbon is deposited. Since it is quite small, it is more convenient to use a colour code that indicates the resistance value than to imprint the numerical value on the case. These resistors are cost-effective, reliable and small in size.

Table 17.1: Resistor colour codes.
Colour	Number	Multiplier	Tolerance
Black	0	$1 0^{0}$
Brown	1	$1 0^{1}$
Red	2	$1 0^{2}$
Orange	3	$1 0^{3}$
Yellow	4	$1 0^{4}$
Green	5	$1 0^{5}$
Blue	6	$1 0^{6}$
Violet	7	$1 0^{7}$
Grey	8	$1 0^{8}$
White	9	$1 0^{9}$
Gold		$1 0^{- 1}$	$\pm 5 %$
Silver		$1 0^{- 2}$	$\pm 10 %$
Colourless (no band)			$\pm 20 %$

Figure 17.4: A carbon resistor. The first two bands give the digits, the third is the decimal multiplier and the fourth is the tolerance. Green, blue, orange and gold read as

56 \times 1 0^{3} Ω \pm 5 %

.

Colour bands represent the resistance value according to Table 17.1. Usually three colour bands signify the resistance: the first two bands are the significant figures, and the third band is the decimal multiplier that follows them. The fourth band, gold or silver, indicates the tolerance of the resistor. If there is no fourth band, the tolerance is 20 %.

R = (10 a + b) \times 1 0^{c} Ω (a, b : first and second digits; c : multiplier power)

the way the three bands are combined

Reading Figure 17.4

For the resistor of Figure 17.4 the first band is green, so the first digit is 5.
The second band is blue, so the second digit is 6.
The third band is orange, so the multiplier is $1 0^{3}$ .
$R = 56 \times 1 0^{3} Ω = 56 kΩ$
The fourth band is gold, so the tolerance is $5 %$ (taking gold as $\pm 5 %$ ). The value is written collectively as $56 \times 1 0^{3} \pm 5 % Ω$ .

Worked example: reading a three-band resistor

A resistor has three bands, yellow, violet and orange, and no fourth band. Find its resistance and the range in which its true value may lie.

Given

$a = 4 (yellow)$
$b = 7 (violet)$
$c = 3 (orange)$

To find

resistance $R$
range of possible values

Read the first two bands as digits: yellow is 4 and violet is 7, so the significant figures are $47$ .
$47$
The third band is the multiplier, the power of ten: orange means $1 0^{3}$ , not three zeros typed after a digit you guess.
$R = 47 \times 1 0^{3} Ω = 47 kΩ$
There is no fourth band, so the tolerance is $20 %$ (Table 17.1).
$Δ R = 20 % of 47000 = 0.20 \times 47000 = 9400 Ω$
The true value lies between $R - Δ R$ and $R + Δ R$ .
$47000 - 9400 = 37600 Ω 47000 + 9400 = 56400 Ω$
Check: $56400$ is $1.2$ times $47000$ and $37600$ is $0.8$ times $47000$ , which is $\pm 20 %$ .

The three bands of this resistor and its reading.

R = 47 kΩ (\pm 20 %) : 37.6 kΩ to 56.4 kΩ

17.3.1Linear resistors: fixed and variable

Normally resistors are classified into two types: linear resistors and non-linear resistors.

Linear resistors

Linear resistor: A resistor whose current is precisely proportional to the applied voltage, that is, one that obeys Ohm's law. There are two types of linear resistors: fixed and variable.

A. Fixed resistors

A fixed resistor has a single resistance value. Fixed resistors are made from a variety of resistor materials; the most common material is carbon. A fixed resistor may be readily attached to the circuit and can handle high voltages. However, it is less stable, which means its temperature coefficient is quite high.

B. Variable resistors

A variable resistor allows its resistance value to be adjusted. Usually a mechanical movement is used to modify the resistance. High-stability and high-accuracy resistors are always wire wound.

A wire of a certain length has a fixed resistance, and increasing the length of the wire increases the resistance.
So a wire of known length and resistance is wound around and enclosed in an insulating cover to make the resistor.
It is made variable by putting a sliding contact on it, which varies the length of wire in the circuit and thus the resistance (Figure 17.5).
Nickel and chromium are generally used as the wire material because of their low temperature coefficient of resistance.

Figure 17.5 (drawn schematically): A wire-wound variable resistor. Moving the sliding contact

C

changes the length of wire between

A

and

C

, and so changes the resistance. The book shows photographs of a rheostat and a potentiometer.

Two uses of a wire-wound variable resistor.
Name	Used as	Purpose
Rheostat	A variable resistance	To regulate the current in a circuit
Potentiometer	A variable voltage divider	To obtain a variable potential difference from a fixed voltage applied

A potential divider is a circuit that divides a voltage into two or more smaller voltages. Using it we can obtain a variable potential difference from a fixed voltage applied. It is described in the next section.

17.3.1Potential divider

Potential divider: A circuit that produces an output voltage $V_{o u t}$ that is a fraction of its input voltage $V_{in}$ . It is a simple and fundamental circuit in electronics, used to generate a voltage that is a fraction of the input voltage.

Derivation Output voltage of a potential divider

Aim. Find the potential difference $V_{2}$ that is tapped off, in terms of the applied voltage $V$ and the resistances.

Assumptions.

The output is taken between the sliding contact C and the end B, with nothing drawn from it (no current leaves the output).

Figure 17.6: A potential divider. The battery

V

is applied across the whole wire AB of resistance

R = R_{1} + R_{2}

; the output

V_{2}

is taken between the sliding contact C and B.

A fixed voltage $V$ is applied across the ends A and B of a wire-wound variable resistor of total resistance $R$ . By Ohm's law the current is $I = V / R$ .
$I = \frac{V}{R} (1)$
The sliding contact C divides the wire into two parts. Let $R_{2}$ be the resistance of the portion between C and B. The same current $I$ flows through it, so the potential difference across it is
$V_{2} = I R_{2} (2)$
Similarly, if $R_{1}$ is the resistance of the portion AC, the potential difference across it is
$V_{1} = I R_{1} (3)$
Divide equation (3) by equation (2). The current $I$ cancels.
$\frac{V _{1}}{V _{2}} = \frac{R _{1}}{R _{2}}$
Put equation (1) in equation (2).
$V_{2} = \frac{R _{2}}{R} V$

Result

\frac{V _{1}}{V _{2}} = \frac{R _{1}}{R _{2}} (17.1) V_{2} = \frac{R _{2}}{R} V (17.2)

The potential difference $V_{2}$ is some fraction of the total potential applied, depending on the position of the sliding contact. Moving the contact varies the potential difference from zero to $V$ .

Since $V_{2}$ is the output voltage $V_{o u t}$ , $V$ is the input voltage $V_{in}$ and $R = R_{1} + R_{2}$ , equation 17.2 can also be written as

V_{o u t} = \frac{R _{2}}{R _{1} + R _{2}} V_{in}

As a special case, if $R_{1} = R_{2} = r$ :

V_{o u t} = \frac{r}{r + r} V_{in} = \frac{r}{2 r} V_{in} so V_{o u t} = \frac{1}{2} V_{in}

This means that the input voltage is split into two equal halves at the output.

Output of a potential divider

0.25

0.5

0.75

1

2

3

4

5

6

O

\frac{R _{2}}{R}

V_{2} (V)

\frac{1}{2} V

V

V_{2} = \frac{R _{2}}{R} V

For a uniform wire the resistance

R_{2}

is proportional to its length, so

V_{2} = (R_{2} / R) V

rises in a straight line as the slider moves from B to A (here

V = 6 V

), where the horizontal axis is the fraction

R_{2} / R

(the slider position measured from B). At the middle the output is half the input.

Potential dividers are commonly used in electronics to adjust the level of a voltage signal or to measure voltage accurately. They are used in the volume control knob of music systems. Sensory circuits using light-dependent resistors and thermistors also use potential dividers.

Worked example: a potential divider with an LDR

A

12 V

supply is joined across a fixed resistor

R_{1} = 6 kΩ

in series with a light dependent resistor (LDR). The output is taken across the LDR. In bright light the LDR has

2 kΩ

and in the dark it has

18 kΩ

. (a) Draw the circuit diagram. (b) Find the output voltage in bright light and in the dark. (c) Explain why the output rises when it gets dark.

Given

$V_{in} = 12 V$
$R_{1} = 6 kΩ$
$R_{L D R} (bright) = 2 kΩ$
$R_{L D R} (dark) = 18 kΩ$

To find

circuit diagram
$V_{o u t}$ in bright light and in the dark
reason for the change

(a) The circuit: the

12 V

battery, the fixed resistor

R_{1}

and the LDR in series. The output is taken across the LDR.

The LDR is the resistor across which the output is taken, so it plays the part of $R_{2}$ in $V_{o u t} = \frac{R _{2}}{R _{1} + R _{2}} V_{in}$ . Both resistances can stay in kilohms because they appear as a ratio.
$V_{o u t} = \frac{R _{L D R}}{R _{1} + R _{L D R}} V_{in}$
Bright light.
$V_{o u t} = \frac{2}{6 + 2} \times 12 = \frac{2}{8} \times 12 = 3 V$
Dark.
$V_{o u t} = \frac{18}{6 + 18} \times 12 = \frac{18}{24} \times 12 = 9 V$
Check with equation 17.1: the voltage across $R_{1}$ is the rest of the $12 V$ . In bright light it is $9 V$ and $9/3 = 3 = 6/2$ ; in the dark it is $3 V$ and $3/9 = \frac{1}{3} = 6/18$ . Both agree.

(c) In low light an LDR has a high resistance. A larger share of the $12 V$ is then dropped across it, because the voltages divide in the ratio of the resistances. In bright light its resistance falls, so it takes a smaller share and the output drops. This is how the divider senses light.

V_{o u t} = 3 V (bright) 9 V (dark)

17.3.2Non-linear resistors: thermistors, varistors and LDRs

In non-linear resistors the resistance value fluctuates in response to temperature, light or voltage. These resistors are further divided into thermistors, varistors and photo resistors.

A. Thermistor

Thermistor: A type of resistor whose resistance varies significantly with temperature.

Thermistor is short for thermal resistor: a device with heat-sensitive resistance, so its temperature coefficient is very high. The material used in a thermistor is generally a semiconductor ceramic or polymer (Figure 17.7 shows a thermistor and its circuit symbol, which also appears in Figure 17.3(b)). Thermistors are of two types.

Positive temperature coefficient (PTC) thermistor: Its resistance increases with the increase of temperature.
Negative temperature coefficient (NTC) thermistor: Its resistance decreases with the increase of temperature.

Resistance against temperature (sketch)

Sketch of the two types of thermistor (the shapes are only illustrative, not measured values). The resistance of an NTC thermistor falls as it gets hotter; the resistance of a PTC thermistor rises.

Thermistors of different sizes and shapes are formed according to the need of the industry. Their sensitivity to a change in temperature enables them to perform many functions.

Inrush current limiter. An inrush (input surge) current is the peak current drawn by an electrical device when powered on. For example, incandescent bulbs have high inrush currents until their filaments heat up. NTC thermistors can limit the inrush current in power supply circuits by initially showing high resistance, which restricts large currents during start-up.
Overheating alarm. NTC thermistors can trigger alarms when the temperature of motor, transformer or generator windings rises. They maintain high resistance at lower temperatures, allowing minimal current flow. As the temperature increases, the resistance drops, allowing more current to flow to an alarm system. This principle is also applicable in fire alarms and thermostats.
Temperature measurement. Thermistors provide precise temperature measurements because of their high temperature coefficient. Their resistance changes with temperature, enabling effective detection of and response to temperature variations.

Science tidbit: resistance thermometers

Resistance thermometers, also called Resistance Temperature Detectors (RTDs), are sensors used to measure temperature by correlating the resistance of the RTD element with temperature. Most RTD elements consist of a length of fine coiled wire wrapped around a ceramic or glass core. The element is made from a pure material such as platinum, nickel or copper that has a predictable change in resistance as the temperature changes, and this predictable change is used to determine temperature. Platinum is the best metal for RTDs because it follows a very linear resistance-temperature relationship in a highly repeatable manner over a wide temperature range. Its properties make it the material of choice for temperature standards, with the range from -185^\circ\mathrm{C} to 630^\circ\mathrm{C}.

B. Varistors (VDR)

Varistors (Figure 17.8) are voltage dependent resistors (VDR). They are a type of variable resistor used to protect circuits from destructive voltage spikes. When the voltage across a connected sensitive device or system increases (due to lightning or line faults), the varistor reduces the voltage to a secure level. Varistors are typically made from metal oxide materials, with zinc oxide (ZnO) the most common. They consist of a ceramic body of zinc oxide grains interspersed with other metal oxides such as bismuth, cobalt and manganese. The grain boundaries act as the non-linear junctions that give the varistor its voltage-dependent characteristics.

C. Light dependent resistor (LDR)

Light dependent resistor (LDR): Also known as a photo resistor or photocell: a light-sensitive device that changes resistance with the intensity of light.

In low light conditions an LDR has a high resistance, which decreases as the light intensity increases (Figure 17.9). LDRs are made from materials such as cadmium sulphide and lead sulphide. The surface of the LDR is exposed to light with a serpentine pattern to increase light exposure. LDRs sense light levels and adjust the resistance accordingly, making them useful in applications like automatic lighting control, photography and security systems. They are versatile and cost-effective components used to detect and respond to changes in light.

Resistance of an LDR (sketch)

Sketch of an LDR: the resistance is high in low light and falls as the light intensity increases. The shape is only illustrative.

Non-linear resistors compared.
Resistor	What changes its resistance	Behaviour	Use named in the book
NTC thermistor	Temperature	Resistance falls as temperature rises	Inrush current limiter, overheating alarm, temperature measurement
PTC thermistor	Temperature	Resistance rises as temperature rises	None named in the book
Varistor (VDR)	Voltage	Reduces the voltage to a secure level when it rises	Protects circuits from voltage spikes
LDR (photo resistor)	Light intensity	High resistance in low light, falls as light increases	Automatic lighting control, photography, security systems

17.4.1Combination of resistors: seriesOften examined

For any combination of resistors that obey Ohm's law we can always find a single resistor that could replace the combination and give the same total current and potential difference. The resistance of this single resistor is called the equivalent resistance of the combination. For series and parallel combinations we can derive general equations for the equivalent resistance.

Equivalent resistance (R_{eq}): The resistance of the single resistor that can replace a combination of resistors and result in the same total current and potential difference.

Resistors in series

In a series circuit there is only one path for the charges to follow, so the same current flows in each device. The voltage is divided among the devices: the voltage on the first device plus the voltage on the second device, and so on, equals the voltage of the power supply.

For example, if three light bulbs of the same resistance are connected in series to a 12 V battery, the voltage on each bulb is 4 V. If the bulbs had different resistances, each one's share of the voltage would be proportional to its resistance.

Derivation Equivalent resistance of resistors in series

Aim. Find the single resistance R_{eq} that can replace three resistors R_1, R_2 and R_3 joined in series across a potential difference V.

Assumptions.

The connecting wires have negligible resistance.
The same current flows through every resistor because there is only one path.

Figure 17.10: Series combination of resistors. The same current I flows through R_1, R_2 and R_3, and the potential differences V_1, V_2, V_3 add up to the battery potential difference V.

In a series circuit (Figure 17.10) the same current I flows through every resistor.
I_1=I_2=I_3=I
Apply Ohm's law V=IR to each resistor.
V_1=IR_1,\qquad V_2=IR_2,\qquad V_3=IR_3
The total potential difference V is the sum of the potential differences across the individual resistors.
V=V_1+V_2+V_3
Put the Ohm's law expressions in and take I as a common factor.
V=I(R_1+R_2+R_3)\qquad\text{or}\qquad\dfrac{V}{I}=R_1+R_2+R_3
By definition, the equivalent resistance is the ratio of the total potential difference to the current.
R_{eq}=\dfrac{V}{I}
Compare the last two lines.
R_{eq}=R_1+R_2+R_3
The same reasoning holds for any number n of resistors.
R_{eq}=R_1+R_2+R_3+R_4+\cdots+R_n

ResultR_{eq}=R_1+R_2+R_3\quad(17.3)\qquad R_{eq}=R_1+R_2+\cdots+R_n\quad(17.4)

The equivalent resistance of any number of resistors in series equals the sum of their individual resistances. It is always greater than any individual resistance.

Example 17.1: three resistors in series

Three resistors R_1=10\,\Omega, R_2=20\,\Omega and R_3=30\,\Omega are connected in series across a 20\,\mathrm{V} battery. Find the equivalent resistance and the voltage drop across each resistor.

Given

V = 20\,\mathrm{V}
R_1 = 10\,\Omega
R_2 = 20\,\Omega
R_3 = 30\,\Omega

To find

equivalent resistance R_{eq}
voltage drops V_1, V_2, V_3

For a series combination the equivalent resistance is the sum of the resistances.
R_{eq}=R_1+R_2+R_3=10+20+30=60\,\Omega
The same current flows through all the resistors: I_1=I_2=I_3=V/R_{eq}.
I=\dfrac{20.0\,\mathrm{V}}{60\,\Omega}=0.333\,\mathrm{A}
The voltage drop splits among the resistors. Across R_1:
V_1=IR_1=0.333\times10=3.33\,\mathrm{V}
Across R_2:
V_2=IR_2=0.333\times20=6.67\,\mathrm{V}
Across R_3:
V_3=IR_3=0.333\times30=9.99\,\mathrm{V}\approx10\,\mathrm{V}
Keeping I=\tfrac13\,\mathrm{A} exactly gives 10.0\,\mathrm{V}; the book's 9.99\,\mathrm{V} comes from rounding I to 0.333\,\mathrm{A}.
Check: the drops add up to the battery voltage.
V_1+V_2+V_3=3.33+6.67+9.99\approx20\,\mathrm{V}=V

R_{eq}=60\,\Omega;\quad V_1=3.33\,\mathrm{V},\ V_2=6.67\,\mathrm{V},\ V_3=9.99\,\mathrm{V}

17.4.2Combination of resistors: parallelOften examined

Resistors in parallel

In a parallel circuit the current from the power supply is "shared" among the devices while each has the same voltage. The current flowing in the first device plus the current in the second device, and so on, equals the current output by the power supply. There is more than one path for the charges to follow (three, in Figure 17.11). If one of the devices burns out or is removed, the others still function. The light bulbs in multiple-bulb light fixtures are in parallel so that if one bulb burns out the others remain lit.

Derivation Equivalent resistance of resistors in parallel

Aim. Find the single resistance R_{eq} that can replace three resistors R_1, R_2 and R_3 joined in parallel across a potential difference V.

Assumptions.

The connecting wires have negligible resistance.
All the resistors are joined directly across the same two points, so each has the full potential difference V.

Figure 17.11: Parallel combination of resistors. The same potential difference V is across every branch; the currents I_1, I_2, I_3 add up to the total current I.

For resistors in parallel (Figure 17.11) the potential difference between the terminals of each resistor must be the same and equal to V.
V_1=V_2=V_3=V
Call the currents in the three resistors I_1, I_2, I_3. By Ohm's law,
I_1=\dfrac{V}{R_1},\qquad I_2=\dfrac{V}{R_2},\qquad I_3=\dfrac{V}{R_3}
Charge is neither accumulating nor draining out, so the total current I equals the sum of the three currents.
I=I_1+I_2+I_3=V\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\right)
Divide both sides by V.
\dfrac{I}{V}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}
By definition the equivalent resistance gives I/V=1/R_{eq}.
\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}
For any number n of resistors, the reciprocal of the equivalent resistance equals the sum of the reciprocals.
\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}+\cdots+\dfrac{1}{R_n}

Result\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\quad(17.5)\qquad\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\cdots+\dfrac{1}{R_n}\quad(17.6)

The equivalent resistance is always less than any individual resistance. This is because each new branch gives the current one more path.

Series and parallel in failure

The primary distinction between series and parallel connections lies in how they respond to component failure. In a series circuit all the components are connected along a single current pathway. If one element, such as a lamp, fails, the entire circuit is interrupted and all the devices stop working (Figure 17.12(a)). Parallel circuits offer multiple independent paths for the current. Each device, like a lamp in your home, is directly connected to the power source. If one lamp fails, the current simply passes through the remaining branches, allowing the other lamps to continue shining (Figure 17.12(b)). This redundancy makes parallel circuits ideal for powering multiple devices in homes and buildings.

Figure 17.12: Series and parallel combinations of light bulbs. In (a) one blown bulb breaks the only path, so all the bulbs go off. In (b) the other branches still carry current, so the other bulbs stay on.

Additionally, parallel circuits provide individual control. You can turn off a single lamp without affecting the others, unlike a series circuit where turning off one device disrupts the entire circuit. This independent control adds to the practicality and flexibility of parallel connections. Household appliances are always connected in parallel so that even if one is switched off, the other devices function properly.

Series and parallel combinations compared.
	Series	Parallel
Paths for the current	One	More than one
Current	The same in every resistor	Shared: I=I_1+I_2+\cdots
Potential difference	Shared: V=V_1+V_2+\cdots	The same across every resistor
Equivalent resistance	R_{eq}=R_1+R_2+\cdots+R_n (greater than every one)	\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\cdots+\dfrac{1}{R_n} (less than every one)
If one device fails	All stop	The others continue

Equivalent resistance of identical 10\,\Omega resistors

Adding more identical 10\,\Omega resistors raises R_{eq} in series (R_{eq}=nR) and lowers it in parallel (R_{eq}=R/n). With five resistors the series value is 50\,\Omega and the parallel value is 2\,\Omega.

Example 17.2: three resistors in parallel

Three resistors R_1=5.0\,\Omega, R_2=10.0\,\Omega and R_3=15.0\,\Omega are connected in parallel across a 12.0\,\mathrm{V} battery. Find the equivalent resistance and the current that passes through each resistor.

Given

V = 12\,\mathrm{V}
R_1 = 5.0\,\Omega
R_2 = 10.0\,\Omega
R_3 = 15.0\,\Omega

To find

equivalent resistance R_{eq}
currents I_1, I_2, I_3

For a parallel combination use the sum of reciprocals.
\dfrac{1}{R_{eq}}=\dfrac{1}{5.0}+\dfrac{1}{10.0}+\dfrac{1}{15.0}=\dfrac{6+3+2}{30}=\dfrac{11}{30}\,\Omega^{-1}
Invert to get the equivalent resistance.
R_{eq}=\dfrac{30}{11}=2.73\,\Omega
The voltage across every resistor is the same, V_1=V_2=V_3=12.0\,\mathrm{V}. The total current is
I=\dfrac{V}{R_{eq}}=\dfrac{12.0}{2.73}=4.4\,\mathrm{A}
The current splits among the resistors. Through R_1:
I_1=\dfrac{V}{R_1}=\dfrac{12.0}{5.0}=2.4\,\mathrm{A}
Through R_2:
I_2=\dfrac{V}{R_2}=\dfrac{12.0}{10.0}=1.2\,\mathrm{A}
Through R_3:
I_3=\dfrac{V}{R_3}=\dfrac{12.0}{15.0}=0.8\,\mathrm{A}
Check: the branch currents add up to the total current.
I_1+I_2+I_3=2.4+1.2+0.8=4.4\,\mathrm{A}=I

R_{eq}=2.73\,\Omega;\quad I_1=2.4\,\mathrm{A},\ I_2=1.2\,\mathrm{A},\ I_3=0.8\,\mathrm{A}

Worked example: a series-parallel combination

A 4\,\Omega resistor is joined in series with a parallel pair of 6\,\Omega and 12\,\Omega resistors across a 12\,\mathrm{V} battery, as in the figure. Find (a) the equivalent resistance of the whole circuit, (b) the current drawn from the battery, (c) the potential difference across the parallel pair and the current in each of its resistors.

Given

V = 12\,\mathrm{V}
R_1 = 4\,\Omega\ (\text{in series})
R_2 = 6\,\Omega\ (\text{parallel})
R_3 = 12\,\Omega\ (\text{parallel})

To find

R_{eq} of the circuit
current I
potential difference across the pair and the currents I_2, I_3

The circuit: R_1 in series with the parallel pair R_2 and R_3.

First replace the parallel pair by one resistor R_p (equation 17.6).
\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{2+1}{12}=\dfrac{3}{12}\ \Rightarrow\ R_p=4\,\Omega
The 4\,\Omega resistor and R_p are now in series (equation 17.4).
R_{eq}=R_1+R_p=4+4=8\,\Omega
The current from the battery is the current through the series part.
I=\dfrac{V}{R_{eq}}=\dfrac{12}{8}=1.5\,\mathrm{A}
This current flows through R_1 and then splits. The potential difference across R_1 is IR_1, and the rest of the 12\,\mathrm{V} is across the pair.
V_1=1.5\times4=6\,\mathrm{V}\qquad V_p=12-6=6\,\mathrm{V}
Each resistor of the pair has this 6\,\mathrm{V} across it.
I_2=\dfrac{6}{6}=1\,\mathrm{A}\qquad I_3=\dfrac{6}{12}=0.5\,\mathrm{A}
Check: I_2+I_3=1.5\,\mathrm{A}=I, and R_p=\dfrac{6\times12}{6+12}=4\,\Omega again.

R_{eq}=8\,\Omega;\quad I=1.5\,\mathrm{A};\quad V_p=6\,\mathrm{V};\quad I_2=1\,\mathrm{A},\ I_3=0.5\,\mathrm{A}

17.5Combination of emf sources

A combination of emf sources means that two or more emf sources are connected in such a way that a combined effect of these sources is obtained. Normally there are two possible combinations: series and parallel.

Series combination of emf sources

In a series combination the positive terminal of an emf source, like a battery, is connected to the negative terminal of another source, and so on. Finally the negative terminal of the first source is connected to the positive terminal of the last source through other circuit elements such as bulbs and fans. In Figure 17.13 four batteries are connected in series, so the emfs of all the batteries add and give the resultant emf (6\,\mathrm{V}) to the external circuit.

Figure 17.13: Series connection of batteries. Four 1.5\,\mathrm{V} batteries in series give 6.0\,\mathrm{V}.

For n batteries connected in series we can write

\varepsilon_{net}=\varepsilon_1+\varepsilon_2+\varepsilon_3+\cdots+\varepsilon_n

17.7

A series combination of emf sources is used when a higher voltage is required than the voltage of the available sources, and no change in the ampere-hour (Ah) values is needed. If, for example, the above batteries are 1.5\,\mathrm{V}, 2.5\,\mathrm{Ah} each, the equivalent rating when connected in series will be 6.0\,\mathrm{V}, 2.5\,\mathrm{Ah}.

Parallel combination of emf sources

In a parallel combination the positive terminals of all the sources are connected together and the negative terminals of all the sources are connected together. Finally the load is connected by taking one connection from the common positive terminal and one from the common negative terminal (Figure 17.14). When identical sources are connected in parallel, the equivalent emf is equal to the emf of one of the sources.

Figure 17.14: Parallel connection of batteries. All the positive terminals are joined to one rail and all the negative terminals to the other, so the output is 1.5\,\mathrm{V}, the emf of one battery.

For n identical batteries connected in parallel we can write

\varepsilon_{net}=\varepsilon_1=\varepsilon_2=\varepsilon_3=\cdots=\varepsilon_n

17.8

A parallel combination of emf sources is used when a higher ampere-hour value is required than that of the available sources, and no change in the voltage is needed.

Combining emf sources.
Connection	Net emf	Ampere-hour rating	Used when
Series (+ to −)	Sum: \varepsilon_1+\varepsilon_2+\cdots+\varepsilon_n	Unchanged (that of one battery)	A higher voltage is required
Parallel (+ to +, − to −), identical sources	Equal to the emf of one source	Increases (adds)	A higher ampere-hour value is required

Example 17.3: cells in series

A series combination consists of n identical cells. If each cell has an emf of 0.4\,\mathrm{V} and the total emf of the series combination is 12\,\mathrm{V}, find n.

Given

\varepsilon = 0.4\,\mathrm{V}\ \text{(each cell)}
\varepsilon_{net} = 12\,\mathrm{V}

To find

number of cells n

In a series combination the emfs of the individual cells add.
\varepsilon_{net}=\varepsilon_1+\varepsilon_2+\varepsilon_3+\cdots+\varepsilon_n
All the cells are identical, so each term is \varepsilon and there are n terms.
\varepsilon_{net}=n\varepsilon\quad\text{or}\quad n=\dfrac{\varepsilon_{net}}{\varepsilon}
Put in the values.
n=\dfrac{12\,\mathrm{V}}{0.4\,\mathrm{V}}=30
So 30 identical cells of emf 0.4\,\mathrm{V} arranged in series give a total emf of 12\,\mathrm{V}. Check: 30\times0.4=12\,\mathrm{V}.

n=30

17.6Electricity and its uses

Electricity is a versatile form of energy that serves numerous purposes in our daily lives. From providing illumination to powering electronic devices, its applications are vast and diverse.

A. Heating: Electricity keeps us warm during colder months. Electric heaters, from portable space heaters to central heating systems, work by converting electrical energy into heat, which is distributed to keep a comfortable temperature. Electric stoves cook our meals by generating heat through electrical resistance in the stove's heating elements.
B. Lighting: Electric lighting brightens our surroundings indoors and outdoors, at any time of the day. In homes, electric light bulbs illuminate our living spaces for reading, cooking and socialising. Streetlights illuminate roads and pathways, enhancing the safety and visibility of pedestrians and motorists during the night.
C. Battery charging: Portable electronic devices such as smartphones, laptops and tablets rely on rechargeable batteries. Electricity powers the charging process, replenishing the energy stored in these batteries so that the devices keep running smoothly at home, at work or on the go.
D. Powering motors: Electric motors are the driving force behind a wide range of machinery, appliances and transportation vehicles: household appliances such as washing machines and vacuum cleaners, industrial equipment such as conveyor belts and manufacturing machinery, and electric vehicles (EVs), a cleaner and more sustainable alternative to traditional internal combustion engines.
E. Powering electronic systems: Computers enable us to work, communicate and access information. Televisions entertain us. Refrigerators keep food fresh while air conditioners maintain comfortable indoor temperatures. Smartphones, gaming consoles and home automation devices also rely on electricity.

Uses of electricity (Figure 17.15 of the book shows a heater, lighting, power lines, a motor and an electric vehicle).
Use	Device or system named in the book	Energy change
Heating	Space heaters, central heating, electric stoves	Electrical energy to heat
Lighting	Light bulbs, streetlights	Electrical energy to light
Battery charging	Smartphones, laptops, tablets	Electrical energy stored in the battery
Powering motors	Washing machines, vacuum cleaners, conveyor belts, electric vehicles	Electrical energy to mechanical energy
Powering electronic systems	Computers, televisions, refrigerators, air conditioners	Electrical energy to run the device

17.7Electrical energyOften examined

When a voltage source such as a battery is connected to a device like a bulb (or any other device), the source delivers some electrical energy to it. There is a potential difference across the two ends of the device (Figure 17.16 of the book shows a battery, a switch and a bulb). The energy delivered is found from the definition of potential difference.

Derivation Electrical energy consumed by a device

Aim. Find the energy E delivered to a device in terms of the current I, the potential difference V and the time t.

Assumptions.

All of the energy supplied is delivered to the device (no energy is lost in the connecting wires).

The potential difference across the device is the work (energy) per unit charge, V=W/q=E/q. Make E the subject.
E=qV\qquad(1)
From the definition of electric current, the charge that passes in time t is
q=It\qquad(2)
Put equation (2) in equation (1).
E=ItV
Replace V by IR (Ohm's law) to get the energy in terms of the current and resistance.
E=I\,t\,(IR)=I^{2}Rt
Replace I by V/R to get the energy in terms of the voltage and resistance.
E=\dfrac{V}{R}\,t\,V=\dfrac{V^{2}}{R}\,t

ResultE=IVt\ \ (17.9)\qquad E=I^{2}Rt\ \ (17.10)\qquad E=\dfrac{V^{2}}{R}\,t\ \ (17.11)

All three equations (17.9, 17.10 and 17.11) give relations for calculating the electrical energy consumed. The SI unit of energy is the joule.

Which form of the energy equation to choose.
Form	Use it when you know	Handy for
E=IVt	I, V and t	Any device
E=I^{2}Rt	I, R and t	Resistors in series (the same current)
E=\dfrac{V^{2}}{R}\,t	V, R and t	Resistors in parallel (the same voltage)

Example 17.4: energy used by an electric iron

An electric iron working at a potential difference of 200\,\mathrm{V} draws 7\,\mathrm{A} of current. How much electrical energy will it consume in 1 hour?

Given

V = 200\,\mathrm{V}
I = 7\,\mathrm{A}
t = 1\ \mathrm{h}=60\times60\,\mathrm{s}=3600\,\mathrm{s}

To find

electrical energy E

Time must be in seconds for the answer to be in joules. The energy is given by equation 17.9.
E=ItV
Put in the values.
E=(7\,\mathrm{A})\times(3600\,\mathrm{s})\times(200\,\mathrm{V})=5\,040\,000\,\mathrm{J}
Write it in standard form.
E=5.04\times10^{6}\,\mathrm{J}=5.04\,\mathrm{MJ}

E=5.04\times10^{6}\,\mathrm{J}

17.8Electric powerOften examined

Electric power: The rate at which electric work is done by the voltage source (that is, the battery) in maintaining the electric current in the circuit. Equivalently, the electrical energy consumed by a device per unit time.

Mathematically,

P=\dfrac{E}{t}

1

Derivation Three forms of electric power

Aim. Write the power in terms of the current, voltage and resistance.

Use the electrical energy E=ItV in P=E/t.
P=\dfrac{ItV}{t}
The time cancels.
P=IV
Put V=IR (Ohm's law) in P=IV.
P=I\,(IR)=I^{2}R
Put I=V/R in P=IV.
P=\dfrac{V}{R}\,V=\dfrac{V^{2}}{R}

ResultP=IV\ \ (17.12)\qquad P=I^{2}R\ \ (17.13)\qquad P=\dfrac{V^{2}}{R}\ \ (17.14)

All three equations are equally valid for calculating electric power. The SI unit of electrical power is the watt (W). A power of 1\,\mathrm{W} is consumed in an electric circuit if a potential difference of 1\,\mathrm{V} causes a current of 1\,\mathrm{A} to flow.

Example 17.5: power of an electric toaster

What is the power consumption of an electric toaster if its resistance is 19\,\Omega and it operates on a household circuit of 200\,\mathrm{V}?

Given

R = 19\,\Omega
V = 200\,\mathrm{V}

To find

power P

By Ohm's law find the current.
I=\dfrac{V}{R}=\dfrac{200}{19}=10.5\,\mathrm{A}
The power is the product of current and voltage (equation 17.12).
P=IV=10.5\times200=2100\,\mathrm{W}
Using the exact I=10.526\,\mathrm{A}, or P=V^{2}/R=40\,000/19, gives 2105\,\mathrm{W}; the book rounds to 2100\,\mathrm{W}.

P\approx2100\,\mathrm{W}

Example 17.6: will the resistor be damaged?

A resistor has a value of 100\,\Omega and a power rating of \tfrac12\,\mathrm{W}. If the resistor is connected to a 10\,\mathrm{V} power supply, will it be damaged?

Given

R = 100\,\Omega
P_R = 0.5\,\mathrm{W}\ \text{(rated power)}
V = 10\,\mathrm{V}

To find

power dissipated P, and whether it exceeds the rating

Find the heat dissipated in the resistor with the power formula that uses V and R.
P=\dfrac{V^{2}}{R}=\dfrac{(10\,\mathrm{V})^{2}}{100\,\Omega}=1\,\mathrm{W}
Compare with the rating: 1\,\mathrm{W} is twice the rated 0.5\,\mathrm{W}. The resistor will be damaged.

P=1\,\mathrm{W}>0.5\,\mathrm{W}\ \Rightarrow\ \text{the resistor is damaged}

Power in a 10\,\Omega resistor

P=I^{2}R for R=10\,\Omega is a parabola. Doubling the current (from 1\,\mathrm{A} to 2\,\mathrm{A}, or 2\,\mathrm{A} to 4\,\mathrm{A}) makes the power four times as large.

Worked example: the same two resistors in series and in parallel

Two identical 10\,\Omega resistors are connected to the same 12\,\mathrm{V} battery, first in series and then in parallel. Find the total power dissipated in each arrangement and compare them.

Given

R = 10\,\Omega\ \text{(each)}
V = 12\,\mathrm{V}

To find

total power in series P_s
total power in parallel P_p
the ratio P_p/P_s

(a) The two resistors in series and (b) in parallel across the same battery.

Series. Add the resistances (17.4).
R_s=10+10=20\,\Omega
The battery voltage is across the whole series pair, so use P=V^{2}/R.
P_s=\dfrac{V^{2}}{R_s}=\dfrac{12^{2}}{20}=\dfrac{144}{20}=7.2\,\mathrm{W}
Check with the current: I=12/20=0.6\,\mathrm{A} and P_s=I^{2}R_s=0.36\times20=7.2\,\mathrm{W}.
Parallel. For two equal resistors (17.6) R_p=R/2.
R_p=\dfrac{10}{2}=5\,\Omega
Each resistor has the full 12\,\mathrm{V} across it, so each dissipates 144/10=14.4\,\mathrm{W}. The total is
P_p=\dfrac{V^{2}}{R_p}=\dfrac{144}{5}=28.8\,\mathrm{W}
Compare.
\dfrac{P_p}{P_s}=\dfrac{28.8}{7.2}=4

P_s=7.2\,\mathrm{W},\quad P_p=28.8\,\mathrm{W},\quad P_p=4P_s

17.8Kilowatt-hour and the cost of electricityOften examined

Does the unit kilowatt-hour (kWh) measure electric energy or electric power? The unit has "watt", a unit of power, in it, which suggests that it might be a unit of power. However, the kilowatt-hour measures electric energy, not electric power.

Electric power is the rate at which electrical energy is transferred or consumed. It is measured in watts (W) or kilowatts (kW), where 1\,\mathrm{kW}=1000\,\mathrm{W}. Power is an instantaneous measurement: it shows how much energy is being used or generated at a specific moment. Electric energy, on the other hand, is the total amount of electrical work done or energy consumed over a period of time. It is measured in kilowatt-hours because electrical energy is consumed in homes and industry in large amounts, for which the joule is a small unit.

Kilowatt-hour (kWh): The unit of electrical energy commonly used to measure consumption or production over time, equal to the energy used by one kilowatt of power in one hour. It is also called a commercial unit.

1\,\mathrm{kWh}=1000\,\mathrm{W}\times1\,\mathrm{h}=1000\,\mathrm{J\,s^{-1}}\times3600\,\mathrm{s}=3.6\times10^{6}\,\mathrm{J}

One kilowatt-hour is the energy consumed when a power of one kilowatt is used for one hour. For example, a device rated at 500\,\mathrm{W} (0.5\,\mathrm{kW}) operating for 2 hours consumes 1\,\mathrm{kWh} of energy (0.5\,\mathrm{kW}\times2\,\mathrm{h}=1\,\mathrm{kWh}).

Energy used at a steady power

E=Pt: the energy in kWh grows in a straight line with time and the gradient of each line is the power. The 0.5\,\mathrm{kW} device of the book's example reaches 1\,\mathrm{kWh} after 2 hours.

Cost of electricity: the energy used (in kWh, the commercial unit) is multiplied by the tariff rate, the price per kWh.

\text{Cost}=E\ (\mathrm{kWh})\times\text{Tariff rate (price per kWh)}

Example 17.7: the monthly cost of an air conditioner

An air conditioner of power 2200\,\mathrm{W} is used for an average of 24 hours daily. How much electrical energy will it consume, and what will be the cost of electricity for a month (30 days), if the cost of electricity is Rs. 25.0 per kWh?

Given

P = 2200\,\mathrm{W}=2.2\,\mathrm{kW}
t = 30\times24=720\,\mathrm{h}
\text{tariff} = \text{Rs. }25.0\ \text{per kWh}

To find

energy E in kWh
cost

Power is energy per time, so E=Pt. Use kilowatts and hours so that the energy comes out in kWh.
E=Pt=2.2\,\mathrm{kW}\times720\,\mathrm{h}=1584\,\mathrm{kWh}
The total energy consumed is 1584 kWh (or 1584 commercial units). Multiply by the tariff to get the cost.
\text{Cost}=1584\times25.0=39\,600\ \text{rupees}

E=1584\,\mathrm{kWh};\quad\text{Cost}=\text{Rs. }39\,600

Worked example: a household electricity bill

In a house a 1.5\,\mathrm{kW} heater runs for 4 hours every day and three 60\,\mathrm{W} lamps are each on for 5 hours every day. For a month of 30 days find (a) the energy used by the heater and by the lamps in kWh, (b) the cost at Rs. 30 per kWh, (c) the total energy in joules.

Given

P_h = 1.5\,\mathrm{kW}
t_h = 4\,\mathrm{h/day}
P_l = 3\times60\,\mathrm{W}=180\,\mathrm{W}=0.18\,\mathrm{kW}
t_l = 5\,\mathrm{h/day}
\text{tariff} = \text{Rs. }30\ \text{per kWh}

To find

energy in kWh for 30 days
cost
energy in J

Change the lamps' power to kilowatts before using E=Pt (60\,\mathrm{W} is 0.06\,\mathrm{kW}, three lamps are 0.18\,\mathrm{kW}).
Heater, energy per day then for the month.
E_h=1.5\times4=6\,\mathrm{kWh/day}\ \Rightarrow\ 6\times30=180\,\mathrm{kWh}
Lamps, energy per day then for the month.
E_l=0.18\times5=0.9\,\mathrm{kWh/day}\ \Rightarrow\ 0.9\times30=27\,\mathrm{kWh}
Total energy for the month.
E=180+27=207\,\mathrm{kWh}
Cost.
\text{Cost}=207\times30=\text{Rs. }6210
Convert to joules with 1\,\mathrm{kWh}=3.6\times10^{6}\,\mathrm{J}.
E=207\times3.6\times10^{6}=7.45\times10^{8}\,\mathrm{J}

E=207\,\mathrm{kWh},\quad\text{Cost}=\text{Rs. }6210,\quad E=7.45\times10^{8}\,\mathrm{J}

17.9Household circuits and the wiring systemOften examined

Even though household circuits use alternating current, which has not been discussed yet, their functioning (along with numerous practical applications) can be understood by applying the same circuit principles just studied. In household wiring the electrical elements should be connected in parallel. For example, when a bulb in a lamp in your kitchen burns out, other appliances on the same circuit, like the coffee maker, should still work. Additionally, household appliances and lights are typically designed to work at around 220\,\mathrm{V}. If they were connected in series, each appliance would only receive a fraction of 220\,\mathrm{V}.

Household wiring system

A mains circuit is composed of three wires: a live wire (line wire), a neutral wire and an earth wire (Figure 17.17). Each wire has a distinct function in guaranteeing the secure distribution and use of electricity.

The three wires of a mains circuit.
Wire	Function
Live (line) wire	Carries the electrical current from the power source to the electrical devices or appliances. It carries the potential difference (voltage) that powers the devices.
Neutral wire	Completes the electrical circuit. It provides a return path for the electrical current back to the power source. It carries the same current as the live wire, but with opposite polarity.
Earth (grounding) wire	A safety measure in case of electrical faults or malfunctions. It provides a low-resistance path for the current to flow into the ground, preventing electric shocks and reducing the risk of fire. It acts as a safety backup, redirecting the current away from people or objects in case of a fault.

Figure 17.17 (simplified): A mains circuit. The live wire reaches the appliance through the fuse and switch, the neutral wire returns the current, and the earth wire joins the metal case of the appliance to the ground.

The book's Figure 17.17 shows the whole house: the supply cable (live and neutral wires) passes through a main fuse (marked 60\,\mathrm{A}), the kWh meter and the main switch to the breaker box. The breaker box has three circuits protected at 5\,\mathrm{A} (the bulb and fan, each with its own switch), 15\,\mathrm{A} (a heater) and 30\,\mathrm{A} (a ring main circuit with sockets), and the earth connection is marked E.

17.9.2Fuses, circuit breakers and the earth wireOften examined

Fuses

To avoid overloading, the current is restricted by using fuses or circuit breakers. Fuses (Figure 17.18) are found in older homes. Inside the fuse is a metal strip that melts when the current is larger than the rated value (the threads of a fuse, for example, have specific current ratings). The melting of the strip opens the circuit, and the current drops to zero.

Fuse: A safety device containing a metal strip that melts when the current exceeds its rated value, opening the circuit.

Circuit breakers

Newer homes use circuit breakers exclusively. One type uses a bimetallic strip (Figure 17.19). When the current flowing through the strip increases, the strip heats up and bends. Once the current reaches its rated value, the strip bends enough to open the circuit (trip elements using magnetic principles also exist). Afterwards the strip cools down, allowing the breaker to be reset.

Figure 17.19(b): A circuit breaker with a bimetallic strip, in two states. Left: normal current, the strip is straight and the contact is closed. Right: an overload heats the strip, it bends and the contact is broken. Figure 17.19(a) of the book is a photograph of a bank of household breakers.

If a fuse blows or a circuit breaker trips, it indicates that the circuit is drawing excessive current. It is important to identify and fix the problem before replacing the fuse or resetting the circuit breaker. It is crucial not to temporarily replace a blown fuse with one that has a higher current rating, because this can lead to further damage or even fire hazards.

The earth wire and the metal case

Adding a third (grounding) wire to the circuit helps prevent electric shocks by providing a low-resistance path through the metal casing of appliances. If a hot (live) wire touches the casing, the circuit is completed through this wire, blowing the fuse or tripping the circuit breaker. In case of contact, the current would flow through the ground wire, minimising harm to you. For this reason the outer casing of an appliance must be either non-conducting (double-insulated) or earthed.

Worked example: choosing a fuse and the earth fault

A heater of power 2000\,\mathrm{W} and a lamp of power 200\,\mathrm{W} are on one circuit of a house supplied at 220\,\mathrm{V}. (a) Which of the protective ratings 5\,\mathrm{A}, 15\,\mathrm{A} and 30\,\mathrm{A} of the breaker box should protect it? (b) The live wire inside the metal-cased heater comes loose and touches the case. Explain what happens when the case is earthed and when it is not.

Given

P_{heater} = 2000\,\mathrm{W}
P_{lamp} = 200\,\mathrm{W}
V = 220\,\mathrm{V}

To find

the rating of the fuse
what happens in the fault

Both appliances are in parallel across 220\,\mathrm{V}, so their powers add.
P=2000+200=2200\,\mathrm{W}
The current drawn is found from P=IV.
I=\dfrac{P}{V}=\dfrac{2200}{220}=10\,\mathrm{A}
The rating must be above the normal current of 10\,\mathrm{A}, or it would blow every time the circuit is used, so 5\,\mathrm{A} is too small. The next rating, 15\,\mathrm{A}, is enough. 30\,\mathrm{A} is unnecessarily high: it would allow a dangerous overload to flow before the circuit is cut off. Choose 15\,\mathrm{A}.

(b) The earthed heater with a fault: the live wire touches the metal case, and the fault current flows through the earth wire.

(b) With the earth wire connected, the casing joins the live wire to the ground through a path of very low resistance. A large current flows, which is far above the rating, so the fuse melts (or the breaker trips) at once and the live wire is cut off. The case is then safe to touch. Without the earth wire the case simply becomes live and stays at a high potential, with the fuse not blowing, until someone touches it; the current then passes through that person to the ground.

\text{fuse rating}=15\,\mathrm{A}\ (\text{normal current}=10\,\mathrm{A})

17.9.4Residual-current devices (RCDs)

Residual-current devices (RCDs), also known as ground fault circuit interrupters (GFCIs), are modern safety devices designed to protect against electric shock and to reduce the risk of electrical fires. They do this by detecting differences in the electrical current between the live (hot) and neutral wires of a circuit. Ideally the current in both wires should be equal but opposite in direction.

Figure 17.20: Ground fault circuit interrupter (GFCI). In (a) the live and neutral currents are the same but in opposite directions. In (b) a ground fault leaks 0.1\,\mathrm{A} through the person, so the live wire carries more current than the neutral wire.

For example, in Figure 17.20 a current of 10 amperes flows through both the live and the neutral wires passing through an electric kettle. A ground fault happens when there is an unintended path for the electric current to flow to the ground. Imagine someone accidentally touching a faulty appliance (the kettle). A new path for the current can be created through the body. Let us assume there is a current of 0.1\,\mathrm{A} flowing through the person. This means that the live wire now carries 10.1\,\mathrm{A}, while the neutral wire still has 10\,\mathrm{A}.

The figures in the example

The current that goes in by the live wire normally comes back by the neutral wire.
I_{live}=I_{neutral}=10\,\mathrm{A}
With a ground fault, some current leaks to the earth through the person.
I_{leak}=I_{live}-I_{neutral}=10.1-10=0.1\,\mathrm{A}=100\,\mathrm{mA}
The GFCI is designed to detect a difference of about 5\,\mathrm{mA} (0.005\,\mathrm{A}) and interrupt the circuit. The leak here is 100/5=20 times that, so the device cuts the circuit at once.
100\,\mathrm{mA}\gg5\,\mathrm{mA}

To protect people from such situations, GFCIs are designed to detect this difference in current and interrupt the circuit when they detect a difference of around 5 milliamperes (0.005 ampere). This helps to prevent electric shocks and potential harm to individuals.

17.9.5Electrical hazards and the effects of current

Electric hazards are often the result of negligence or lack of maintenance. Damaged insulation, overheating cables, damp conditions and overloading of electrical outlets are among the common hazards. Identifying these hazards and addressing them promptly through regular inspection and maintenance is crucial to prevent accidents and ensure safety.

Damaged insulation
Overheating cables
Damp conditions
Overloading of electrical outlets

Electric shock

Electric shock can have severe consequences on the human body, including burns, cardio-respiratory failure and seizures. Understanding these effects underscores the importance of adhering to safety protocols and implementing preventive measures. First aid procedures should be known and readily available to mitigate the impact of electric shock incidents.

The amount of current passing through our body at a specific voltage is a complex question with no simple answer. The current depends on the resistance of the path it takes. Since different individuals have varying levels of resistance, it is difficult to give a definitive answer. Moreover, factors such as sweating on a hot day, which enhances conductivity because of salt water, as well as one's diet and the length of the current path, also influence the body's resistance. It is therefore crucial to consider all these factors when assessing body resistance. Table 17.2 illustrates the effects of different amounts of current on the body. This data is general, as some people may have less tolerance to electricity than others.

Table 17.2: The effects of electric current on the body.
Current in milliamperes (mA)	Effect on the body
2 to 3	A slight tingling sensation
9 to 10	Moderately painful
10 to 20	Highly painful
20 to 30	A person seizes the line and is unable to let go of the circuit
30 to 40	Causes muscular paralysis
40 to 60	Causes breathing difficulty
60 to 100	Breathing becomes extremely difficult
100 to 200	Generally causes death, because the heart usually goes into fibrillation, a condition in which the heart begins to quiver and the pumping action stops

Formula sheet

Resistor colour code

R=(10a+b)\times10^{c}\ \Omega

a, b: digits of the first two bands; c: power of ten of the third band (multiplier); the fourth band gives the tolerance

Unit: \Omega

Ratio of voltages in a potential divider

\dfrac{V_1}{V_2}=\dfrac{R_1}{R_2}\quad(17.1)

V_1, V_2: potential differences across R_1, R_2 in series

no unit (ratio)

Output of a potential divider

V_2=\dfrac{R_2}{R}\,V\quad(17.2),\qquad V_{out}=\dfrac{R_2}{R_1+R_2}\,V_{in}

V or V_{in}: input voltage; R=R_1+R_2; R_2: resistance across the output

Unit: \mathrm{V}

Equivalent resistance in series

R_{eq}=R_1+R_2+R_3+\cdots+R_n\quad(17.3,\ 17.4)

R_1,\dots,R_n: the resistances joined end to end; the same current flows in each

Unit: \Omega

Equivalent resistance in parallel

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\cdots+\dfrac{1}{R_n}\quad(17.5,\ 17.6)

R_1,\dots,R_n: the resistances across the same potential difference

Unit: \Omega

Net emf of sources in series

\varepsilon_{net}=\varepsilon_1+\varepsilon_2+\cdots+\varepsilon_n\quad(17.7)

\varepsilon_k: emf of each source

Unit: \mathrm{V}

Net emf of identical sources in parallel

\varepsilon_{net}=\varepsilon_1=\varepsilon_2=\cdots=\varepsilon_n\quad(17.8)

\varepsilon_k: emf of each (identical) source

Unit: \mathrm{V}

Electrical energy

E=IVt\quad(17.9)=I^{2}Rt\quad(17.10)=\dfrac{V^{2}}{R}\,t\quad(17.11)

I: current; V: potential difference; R: resistance; t: time

Units: \mathrm{J} (or kWh)

Electric power

P=\dfrac{E}{t}=IV\quad(17.12)=I^{2}R\quad(17.13)=\dfrac{V^{2}}{R}\quad(17.14)

E: energy; t: time; I, V, R as above

Unit: \mathrm{W} (\mathrm{J\,s^{-1}})

Kilowatt-hour

1\,\mathrm{kWh}=1000\,\mathrm{W}\times3600\,\mathrm{s}=3.6\times10^{6}\,\mathrm{J}

the commercial unit of electrical energy

Unit: \mathrm{J}

Cost of electricity

\text{Cost}=E\ (\mathrm{kWh})\times\text{tariff rate}

E: energy in kWh; tariff rate: price per kWh

Unit: rupees

Key terms

Electric circuit: A closed pathway through which electricity flows, made of an interconnection of electrical elements such as a power source, wires and components like resistors.
Active element: A circuit element that can supply or generate energy, for example a battery or a generator.
Passive element: A circuit element that cannot generate energy but absorbs, stores or dissipates it, for example a resistor, capacitor or inductor.
Circuit diagram: A simplified map in which components are drawn as symbols and wires as straight lines, showing how the components are interconnected.
Resistor: The component of an electric circuit whose function is to provide a specified value of resistance; it limits the current and dissipates energy as heat.
Carbon resistor: A resistor with a ceramic core on which a thin layer of crystalline carbon is deposited; its value is marked by colour bands.
Tolerance: The allowed variation in the marked value of a resistor, given by the fourth colour band (no fourth band means \pm20\,\%).
Linear resistor: A resistor whose current is proportional to the applied voltage (it obeys Ohm's law); fixed or variable.
Fixed resistor: A resistor with a single resistance value, most commonly made of carbon.
Variable resistor: A resistor whose resistance can be adjusted, usually by a sliding contact on a wire-wound resistor.
Rheostat: A variable resistor used to regulate the current in a circuit.
Potentiometer: A variable resistor used as a variable voltage divider.
Potential divider: A circuit that produces an output voltage V_{out} that is a fraction of its input voltage V_{in}: V_{out}=\dfrac{R_2}{R_1+R_2}V_{in}.
Non-linear resistor: A resistor whose resistance changes with temperature, light or voltage: thermistors, varistors and photo resistors.
Thermistor: A resistor whose resistance varies significantly with temperature; NTC (resistance falls as temperature rises) or PTC (resistance rises as temperature rises).
Varistor (VDR): A voltage dependent resistor used to protect circuits from destructive voltage spikes.
Light dependent resistor (LDR): A photo resistor whose resistance is high in low light and falls as the light intensity increases.
Equivalent resistance: The resistance of the single resistor that can replace a combination of resistors and give the same total current and potential difference.
Series combination: Components connected end to end in one path so that the same current flows through each: R_{eq}=R_1+R_2+\cdots+R_n.
Parallel combination: Components connected side by side across the same voltage, giving more than one current path: \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\cdots+\dfrac{1}{R_n}.
Electrical energy: The ability of electricity to do work or produce heat; E=IVt, measured in joules or kilowatt-hours.
Electric power: The rate at which electrical energy is transferred or consumed in a circuit; P=IV, measured in watts.
Watt: The SI unit of power: 1\,\mathrm{W} is consumed when a potential difference of 1\,\mathrm{V} causes a current of 1\,\mathrm{A} to flow.
Kilowatt-hour (kWh): The unit of electrical energy equal to one kilowatt of power used for one hour: 3.6\times10^{6}\,\mathrm{J}.
Live (line) wire: The wire of a mains circuit that carries current from the power source to the appliance at the high potential.
Neutral wire: The wire that completes the circuit and returns the current to the power source.
Earth wire: The safety wire that gives the current a low-resistance path to the ground in case of a fault.
Fuse: A safety device with a metal strip that melts when the current exceeds its rating, opening the circuit.
Circuit breaker: A safety switch (for example with a bimetallic strip) that opens the circuit when the current exceeds its rating and can be reset.
Residual-current device (RCD / GFCI): A safety device that detects a difference between the live and neutral currents (about 5\,\mathrm{mA}) and interrupts the circuit.
Inrush current: The peak current drawn by a device when it is first switched on; an NTC thermistor can limit it.

Multiple choice questionsp. 230–231

1

Two resistances of 4\,\Omega are connected in parallel and this combination is connected to another 6\,\Omega resistance in series. The equivalent resistance will be:

A4\,\Omega
B6\,\Omega
C8\,\OmegaAnswer
D10\,\Omega

Why

\dfrac{1}{R_p}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{2}{4}\quad\Rightarrow\quad R_p=2\,\Omega

The parallel pair is in series with the 6\,\Omega resistor. Series resistances add: R_{eq}=2+6=8\,\Omega.

A (4\,\Omega) is the resistance of one single resistor, not of the combination. D (10\,\Omega) is 4+6, which ignores one of the two 4\,\Omega resistors.

2

In series connection of resistors, the current across each resistor:

Aincreases
Bdecreases
Cremain the sameAnswer
Dfirst increases then decreases

Why

In a series circuit there is only one path for the charges, so the same current flows through every resistor. Charge is not stored or lost along the path.

The voltage is shared among the resistors instead: each resistor gets V_n=IR_n, and the sum of these is the supply voltage. Options A, B and D describe a current that changes along one path, which does not happen.

3

A resistor's first three colours are brown, black and red, its value is:

A10\,\Omega
B10\,\mathrm{k\Omega}
C200\,\Omega
D1000\,\OmegaAnswer

Why

From Table 17.1: brown is 1, black is 0 and red is the multiplier 10^{2}. The first two digits give 10 and the multiplier gives:

R=10\times10^{2}\,\Omega=1000\,\Omega=1\,\mathrm{k\Omega}

Option B would need an orange multiplier (10^{3}), and option A would need a multiplier of 10^{0}. Option C ignores the multiplier altogether.

4

The relation for electric power in a circuit is P=

AIV
BI^{2}R
C\dfrac{V^{2}}{R}
Dall of theseAnswer

Why

Equations 17.12, 17.13 and 17.14 give the three forms of electric power: P=IV, P=I^{2}R (from V=IR) and P=V^{2}/R (from I=V/R). The book says all three are equally valid.

Because options A, B and C are each correct, no single one of them is the complete answer. The correct option is D.

5

In the given figure, the current passing through 6\,\Omega resistor is:

The 1.2 A current divides between the 6\,\Omega and 4\,\Omega resistors, which are in parallel.

A0.40\,\mathrm{A}
B0.48\,\mathrm{A}Answer
C0.72\,\mathrm{A}
D80\,\mathrm{A}

Why

R_{eq}=\dfrac{6\times4}{6+4}=2.4\,\Omega

The same voltage is across both parallel branches, so the voltage is found from the total current: V=IR_{eq}=1.2\,\mathrm{A}\times2.4\,\Omega=2.88\,\mathrm{V}.

I_{6\,\Omega}=\dfrac{V}{6}=\dfrac{2.88}{6}=0.48\,\mathrm{A}

Check: I_{4\,\Omega}=2.88/4=0.72\,\mathrm{A} and 0.48+0.72=1.2\,\mathrm{A}. C (0.72 A) is the current in the 4\,\Omega branch, and D is far too large for a 1.2 A supply.

6

When four cells, each with an emf of 0.5\,\mathrm{V}, are connected in parallel, the net emf is:

A4.5\,\mathrm{V}
B2.0\,\mathrm{V}
C0.75\,\mathrm{V}
D0.5\,\mathrm{V}Answer

Why

For identical cells connected in parallel the net emf equals the emf of one cell (Equation 17.8), so \varepsilon_{net}=0.5\,\mathrm{V}.

B (2.0\,\mathrm{V}) is the answer for four cells in series (4\times0.5), which adds the emfs instead. A and C do not follow from either rule.

7

The purpose of grounding electrical systems is to:

Aprovide a path for current to flow
Bprotect against electric shocksAnswer
Creduce electrical resistance
Dincrease energy efficiency

Why

The earth (ground) wire gives a low-resistance path into the ground. If a live wire touches a metal case, the fault current flows away through the earth wire instead of through a person, so the shock risk is reduced.

Option A is only what the earth wire does during a fault, not its purpose. Grounding does not reduce the resistance of the circuit (C) and does not make appliances more efficient (D).

8

The purpose of a circuit breaker is to:

Aincrease electrical resistance
Bprevent electrical shock
Cregulate voltage
Dinterrupt the flow of electricity in case of a faultAnswer

Why

A circuit breaker contains a bimetallic strip. When the current exceeds its rated value the strip heats, bends and opens the contact, which stops the current (Figure 17.19).

A circuit breaker does not change resistance (A) or voltage (C). Protection against shock is the job of the earth wire and residual current devices (B), not of the breaker.

9

Which electrical element controls the flow of the electric current in a circuit?

Aconductor
Binsulator
CresistorAnswer
Dcapacitor

Why

A resistor is the component whose function is to provide a specified value of resistance, so it limits the current in the circuit (Section 17.2).

A conductor lets current flow freely, an insulator blocks it, and a capacitor stores charge. None of these is designed to control the current.

10

A and B are two bulbs connected in parallel. A is glowing brighter than B, then the relation between R_A and R_B is:

AR_A>R_B
BR_A<R_BAnswer
CR_A=R_B
DR_A=0

Why

Bulbs in parallel have the same voltage V across them. The power of each bulb is P=V^{2}/R, and brightness follows the power dissipated.

A is brighter, so P_A>P_B. With the same V, a larger power needs a smaller resistance: R_A<R_B. Option A reverses this relation.

11

The S.I. unit for electrical power is:

AwattAnswer
Bjoule
Cvolt
DkWh

Why

Power is the rate of energy transfer, P=E/t, so its S.I. unit is joule per second. This unit is called the watt (W).

The joule (B) is the unit of energy, the volt (C) is the unit of potential difference and the kilowatt-hour (D) is a commercial unit of energy.

12

A wire of resistance R is divided in 10 equal parts. These parts are connected in parallel, the equivalent resistance of such connection will be:

A10R
BR
C0.1R
D0.01RAnswer

Why

Each part has resistance R/10. Ten identical resistances in parallel give:

\dfrac{1}{R_{eq}}=10\times\dfrac{10}{R}\quad\Rightarrow\quad R_{eq}=\dfrac{R}{100}=0.01\,R

C (0.1R) is the resistance of one part. It forgets that the ten parts in parallel lower the resistance further.

Constructed response questionsp. 231–232

1

A battery of voltage V is connected to a 30\,\mathrm{W} bulb and a 60\,\mathrm{W} bulb as shown in the figure. (a) Identify the brightest bulb. (b) Which bulb has the greater resistance? (c) Suppose the two bulbs are connected in series. Which bulb will glow brighter?

The two bulbs are in parallel across the battery.

Solution

(a) The brightest bulb.

Both bulbs are connected across the same battery, so each bulb has the voltage V.
The power a bulb uses is P=V^{2}/R, and the brightness depends on this power. At the same voltage the bulb with the larger power rating uses more power, so it glows more brightly.
The 60\,\mathrm{W} bulb has the larger power rating, so it is the brighter bulb.

\text{The } 60\,\mathrm{W} \text{ bulb is the brightest.}

(b) The bulb with the greater resistance.

Rearrange P=V^{2}/R to get the resistance of each bulb:
R=\dfrac{V^{2}}{P}
For the two bulbs:
R_{30}=\dfrac{V^{2}}{30}\qquad R_{60}=\dfrac{V^{2}}{60}
Divide the first by the second. The 30\,\mathrm{W} bulb has twice the resistance of the 60\,\mathrm{W} bulb:
\dfrac{R_{30}}{R_{60}}=\dfrac{60}{30}=2

\text{The } 30\,\mathrm{W} \text{ bulb has the greater resistance, } R_{30}=2R_{60}.

(c) The brighter bulb when the two are in series.

In a series circuit the same current I flows through both bulbs.
The power in each bulb is P=I^{2}R. With the same current, the bulb with the larger resistance dissipates more power.
P_{30}=I^{2}R_{30}=2I^{2}R_{60}\qquad P_{60}=I^{2}R_{60}
The 30\,\mathrm{W} bulb has the larger resistance, so it glows brighter in series. This is the reverse of part (a).

\text{The } 30\,\mathrm{W} \text{ bulb glows brighter in series.}

Note. Assumes each bulb works at the battery voltage V (its rated voltage), so its rated power is the power it uses when connected in parallel.

2

What is the value of resistance of the following resistors? (Six resistors are drawn with blank value boxes. They are labelled (i) to (vi) here, from the top row to the bottom row, left column then right column.)

Solution

Band colours read from the figure and the resulting values
Part	Position	Bands (left to right)	Value	Tolerance
(i)	Top left	grey, black, green, silver	8.0\,\mathrm{M\Omega}	\pm5\,\%
(ii)	Top right	orange, grey, yellow, gold	380\,\mathrm{k\Omega}	\pm10\,\%
(iii)	Middle left	yellow, green, yellow, gold	540\,\mathrm{k\Omega}	\pm10\,\%
(iv)	Middle right	yellow, orange, yellow (three bands)	430\,\mathrm{k\Omega}	\pm20\,\% (no band)
(v)	Bottom left	yellow, yellow, green, silver	4.4\,\mathrm{M\Omega}	\pm5\,\%
(vi)	Bottom right	orange, yellow, gold (three bands)	0.34\,\Omega	\pm20\,\% (no band)

(i) Top left resistor.

Colour bands from the left: grey, black, green, silver.
Read the digits and multiplier from Table 17.1: 8 and 0, multiplier 10^{5}, tolerance \pm5\,\%.
Write the value and multiply:
8\times10^{5}\,\Omega=8\,\mathrm{M\Omega}

8\,\mathrm{M\Omega}\pm5\,\%

(ii) Top right resistor.

Colour bands from the left: orange, grey, yellow, gold.
Read the digits and multiplier from Table 17.1: 3 and 8, multiplier 10^{4}, tolerance \pm10\,\%.
Write the value and multiply:
38\times10^{4}\,\Omega=380\,\mathrm{k\Omega}

380\,\mathrm{k\Omega}\pm10\,\%

(iii) Middle left resistor.

Colour bands from the left: yellow, green, yellow, gold.
Read the digits and multiplier from Table 17.1: 5 and 4, multiplier 10^{4}, tolerance \pm10\,\%.
Write the value and multiply:
54\times10^{4}\,\Omega=540\,\mathrm{k\Omega}

540\,\mathrm{k\Omega}\pm10\,\%

(iv) Middle right resistor.

Colour bands from the left: yellow, orange, yellow (three bands).
Read the digits and multiplier from Table 17.1: 4 and 3, multiplier 10^{4}, no tolerance band.
Write the value and multiply:
43\times10^{4}\,\Omega=430\,\mathrm{k\Omega}

430\,\mathrm{k\Omega}\pm20\,\%

(v) Bottom left resistor.

Colour bands from the left: yellow, yellow, green, silver.
Read the digits and multiplier from Table 17.1: 4 and 4, multiplier 10^{5}, tolerance \pm5\,\%.
Write the value and multiply:
44\times10^{5}\,\Omega=4.4\,\mathrm{M\Omega}

4.4\,\mathrm{M\Omega}\pm5\,\%

(vi) Bottom right resistor.

Colour bands from the left: orange, yellow, gold (three bands).
Read the digits and multiplier from Table 17.1: 3 and 4, multiplier 10^{-2}.
Write the value and multiply:
34\times10^{-2}\,\Omega=0.34\,\Omega

0.34\,\Omega\pm20\,\%

Note. The bands in the figure are small, and grey, brown, orange and yellow look alike, so the colour readings given are the most likely ones; check each against Table 17.1. The book does not print answers for this question. Table 17.1 gives gold as a 10^{-2} multiplier with \pm10\,\% tolerance, while the text on page 210 gives the tolerance of a gold band as 5\,\%; the table is used here because the exam follows the book. With the standard colour code explained in the notes (gold multiplier 10^{-1}, gold tolerance \pm5\,\%), parts (ii) and (iii) would be \pm5\,\% and part (vi) would be 3.4\,\Omega.

3

Calculate the equivalent resistance between points A and B in the given circuit.

Each block has two equal resistors in parallel. The three blocks are in series between A and B.

Solution

Each block is a pair of equal resistors in parallel. Two equal resistances R in parallel give R/2, so the blocks reduce as follows.

Block 1: two 2\,\Omega resistors in parallel.
\dfrac{1}{R_1}=\dfrac{1}{2}+\dfrac{1}{2}=1\quad\Rightarrow\quad R_1=1\,\Omega
Block 2: two 4\,\Omega resistors in parallel.
\dfrac{1}{R_2}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\quad\Rightarrow\quad R_2=2\,\Omega
Block 3: two 6\,\Omega resistors in parallel.
\dfrac{1}{R_3}=\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{1}{3}\quad\Rightarrow\quad R_3=3\,\Omega
The three blocks are joined end to end, so they are in series. Series resistances add.
R_{AB}=R_1+R_2+R_3=1+2+3=6\,\Omega

R_{AB}=6\,\Omega

4

Consider a circuit in which a heater is in series with the supply voltage. Which device out of the following should be placed to present a higher resistance initially, preventing large currents from flowing at turn-on?

The options are: (a) a thermistor symbol, a resistor with a diagonal line through it; (b) a photodiode, a diode symbol with arrows of light pointing at it; (c) a motor, a circle marked M; (d) a lamp, a circle with a cross inside.

Solution

When the circuit is switched on, the device must have high resistance while it is cold, so that the current does not surge. Its resistance should then fall as it warms up.
A negative temperature coefficient (NTC) thermistor does this. Its resistance is high when it is cold and drops as its temperature rises, so it limits the inrush current at turn-on (Section 17.3.2).

The other options do not do this. A photodiode (b) is a diode that passes current in one direction and is controlled by light, not by a surge. A motor (c) draws a large starting current. A bulb (d) has a low-resistance cold filament, so it allows a large inrush current, which is the problem the question describes.

\text{(a), the thermistor (NTC type).}

Note. Option (a) is drawn as a resistor with a diagonal line through it, which is the thermistor symbol in Figure 17.3.

Short response questionsp. 232–233

13 marks

How does the color code on a carbon resistor provide information about its properties, and why is this method chosen over other labeling techniques?

Solution

The colour bands give the resistance and the tolerance. The first two bands are the significant figures, the third band is the multiplier (a power of ten), and the fourth band (silver or gold) gives the tolerance. A band with no fourth colour means the tolerance is \pm20\,\%.

Colour codes are used because a carbon resistor is small, so it is more convenient to show its value as coloured bands than as printed numbers on the body (Section 17.2). Carbon resistors are also cost-effective, reliable and small in size.

Band 1 and band 2: the first two digits, from Table 17.1.
Band 3: the multiplier, for example red is 10^{2}.
Band 4: the tolerance, silver \pm5\,\% or gold \pm10\,\% in Table 17.1.

23 marks

Why are electrical devices in homes connected in parallel rather than in series? How would household functionality be impacted if they were connected in series?

Solution

Household devices are connected in parallel so that each one has the full supply voltage of about 220\,\mathrm{V}. Each device then works at its rated power, and it works whether or not the others are switched on.

In a series connection the supply voltage is shared among the devices, so each gets only a fraction of 220\,\mathrm{V} and runs dimly. The devices also depend on each other. If one device fails or is switched off, the current stops and every other device stops working.

Parallel: each appliance has the full voltage; each can be switched on or off on its own.
Series: the voltage divides, so appliances run dimly; one open switch or failed element switches off all of them.

33 marks

When resistors are connected in parallel, why does the total resistance decrease? How does this principle relate to the efficiency of electrical systems?

Solution

In a parallel connection the current has more than one path to follow. Adding a branch lets more charge flow for the same voltage, so the total current increases and the equivalent resistance falls. For resistors in parallel:

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}

The equivalent resistance is always smaller than the smallest single resistance in the combination.

Parallel connection makes electrical systems efficient in use. Each appliance receives the full voltage, so it operates at its rated power, and a fault in one branch does not stop the others. The total current drawn from the supply is the sum of the branch currents, so it grows as more branches are added.

43 marks

How incorrect parallel placement of ammeters can damage circuits, suggesting ways to prevent this.

Solution

An ammeter must be connected in series, so that the whole current of the branch passes through it. Its resistance is very low. If it is connected in parallel across a component, it provides an almost direct path for the current. The current through the ammeter then becomes very large, which can burn out its coil and the circuit, and the component it is across is short-circuited.

To prevent this:

Always connect the ammeter in series with the component being measured, as in the mini-lab on page 214.
Check the circuit diagram before closing the switch.

53 marks

Why is it unsafe to touch electrical switches with wet hands? What principles of conductivity and human safety does this advice rely on?

Solution

Water with dissolved salts or minerals conducts electricity, and sweat on the hands is such a solution. Wet hands therefore have a much lower resistance than dry hands. Because I=V/R, a low resistance in the path lets a large current pass through the body when the switch carries a live voltage.

The advice relies on two principles. The first is that the body can carry current, and the current depends on the path resistance. The second is that dangerous effects begin at very small currents: Table 17.2 shows that 10 to 20\,\mathrm{mA} is highly painful, and 100 to 200\,\mathrm{mA} can cause death.

63 marks

What makes touching a live wire while barefoot on the ground particularly dangerous? How do the factors involved influence the severity of electrical shock?

Solution

Being barefoot on the ground connects the body to the earth. If the body touches a live wire, the circuit is completed through the body and down into the ground, so a current flows through the person. Dry insulating footwear would reduce this current, but bare, damp feet on wet ground give a low-resistance path.

The severity of a shock depends on these factors:

Size of the current: Table 17.2 links currents of a few milliamperes to muscle paralysis and breathing difficulty, and currents of 100 to 200\,\mathrm{mA} generally cause death because the heart can go into fibrillation.
Resistance of the body: sweat and damp skin lower it, which increases the current.
Path through the body: a current that passes through the heart is the most dangerous.
Time of contact: the longer the contact, the more harm is done.

73 marks

How does a circuit breaker protect a household's electrical system? What might happen in the absence of a circuit breaker during a power surge or overload?

Solution

A circuit breaker has a bimetallic strip in series with the circuit. When the current exceeds the rated value, the strip heats up, bends and opens the contact, so the current stops (Figure 17.19). The breaker can be reset after the strip cools, and it protects the wiring from overheating when the circuit is overloaded.

Without a circuit breaker, a power surge or overload would keep the current high. The wires and the insulation would overheat, which could damage the appliances and start a fire.

83 marks

In what ways does grounding an appliance's case enhance user safety? How does grounding mitigate potential electrical hazards?

Solution

The earth wire is connected to the metal case of the appliance. If a live wire becomes loose and touches the case, the current flows through the earth wire, which has a low resistance, instead of through a person who touches the case. This large current blows the fuse or trips the circuit breaker, which cuts off the supply.

In this way grounding keeps the metal case at a safe potential, so the user receives little or no current. It mitigates the hazard by providing a safe path for the fault current and by switching the supply off quickly.

93 marks

How does an appliance's wattage influence its energy consumption and operational cost? How can understanding this relationship help in making energy-efficient choices?

Solution

Energy used is E=Pt, so an appliance with a higher wattage uses more energy in the same time. The cost is the energy in kilowatt-hours multiplied by the tariff. For example, a 2200\,\mathrm{W} air conditioner running for 720\,\mathrm{h} in a month uses E=2.2\,\mathrm{kW}\times720\,\mathrm{h}=1584\,\mathrm{kWh}, which costs 1584\times25.0=\mathrm{PKR}\,39{,}600 at PKR 25.0 per kWh (Example 17.7).

Knowing this relationship helps you choose an appliance with a lower wattage for the same task and to switch off appliances when they are not needed. Their running time matters as much as their power.

103 marks

In what ways do electrical energy and electrical power differ in function and application? How do these differences impact our understanding and measurement of electricity?

Solution

Electrical power is the rate at which electrical energy is transferred or used. It is measured in watts (W) and tells how fast a device uses energy. Electrical energy is the total amount of work done over a period of time. It is measured in joules (J) or kilowatt-hours (kWh).

The difference matters for measurement. A power rating on an appliance tells you how strong it is, while the electricity meter in the home records energy in kWh, which is the quantity the supplier charges for. Power is an instant value, but energy depends on how long the appliance runs.

113 marks

How does the kilowatt-hour serve as a unit of energy consumption, and what steps can you take to convert energy usage from joules to kilowatt-hours?

Solution

One kilowatt-hour is the energy used by a power of 1\,\mathrm{kW} in one hour. It is a large unit that suits household and industrial use, where energy is counted over many hours.

Conversion steps:

Write the energy in joules.
Divide by 3.6\times10^{6}, because 1\,\mathrm{kWh}=1000\,\mathrm{W}\times3600\,\mathrm{s}=3.6\times10^{6}\,\mathrm{J}.
Attach the unit kWh to the result.

E=\dfrac{1.8\times10^{7}\,\mathrm{J}}{3.6\times10^{6}\,\mathrm{J\,kWh^{-1}}}=5\,\mathrm{kWh}

123 marks

Describe a situation in your daily life where electrical energy is essential. How does this energy transform to allow the device to perform its function?

Solution

A ceiling fan is a good example. It cannot work without electrical energy supplied by the mains.

Inside the fan, the motor receives electrical energy from the supply.
The motor converts electrical energy into mechanical (kinetic) energy, which turns the blades.
Heat is also produced in the appliance as the energy is transferred.

Other examples are the electric heater, which converts electrical energy into heat, and the light bulb, which converts it into light and heat.

133 marks

What are three common sources of electrical energy, and how do their advantages and disadvantages influence our choice of energy sources?

Solution

The book describes three sources that supply electrical energy in a circuit: the cell or battery, the DC generator and the AC generator.

Cell or battery: converts chemical energy into electrical energy. It is portable and needs no moving parts, but it stores only a limited amount of energy and must be replaced or recharged.
DC generator: supplies a steady direct current. It can deliver more power than a cell, but it needs a continuous supply of mechanical energy and regular maintenance.
AC generator: supplies the alternating voltage used by the mains. It is the source for household and industrial supply, but it is not portable.

The choice depends on the job. Portable devices such as torches and phones use cells or batteries, while homes and factories use the mains supply from generators.

143 marks

Examine the effects and risks of connecting different emf sources in parallel, and suggest safe ways to combine power sources.

Solution

When sources with different emfs are connected in parallel, the source with the higher emf drives a current through the source with the lower emf. This current is wasted inside the sources, which heat up and can be damaged. The weaker source may also be drained or overworked, which reduces the performance of the whole system.

Safe ways to combine sources:

Use identical sources, with the same emf and the same age and charge, when they are connected in parallel (Equation 17.8).
Use series connections when a higher voltage is needed, and balance the cells so that one weak cell does not limit the others (Science Tidbits, page 219).
Use a fuse or a protective circuit, and monitor each battery's charge if sources are combined.

Long response questionsp. 233

1

What is the significance of different types of resistors in electronic circuits, and how do their characteristics influence circuit design?

Solution

Types of resistors

Resistors are divided into two groups (Section 17.3). A linear resistor obeys Ohm's law, so its current is directly proportional to the applied voltage, and its value changes only with voltage and temperature. A non-linear resistor changes its resistance in response to temperature, light or voltage. Each type has a job that the circuit needs.

Types of resistors and their uses
Type	What changes the resistance	Use given in the book
Fixed resistor (carbon)	Nothing: it has one fixed value	Sets a single value in a circuit; can handle high voltages, but is less stable
Variable resistor (rheostat, potentiometer)	A sliding contact or mechanical movement	Regulates the current (rheostat) or divides a voltage (potentiometer)
Thermistor (NTC, PTC)	Temperature	Inrush current limiter, overheating alarm, temperature measurement
Varistor (VDR)	Voltage	Reduces high voltage spikes to protect circuits
Light dependent resistor (LDR)	Light intensity	Automatic lighting control, photography, security systems

How the characteristics shape the design

Required behaviour: a circuit that must sense a quantity uses a resistor that responds to it. A light-sensing circuit uses an LDR and a temperature-sensing circuit uses a thermistor.
Adjustability: a potential divider needs a variable resistor so that the output voltage can be changed. A fixed resistor gives one value only.
Stability and accuracy: high stability and high accuracy resistors are wire wound. A carbon resistor is cheaper but less stable, and its temperature coefficient is quite high.
Power rating: the resistor must be able to dissipate the power it receives. Example 17.6 shows that a 100\,\Omega resistor rated at \tfrac12\,\mathrm{W} is damaged on 10\,\mathrm{V}, because it would dissipate 1\,\mathrm{W}.

2

Explain the principles behind the series combination of resistances and illustrate how the total resistance changes when more resistors are added?

Solution

Principles of series combination

In a series circuit there is only one path for the charges, so the same current I flows through every resistor. The supply voltage V is shared among the resistors, and the voltages across them add up to V.

Resistors in series: the same current I flows through R_1, R_2 and R_3 (Figure 17.10).

Derivation Equivalent resistance of resistors in series

Aim. Find the single resistor R_{eq} that would draw the same current I from the same supply V.

Assumptions.

The connecting wires have no resistance.
The current is the same through every resistor.

Apply Ohm's law to each resistor.
V_1=IR_1,\qquad V_2=IR_2,\qquad V_3=IR_3
The supply voltage is the sum of the voltages across the resistors.
V=V_1+V_2+V_3
Substitute the three voltages and take out the common factor I.
V=I\,(R_1+R_2+R_3)
For the single equivalent resistor, V=IR_{eq}. Compare the two expressions.
R_{eq}=\dfrac{V}{I}=R_1+R_2+R_3

ResultR_{eq}=R_1+R_2+\cdots+R_n

Equivalent resistance in series is the sum of the individual resistances (Equation 17.4). It is always greater than any one of them.

How the total resistance changes as resistors are added

Each resistor added in series adds its resistance to the total. Using the values of Example 17.1 (10\,\Omega, 20\,\Omega and 30\,\Omega):

Resistors in the circuit	Sum	Total resistance R_{eq}
R_1 only	10\,\Omega	10\,\Omega
R_1 and R_2	10+20	30\,\Omega
R_1, R_2 and R_3	10+20+30	60\,\Omega

For n identical resistors of value R the total is nR, so the total rises in equal steps of R each time a resistor is added. The graph shows this for R=10\,\Omega.

Total resistance of identical resistors in series

Each added 10\,\Omega resistor raises the total by 10\,\Omega.

Adding a resistor in series never lowers the total resistance. The same supply voltage now drives its current through a larger total resistance, so the current falls, as I=V/R_{eq} shows.

3

How does the concept of equivalent resistance in parallel circuits differ from that in series circuits, and what practical applications can you identify for each?

Solution

Comparison of the two arrangements

Feature	Series	Parallel
Path for the current	One path	More than one path
Current	The same through every resistor	Divides between the branches; the branch currents add to the total
Voltage	Shared among the resistors	The same across every resistor
Equivalent resistance	R_{eq}=R_1+R_2+R_3, always greater than the largest resistance	\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}, always less than the smallest resistance
If one element fails	The whole circuit stops	The other branches keep working

Equivalent resistance in parallel

Resistors in parallel: each has the same voltage V, and the total current I divides into I_1, I_2 and I_3 (Figure 17.11).

Derivation Equivalent resistance of resistors in parallel

Aim. Find the single resistor R_{eq} that would draw the same total current I from the same supply V.

Assumptions.

The connecting wires have no resistance.
Each resistor is connected across the same two points, so each has the voltage V.

Apply Ohm's law to each branch.
I_1=\dfrac{V}{R_1},\qquad I_2=\dfrac{V}{R_2},\qquad I_3=\dfrac{V}{R_3}
Charge is not stored or lost at the junctions, so the total current is the sum of the branch currents.
I=I_1+I_2+I_3
Substitute the branch currents and take out V.
I=V\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\right)
For the single equivalent resistor I=V/R_{eq}. Divide both sides by V and take the reciprocal.
\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}

Result\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\cdots+\dfrac{1}{R_n}

The reciprocal of the equivalent resistance is the sum of the reciprocals (Equation 17.6).

Check with Example 17.2: for 5.0\,\Omega, 10.0\,\Omega and 15.0\,\Omega in parallel, \dfrac{1}{R_{eq}}=\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{11}{30}, so R_{eq}=2.73\,\Omega. This is less than the smallest resistor, 5.0\,\Omega.

Practical applications

Series, potential divider: two resistors in series share a fixed voltage and give an output that is a fraction of the input. This is used in the volume control knob of music systems (Section 17.3.1).
Series, protection: fuses and circuit breakers are connected in series with the live wire so that the whole circuit can be switched off safely (Section 17.9).
Parallel, household wiring: every appliance is connected in parallel, so each receives the full mains voltage and can be switched on or off independently (Section 17.4.2).
Parallel, lighting: the bulbs of a multiple-bulb fixture are in parallel, so if one bulb fails the others stay lit (Figure 17.12).

4

Discuss the effects of connecting emf sources in series or parallel and their practical uses.

Solution

Series combination of emf sources

In a series combination the positive terminal of one source is joined to the negative terminal of the next, and so on. The emfs add (Figure 17.13). For n sources:

\varepsilon_{net}=\varepsilon_1+\varepsilon_2+\varepsilon_3+\cdots+\varepsilon_n

Four 1.5\,\mathrm{V}, 2.5\,\mathrm{Ah} batteries in series give 6.0\,\mathrm{V} and a rating of 6.0\,\mathrm{V}\;2.5\,\mathrm{Ah}. The ampere-hour rating does not change, because the same charge passes through every source.

Parallel combination of emf sources

In a parallel combination all the positive terminals are joined together and all the negative terminals are joined together. For identical sources the net emf equals the emf of one source (Equation 17.8):

\varepsilon_{net}=\varepsilon_1=\varepsilon_2=\cdots=\varepsilon_n

The ampere-hour rating increases, because each source supplies part of the charge. The book's example of 12\,\mathrm{V}, 2\,\mathrm{Ah} batteries in parallel gives a rating of 12\,\mathrm{V}, 6\,\mathrm{Ah} (see the note).

(a) Cells in series: the emfs add. (b) Identical cells in parallel: the net emf is the emf of one cell (Figures 17.13 and 17.14).

Effects and risks

Series: the voltage increases, so a higher voltage is available from weaker sources. The ampere-hour rating stays the same as one source, so the supply does not last longer than one source.
Parallel with identical sources: the voltage stays the same and the time the supply lasts increases, because the load is shared.
Parallel with different emfs: the source with the higher emf drives a current through the source with the lower emf. This current is wasted inside the sources, which can heat and be damaged (Section 17.5 and the Science Tidbits on page 219).
Series with an unbalanced set: a weaker cell drains faster and may be overworked, which limits the performance of the whole set.

Practical uses

	Series	Parallel
Net emf	Sum of the emfs	Same as one source (identical sources)
Ampere-hour rating	Same as one source	Sum of the ratings
Use	When a higher voltage than one source is needed	When a longer supply from the same voltage is needed

Note. The book's parallel example (page 219) gives 12 V, 6 Ah for batteries that are 12 V, 2 Ah each. The figure shows four batteries, which would give 8 Ah; the value 6 Ah appears to be a misprint or refers to three batteries.

5

In what ways can electric power be measured, and why is it important to understand the different formulas for calculating power in electrical systems?

Solution

Definition of electric power

Electric power is the rate at which electrical energy is transferred or used. It is measured in watts (W), and it is defined as:

P=\dfrac{E}{t}

Ways to find the power

From energy and time: read the energy used from the electricity meter (in kWh) and divide by the time. This gives the average power.
From voltage and current: measure the voltage across the device with a voltmeter and the current through it with an ammeter. Then P=IV.
From current and resistance: if the current and the resistance are known, P=I^{2}R.
From voltage and resistance: if the voltage and the resistance are known, P=V^{2}/R.

Formulas for electric power
Formula	Use it when you know	Source in the book
P=IV	Voltage and current	Equation 17.12
P=I^{2}R	Current and resistance	Equation 17.13
P=V^{2}/R	Voltage and resistance	Equation 17.14
P=E/t	Energy used and time	Equation 1, Section 17.8

Why the choice of formula matters

The three forms of P=IV, P=I^{2}R and P=V^{2}/R give the same answer for the same device, because V=IR links them. The right formula is the one that uses the quantities you have been given, so the problem can be solved without first finding another quantity.

Power of an electric toaster (Example 17.5)

What is the power consumption of an electric toaster if its resistance is 19\,\Omega, and it operates on a household circuit with 200\,\mathrm{V}?

Find the current with Ohm's law, since the voltage and resistance are known.
I=\dfrac{V}{R}=\dfrac{200\,\mathrm{V}}{19\,\Omega}=10.5\,\mathrm{A}
Use P=IV with the current just found.
P=IV=10.5\,\mathrm{A}\times200\,\mathrm{V}=2100\,\mathrm{W}
Check with P=V^{2}/R, which uses the same voltage and resistance without finding the current first.
P=\dfrac{200^{2}}{19}=2105\,\mathrm{W}\approx2100\,\mathrm{W}

P\approx2100\,\mathrm{W}

Knowing the formulas also lets you check a rating. Example 17.6 uses P=V^{2}/R to find that a 100\,\Omega resistor on 10\,\mathrm{V} dissipates 1\,\mathrm{W}, which is double its \tfrac12\,\mathrm{W} rating, so it would be damaged.

6

What are the common hazards associated with household electricity, and how can individuals effectively mitigate these risks through safety measures?

Solution

Common hazards

Electric hazards are often the result of negligence or lack of maintenance (Section 17.9.5). The common hazards are:

Damaged insulation, which exposes live conductors.
Overheating cables, which can start a fire.
Damp conditions, which lower the resistance of the body and of the path to earth.
Overloading of electrical outlets, which draws more current than the wiring can safely carry.
Faulty appliances, for example a live wire that touches a metal case.

Effects of electric current on the body

Table 17.2 shows the effect of different currents on the body. The effects depend on the person, so the table is general.

Effects of electric current on the body (Table 17.2)
Current (mA)	Effect on the body
2 to 3	A slight tingling sensation
9 to 10	Moderately painful
10 to 20	Highly painful
20 to 30	A person may be unable to let go of the circuit
30 to 40	Muscular paralysis
40 to 60	Breathing difficulty
60 to 100	Breathing becomes extremely difficult
100 to 200	Generally causes death, because the heart can go into fibrillation

Safety measures

Earthing: the third (earth) wire and the earthed metal casing of an appliance give a low-resistance path, so a fault blows the fuse or trips the breaker instead of shocking the user (Section 17.9.1).
Fuses and circuit breakers: connect them in the live wire, and use a fuse with the correct current rating for the appliance (SLO E-60).
Double insulation: appliances with a non-conducting (double-insulated) casing do not need to be earthed to be safe in normal use (SLO E-58).
Residual current devices: these detect a difference of about 5\,\mathrm{mA} between the live and neutral currents and interrupt the circuit (Section 17.9.4).
Maintenance: inspect cables, plugs and outlets regularly, and repair damaged insulation promptly.
Keep water away: do not handle switches or appliances with wet hands or in damp places.
First aid: know the first aid for electric shock.

7

In what various ways is electricity utilized in everyday life, and how has its application evolved over time?

Solution

Everyday uses

Heating: electric heaters convert electrical energy into heat in the room, and electric stoves produce heat in their heating elements for cooking (Section 17.6).
Lighting: electric bulbs light homes, and streetlights make roads and pathways safe at night.
Powering motors: washing machines, vacuum cleaners, conveyor belts and manufacturing machinery, and electric vehicles.
Battery charging: smartphones, laptops and tablets are charged from the mains.
Powering electronic systems: computers, televisions, refrigerators, air conditioners, gaming consoles and home automation devices.

Figure 17.15 in the book collects several of these uses on one page.

How its application has changed

The book presents electricity as moving beyond basic heating and lighting into motors, electronics and transport. Electric vehicles are described as a cleaner and more sustainable alternative to traditional internal combustion engines. In the digital age, portable devices such as phones and tablets depend on rechargeable batteries that are charged from the mains.

Note. The book does not describe the history of electricity use, so the change over time is described by comparing the applications it gives.

8

Describe the working principle of a potential divider and discuss its practical uses in electronic circuits?

Solution

Working principle

A potential divider is a circuit that produces an output voltage which is a fraction of its input voltage. A fixed voltage V is applied across the ends of a wire-wound variable resistor AB, of total resistance R. A sliding contact C divides the wire into two parts, R_1 (from A to C) and R_2 (from C to B), as in Figure 17.6.

Figure 17.6: Potential divider

Derivation Output voltage of a potential divider

Aim. Find the output voltage V_{out}=V_2, which is the voltage across R_2, in terms of V, R_1 and R_2.

Assumptions.

The output is taken without drawing any current, so the same current I flows through the whole wire.

Ohm's law for the whole wire gives the current.
I=\dfrac{V}{R}=\dfrac{V}{R_1+R_2}
The voltage across the part of the wire between C and B is V_2=IR_2.
V_2=IR_2
Substitute for I.
V_2=\dfrac{V}{R_1+R_2}\,R_2
Call V_2 the output and V the input.
V_{out}=\dfrac{R_2}{R_1+R_2}\,V_{in}

ResultV_{out}=\dfrac{R_2}{R_1+R_2}\,V_{in}

Moving the contact C changes R_2, so the output can be set anywhere from 0 (C at B) to V (C at A). If R_1=R_2=r, then V_{out}=\tfrac12V_{in}, so the input is split into two equal halves.

Practical uses in electronic circuits

Volume control: the knob of a music system moves the sliding contact, which changes the fraction of the voltage that is passed on, and so changes the loudness (Section 17.3.1).
Adjusting the level of a voltage signal in electronics, and measuring a voltage accurately, because the output can be set to any fraction of the input.
Sensing circuits: a light dependent resistor or a thermistor is used as one part of a potential divider, so the output voltage changes with light or temperature.

10

What are thermistors, and how do their types (NTC and PTC) differ in terms of functionality and applications in temperature sensing and control?

Solution

What a thermistor is

A thermistor is a resistor whose resistance varies significantly with temperature. The word comes from thermal resistor. It is made from a semiconductor ceramic or polymer, so its temperature coefficient is very high (Section 17.3.2).

NTC and PTC thermistors

Property	NTC thermistor	PTC thermistor
Name	Negative temperature coefficient	Positive temperature coefficient
Change with temperature	Resistance decreases as temperature increases	Resistance increases as temperature increases
Applications given in the book	Inrush current limiter, overheating alarm, temperature measurement	No application is described in this unit

Resistance against temperature (sketch)

A sketch only, not to scale: the NTC resistance falls as the temperature rises, and the PTC resistance rises.

Applications of NTC thermistors

Inrush current limiter: when a device is switched on, an NTC thermistor has a high resistance, which limits the surge of current until the device warms up. This limits the inrush current in power supply circuits (page 212). The book also notes that incandescent bulbs draw a high inrush current until their filaments heat up.
Overheating alarm: NTC thermistors in motors, transformers and generator windings keep a high resistance while the windings are cool. As the temperature rises the resistance drops, more current flows to an alarm system, and this also applies to fire alarms and thermostats.
Temperature measurement: the high temperature coefficient gives a precise change in resistance for a small change in temperature.

Functionality compared

An NTC thermistor is the type used in the applications the book describes. Its resistance falls as temperature rises, so it signals a temperature change through a rise in current. A PTC thermistor works in the opposite direction, because its resistance rises with temperature, as the graph shows. The book does not describe a PTC application in this unit.

Note. The book gives applications only for NTC thermistors. PTC thermistors are described by their behaviour, so no PTC application is given.

Numerical response questionsp. 234

1

A fan, a heater and an LED bulb are connected to a battery as shown in the figure. (a) Find the equivalent resistance. (b) Calculate the current passing through each device.

The fan, the heater and the LED are in parallel across the 12 V battery.

Solution

Given

V = 12\,\mathrm{V} battery voltage
R_1 = 6\,\Omega fan
R_2 = 10\,\Omega heater
R_3 = 18\,\Omega LED bulb

To find

(a) R_{eq}
(b) I_1, I_2, I_3

(a) Equivalent resistance.

The three devices are in parallel across the battery, so use the parallel formula.
\dfrac{1}{R_{eq}}=\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{18}=\dfrac{15+9+5}{90}=\dfrac{29}{90}
Invert the sum.
R_{eq}=\dfrac{90}{29}=3.1\,\Omega

R_{eq}\approx3.1\,\Omega

(b) Current through each device.

Each device has the full battery voltage of 12\,\mathrm{V}, so apply Ohm's law to each one.
I=\dfrac{V}{R}
Fan (6 Ω):
I_1=\dfrac{12}{6}=2\,\mathrm{A}
Heater (10 Ω):
I_2=\dfrac{12}{10}=1.2\,\mathrm{A}
LED bulb (18 Ω):
I_3=\dfrac{12}{18}=0.67\,\mathrm{A}
Check: the total current from the battery is the sum, 2+1.2+0.67=3.87\,\mathrm{A}, which equals 12/3.1.

I_1=2\,\mathrm{A},\quad I_2=1.2\,\mathrm{A},\quad I_3=0.67\,\mathrm{A}

Book answer. (a) 3.1\,\Omega; (b) 2\,\mathrm{A}, 1.2\,\mathrm{A}, 0.67\,\mathrm{A}

2

Three resistors R_1=8\,\Omega, R_2=12\,\Omega and R_3=18\,\Omega are connected in parallel. Calculate the total resistance of the combination.

Solution

Given

R_1 = 8\,\Omega
R_2 = 12\,\Omega
R_3 = 18\,\Omega

To find

R_{eq}

Use the parallel formula.
\dfrac{1}{R_{eq}}=\dfrac{1}{8}+\dfrac{1}{12}+\dfrac{1}{18}=\dfrac{9+6+4}{72}=\dfrac{19}{72}
Invert the sum.
R_{eq}=\dfrac{72}{19}=3.789\,\Omega

R_{eq}=3.789\,\Omega

Book answer. 3.789\,\Omega

3

A device with a resistance of 10\,\Omega is connected to a 120\,\mathrm{V} power source. Calculate the power consumed by the device.

Solution

Given

R = 10\,\Omega resistance of the device
V = 120\,\mathrm{V} supply voltage

To find

P

Use the power formula that contains voltage and resistance.
P=\dfrac{V^{2}}{R}
Substitute the values.
P=\dfrac{(120)^{2}}{10}=\dfrac{14\,400}{10}=1440\,\mathrm{W}

P=1440\,\mathrm{W}

Book answer. 1440\,\mathrm{W}

4

A light bulb with a power rating of 60\,\mathrm{W} is switched on for 5 hours. Calculate the electrical energy consumed by the light bulb.

Solution

Given

P = 60\,\mathrm{W} power rating
t = 5\,\mathrm{h} time switched on

To find

E

Energy is power multiplied by time.
E=Pt
Substitute the values, with the time in hours so that the energy is in watt-hours.
E=60\,\mathrm{W}\times5\,\mathrm{h}=300\,\mathrm{Wh}
Convert to kilowatt-hours if required.
E=0.3\,\mathrm{kWh}

E=300\,\mathrm{Wh}=0.3\,\mathrm{kWh}

Book answer. 300\,\mathrm{Wh}

5

One hundred identical cells are connected in series, each of emf 0.2\,\mathrm{V}. What will the resultant emf be when these identical cells are connected in series? When parallel?

Solution

Given

n = 100 number of identical cells
\varepsilon = 0.2\,\mathrm{V} emf of each cell

To find

\varepsilon_{net} in series
\varepsilon_{net} in parallel

(a) Cells in series.

In series the emfs add (Equation 17.7).
\varepsilon_{net}=n\,\varepsilon=100\times0.2\,\mathrm{V}

\varepsilon_{net}=20\,\mathrm{V}

(b) Cells in parallel.

Identical cells in parallel give the emf of one cell (Equation 17.8).
\varepsilon_{net}=\varepsilon=0.2\,\mathrm{V}

\varepsilon_{net}=0.2\,\mathrm{V}

Book answer. 20\,\mathrm{V} in series, 0.2\,\mathrm{V} in parallel

6

In the circuit shown, what is the value of the total resistance of the circuit between points A and B?

The 6\,\Omega resistor is in parallel with the 2\,\Omega and 3\,\Omega resistors in series, and this combination is in series with the 4\,\Omega resistor.

Solution

Given

R_1 = 4\,\Omega
R_2 = 6\,\Omega
R_3 = 2\,\Omega
R_4 = 3\,\Omega

To find

R_{AB}

The 2\,\Omega and 3\,\Omega resistors are in series in one branch. Add them.
R_{34}=2+3=5\,\Omega
This branch is in parallel with the 6\,\Omega resistor.
\dfrac{1}{R_{p}}=\dfrac{1}{6}+\dfrac{1}{5}=\dfrac{11}{30}\quad\Rightarrow\quad R_p=\dfrac{30}{11}=2.73\,\Omega
The 4\,\Omega resistor is in series with this parallel part, so add it last.
R_{AB}=4+2.73=6.73\,\Omega

R_{AB}\approx6.73\,\Omega

Book answer. 6.73\,\Omega

7

Calculate the (a) power consumed, (b) current flowing through and (c) electrical energy consumed by the bulb in the given figure.

The figure shows a bulb of 50\,\Omega connected to a 120\,\mathrm{V} battery.

Solution

Given

R = 50\,\Omega resistance of the bulb
V = 120\,\mathrm{V} battery voltage
t = 5\,\mathrm{h} time, assumed (see note)

To find

(a) P
(b) I
(c) E

(a) Power consumed.

The bulb is across the full battery voltage, so use P=V^{2}/R.
P=\dfrac{(120)^{2}}{50}=\dfrac{14\,400}{50}=288\,\mathrm{W}

P=288\,\mathrm{W}

(b) Current flowing through the bulb.

Ohm's law:
I=\dfrac{V}{R}=\dfrac{120}{50}=2.4\,\mathrm{A}

I=2.4\,\mathrm{A}

(c) Electrical energy consumed.

Energy is power multiplied by the time the bulb is on. The time is taken as 5 hours (see the note).
E=Pt=288\,\mathrm{W}\times5\,\mathrm{h}=1440\,\mathrm{Wh}

E=1440\,\mathrm{Wh}=1.44\,\mathrm{kWh}

Book answer. 288\,\mathrm{W}, 2.4\,\mathrm{A}, 1440\,\mathrm{Wh}

Note. The question does not state how long the bulb is on. The book's answer of 1440 Wh corresponds to 5 hours, so part (c) assumes t=5\,\mathrm{h}.

8

Calculate the total resistance of the circuit shown in the figure.

The two 12.5\,\Omega resistors are in parallel, and this pair is in series with the 4.5\,\Omega resistor.

Solution

Given

V = 10\,\mathrm{V} battery voltage, not needed for the resistance
R_1 = 4.5\,\Omega
R_2 = 12.5\,\Omega
R_3 = 12.5\,\Omega

To find

R_{total}

The two 12.5\,\Omega resistors are in parallel. Two equal resistances give half of one.
R_p=\dfrac{12.5}{2}=6.25\,\Omega
The parallel pair is in series with the 4.5\,\Omega resistor. Series resistances add.
R_{total}=4.5+6.25=10.75\,\Omega

R_{total}=10.75\,\Omega

Book answer. 10.75\,\Omega

FBISE 2026 questions

Questions from the FBISE SSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (iii) OR3 marks

Calculate the total e.m.f of three cells of 1.5 V when they are connected in: a. In series b. In parallel

Solution

Series. The positive terminal of each cell is joined to the negative terminal of the next, so the emfs add: \varepsilon_{net}=\varepsilon_1+\varepsilon_2+\cdots+\varepsilon_n.

Given

n = 3
\varepsilon = 1.5\,\mathrm{V}\ \text{(each cell)}

To find

(a) \varepsilon_{net} in series
(b) \varepsilon_{net} in parallel

(a) Series: add the emfs of the cells.
\varepsilon_{net}=1.5+1.5+1.5=4.5\,\mathrm{V}
(b) Parallel: all the positive terminals are joined together and all the negative terminals are joined together. For identical cells the emf stays that of one cell.
\varepsilon_{net}=\varepsilon_1=\varepsilon_2=\cdots=1.5\,\mathrm{V}

\text{(a) }4.5\,\mathrm{V}\quad\text{(b) }1.5\,\mathrm{V}

FBISE SSC-II 2026 Annual I, Set B, Q.2 (iv)3 marks

Calculate the cost of using a 1000 W heater for 5 hours if the cost of 1 unit is Rs. 30.

Solution

Given

P = 1000\,\mathrm{W}=1\,\mathrm{kW}
t = 5\,\mathrm{h}
\text{rate} = \text{Rs. }30\ \text{per unit (kWh)}

To find

energy E in kWh
cost

Energy used is power multiplied by time. With the power in kW and the time in hours the energy comes out in kilowatt-hours (units).
E=Pt=1\,\mathrm{kW}\times5\,\mathrm{h}=5\,\mathrm{kWh}
One unit is one kWh, so the cost is the number of units multiplied by the price of one unit.
\text{Cost}=5\times30=\text{Rs. }150

E=5\,\mathrm{kWh},\quad\text{Cost}=\text{Rs. }150

FBISE SSC-II 2026 Annual I, Set B, Q.2 (v) OR3 marks

Calculate the power consumed by an appliance operating at 220 V drawing a current of 2 A.

Solution

Given

V = 220\,\mathrm{V}
I = 2\,\mathrm{A}

To find

power P

Electric power is the rate of doing electrical work. The formula list gives it as the product of voltage and current.
P=VI
Substitute.
P=220\,\mathrm{V}\times2\,\mathrm{A}=440\,\mathrm{W}

P=440\,\mathrm{W}

FBISE SSC-II 2026 Annual I, Set B, Q.5 OR3+2 = 5 marks

With the help of circuit diagram derive the formula for equivalent resistance of two resistors connected in parallel. Calculate the combined resistance of two resistors of 6\,\Omega and 3\,\Omega connected in parallel.

Solution

Derivation (3 marks)

Derivation Equivalent resistance of two resistors in parallel

Aim. Find the single resistance R_{eq} that can replace two resistors R_1 and R_2 joined in parallel across a battery of potential difference V.

Assumptions.

The connecting wires and the battery have negligible resistance.
Both resistors are joined directly across the same two points, so each has the full potential difference V.

Two resistors in parallel. The same potential difference V is across each resistor, and the branch currents I_1 and I_2 add up to the total current I.

In a parallel circuit the potential difference across each resistor is the same and equal to the battery voltage V.
V_1=V_2=V
Let I_1 and I_2 be the currents in the two branches. By Ohm's law, I=V/R for each.
I_1=\dfrac{V}{R_1},\qquad I_2=\dfrac{V}{R_2}
Charge does not pile up at the junction, so the total current I from the battery equals the sum of the branch currents.
I=I_1+I_2=\dfrac{V}{R_1}+\dfrac{V}{R_2}=V\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)
Divide both sides by V.
\dfrac{I}{V}=\dfrac{1}{R_1}+\dfrac{1}{R_2}
The equivalent resistance is the single resistance that draws the same current I at the same V, so I/V=1/R_{eq}.
\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}
Combine the fractions to get R_{eq} itself (product over sum).
R_{eq}=\dfrac{R_1R_2}{R_1+R_2}

Result\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\qquad\text{or}\qquad R_{eq}=\dfrac{R_1R_2}{R_1+R_2}

R_{eq} is less than the smaller of the two resistances, because each new branch gives the current one more path.

Numerical (2 marks)

Given

R_1 = 6\,\Omega
R_2 = 3\,\Omega

To find

combined resistance R_{eq} in parallel

Use the parallel formula for two resistors.
\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}=\dfrac{1}{6}+\dfrac{1}{3}
Put both fractions over the common denominator 6.
\dfrac{1}{R_{eq}}=\dfrac{1}{6}+\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}
Take the reciprocal.
R_{eq}=2\,\Omega
Check with product over sum.
R_{eq}=\dfrac{6\times3}{6+3}=\dfrac{18}{9}=2\,\Omega

R_{eq}=2\,\Omega

FBISE SSC-II 2026 Annual I, Set D, Q.2 (iii) OR3 marks

Calculate the total e.m.f of four cells of 2V when they are connected in: a. In series b. In parallel

Solution

Series. The positive terminal of each cell is joined to the negative terminal of the next, so the emfs add: \varepsilon_{net}=\varepsilon_1+\varepsilon_2+\cdots+\varepsilon_n.

Given

n = 4
\varepsilon = 2\,\mathrm{V}\ \text{(each cell)}

To find

(a) \varepsilon_{net} in series
(b) \varepsilon_{net} in parallel

(a) Series: add the emfs of the cells.
\varepsilon_{net}=2+2+2+2=8\,\mathrm{V}
(b) Parallel: all the positive terminals are joined together and all the negative terminals are joined together. For identical cells the emf stays that of one cell.
\varepsilon_{net}=\varepsilon_1=\varepsilon_2=\cdots=2\,\mathrm{V}

\text{(a) }8\,\mathrm{V}\quad\text{(b) }2\,\mathrm{V}

FBISE SSC-II 2026 Annual I, Set D, Q.2 (iv)3 marks

An electric bulb of 100 W is used for 10 hours daily. Calculate the energy consumed in one day.

Solution

Given

P = 100\,\mathrm{W}
t = 10\,\mathrm{h}

To find

energy E consumed in one day

Energy is power multiplied by time.
E=Pt=100\,\mathrm{W}\times10\,\mathrm{h}=1000\,\mathrm{Wh}
Convert to kilowatt-hours (the commercial unit) by dividing by 1000.
E=\dfrac{1000}{1000}=1\,\mathrm{kWh}
In joules, using 1\,\mathrm{kWh}=3.6\times10^{6}\,\mathrm{J}.
E=3.6\times10^{6}\,\mathrm{J}

E=1\,\mathrm{kWh}=3.6\times10^{6}\,\mathrm{J}

FBISE SSC-II 2026 Annual I, Set D, Q.2 (v) OR1+2 = 3 marks

Find the power of an electric bulb if the current through it is 0.5 A at 220 V.

Solution

Given

V = 220\,\mathrm{V}
I = 0.5\,\mathrm{A}

To find

power P of the bulb

Power is the product of the potential difference and the current (1 mark for the formula).
P=VI
Substitute the values (2 marks).
P=220\,\mathrm{V}\times0.5\,\mathrm{A}=110\,\mathrm{W}

P=110\,\mathrm{W}

FBISE SSC-II 2026 Annual I, Set D, Q.5 OR2.5+2.5 = 5 marks

Derive the formula for equivalent resistance and draw circuit diagrams of two resistors when these are connected: a. In parallel b. In series

Solution

(a) Parallel combination (2.5 marks)

Derivation Equivalent resistance of two resistors in parallel

Aim. Find the single resistance R_{eq} that can replace two resistors R_1 and R_2 joined in parallel across a battery of potential difference V.

Assumptions.

The connecting wires and the battery have negligible resistance.
Both resistors are joined directly across the same two points, so each has the full potential difference V.

Two resistors in parallel. The same potential difference V is across each resistor, and the branch currents I_1 and I_2 add up to the total current I.

In a parallel circuit the potential difference across each resistor is the same and equal to the battery voltage V.
V_1=V_2=V
Let I_1 and I_2 be the currents in the two branches. By Ohm's law, I=V/R for each.
I_1=\dfrac{V}{R_1},\qquad I_2=\dfrac{V}{R_2}
Charge does not pile up at the junction, so the total current I from the battery equals the sum of the branch currents.
I=I_1+I_2=\dfrac{V}{R_1}+\dfrac{V}{R_2}=V\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)
Divide both sides by V.
\dfrac{I}{V}=\dfrac{1}{R_1}+\dfrac{1}{R_2}
The equivalent resistance is the single resistance that draws the same current I at the same V, so I/V=1/R_{eq}.
\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}
Combine the fractions to get R_{eq} itself (product over sum).
R_{eq}=\dfrac{R_1R_2}{R_1+R_2}

Result\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\qquad\text{or}\qquad R_{eq}=\dfrac{R_1R_2}{R_1+R_2}

R_{eq} is less than the smaller of the two resistances, because each new branch gives the current one more path.

(b) Series combination (2.5 marks)

Derivation Equivalent resistance of two resistors in series

Aim. Find the single resistance R_{eq} that can replace two resistors R_1 and R_2 joined in series across a battery of potential difference V.

Assumptions.

The connecting wires and the battery have negligible resistance.
There is only one path, so the same current flows through both resistors.

Two resistors in series. The same current I flows through both, and the potential differences add: V=V_1+V_2.

In a series circuit the same current I flows through each resistor.
I_1=I_2=I
Apply Ohm's law V=IR to each resistor.
V_1=IR_1,\qquad V_2=IR_2
The battery voltage is shared between the resistors, so the potential differences add up to V.
V=V_1+V_2
Substitute and take I as a common factor.
V=IR_1+IR_2=I(R_1+R_2)
The equivalent resistance is the ratio of the total potential difference to the current, R_{eq}=V/I.
R_{eq}=\dfrac{V}{I}=R_1+R_2

ResultR_{eq}=R_1+R_2

In series the resistances add, so R_{eq} is greater than either resistor.

\text{series: }R_{eq}=R_1+R_2;\quad\text{parallel: }\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}

FBISE SSC-II 2026 Annual I, Set M, Q.2 (iii) OR1.5+1.5 = 3 marks

Calculate the total e.m.f of two cells of 3V are connected in: a. In series b. In parallel

Solution

Series. The positive terminal of each cell is joined to the negative terminal of the next, so the emfs add: \varepsilon_{net}=\varepsilon_1+\varepsilon_2+\cdots+\varepsilon_n.

Given

n = 2
\varepsilon = 3\,\mathrm{V}\ \text{(each cell)}

To find

(a) \varepsilon_{net} in series
(b) \varepsilon_{net} in parallel

(a) Series: add the emfs of the cells.
\varepsilon_{net}=3+3=6\,\mathrm{V}
(b) Parallel: all the positive terminals are joined together and all the negative terminals are joined together. For identical cells the emf stays that of one cell.
\varepsilon_{net}=\varepsilon_1=\varepsilon_2=\cdots=3\,\mathrm{V}

\text{(a) }6\,\mathrm{V}\quad\text{(b) }3\,\mathrm{V}

FBISE SSC-II 2026 Annual I, Set M, Q.2 (iv)1x3 = 3 marks

Calculate the electricity bill for using a 2 kW electric iron for 3 hours if the cost per unit is Rs. 25.

Solution

Given

P = 2\,\mathrm{kW}
t = 3\,\mathrm{h}
\text{rate} = \text{Rs. }25\ \text{per unit (kWh)}

To find

the electricity bill (cost)

Energy used is power multiplied by time. The power is already in kW.
E=Pt=2\,\mathrm{kW}\times3\,\mathrm{h}=6\,\mathrm{kWh}
Cost is energy in kWh multiplied by the cost per unit.
\text{Cost}=6\times25=\text{Rs. }150

E=6\,\mathrm{kWh},\quad\text{Bill}=\text{Rs. }150

FBISE SSC-II 2026 Annual I, Set M, Q.2 (v) OR3 marks

An appliance uses 220 V and draws a current of 1 A. Calculate the energy consumed in 5 s.

Solution

Given

V = 220\,\mathrm{V}
I = 1\,\mathrm{A}
t = 5\,\mathrm{s}

To find

energy E consumed

The energy transferred by a current is the work done by the charge that moves through the potential difference. The formula list gives it as
E=W=VIt
Substitute the values (the time is in seconds, so the energy is in joules).
E=220\,\mathrm{V}\times1\,\mathrm{A}\times5\,\mathrm{s}=1100\,\mathrm{J}
Check with power: P=VI=220\,\mathrm{W}, so E=Pt=220\times5=1100\,\mathrm{J}.

E=1100\,\mathrm{J}

FBISE SSC-II 2026 Annual I, Set M, Q.5 OR3+2 = 5 marks

a. Derive the formula for equivalent resistance of two resistors connected in parallel. b. Calculate the total resistance when two identical resistors of 10\,\Omega are connected in parallel.

Solution

(a) Derivation (3 marks)

Derivation Equivalent resistance of two resistors in parallel

Aim. Find the single resistance R_{eq} that can replace two resistors R_1 and R_2 joined in parallel across a battery of potential difference V.

Assumptions.

The connecting wires and the battery have negligible resistance.
Both resistors are joined directly across the same two points, so each has the full potential difference V.

Two resistors in parallel. The same potential difference V is across each resistor, and the branch currents I_1 and I_2 add up to the total current I.

In a parallel circuit the potential difference across each resistor is the same and equal to the battery voltage V.
V_1=V_2=V
Let I_1 and I_2 be the currents in the two branches. By Ohm's law, I=V/R for each.
I_1=\dfrac{V}{R_1},\qquad I_2=\dfrac{V}{R_2}
Charge does not pile up at the junction, so the total current I from the battery equals the sum of the branch currents.
I=I_1+I_2=\dfrac{V}{R_1}+\dfrac{V}{R_2}=V\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)
Divide both sides by V.
\dfrac{I}{V}=\dfrac{1}{R_1}+\dfrac{1}{R_2}
The equivalent resistance is the single resistance that draws the same current I at the same V, so I/V=1/R_{eq}.
\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}
Combine the fractions to get R_{eq} itself (product over sum).
R_{eq}=\dfrac{R_1R_2}{R_1+R_2}

Result\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\qquad\text{or}\qquad R_{eq}=\dfrac{R_1R_2}{R_1+R_2}

R_{eq} is less than the smaller of the two resistances, because each new branch gives the current one more path.

(b) Numerical (2 marks)

Given

R_1 = 10\,\Omega
R_2 = 10\,\Omega

To find

total resistance R_{eq} in parallel

Use the parallel formula.
\dfrac{1}{R_{eq}}=\dfrac{1}{10}+\dfrac{1}{10}=\dfrac{2}{10}=\dfrac{1}{5}
Take the reciprocal.
R_{eq}=5\,\Omega
Check with the rule for n equal resistors, R_{eq}=R/n.
R_{eq}=\dfrac{10}{2}=5\,\Omega

R_{eq}=5\,\Omega

Exam focus

Derive the equivalent resistance of two resistors in parallel with a circuit diagram, then find the combined resistance (for example 6\,\Omega with 3\,\Omega, or 10\,\Omega with 10\,\Omega). (long)
Q.5 OR (5 marks) in all three 2026 sets: Set B and Set M ask for the parallel derivation plus a numerical, and Set D asks for the parallel derivation. Long question 3 of the book covers the same derivation.
Go to answer
Derive R_{eq}=R_1+R_2 for two resistors in series and draw the circuit diagram, together with the parallel case. (long)
Set D, Q.5 OR (2.5+2.5 marks) asks for both series and parallel derivations with circuit diagrams. Long question 2 of the book is the series derivation.
Go to answer
Calculate the cost of running an appliance: change watts to kW, find E=Pt in kWh, then multiply by the price per unit. (short)
Q.2 (iv) (3 marks) in Set B (1000 W heater, 5 h, Rs. 30 per unit), Set D (100 W bulb, 10 h a day, energy only) and Set M (2 kW iron, 3 h, Rs. 25 per unit). Numericals 3, 4 and 7 of the book practise the same E=Pt.
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Find the total emf of identical cells joined in series and in parallel (for example three cells of 1.5 V, four of 2 V, two of 3 V). (short)
Q.2 (iii) OR (3 marks) in all three 2026 sets, always series and parallel for identical cells. Numerical 5 and MCQ 6 of the book use the same two rules.
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Calculate electric power with P=VI (for example 220 V and 2 A, or 220 V and 0.5 A) and electrical energy with E=VIt. (short)
Q.2 (v) OR (3 marks) in all three 2026 sets: power in Set B (440 W) and Set D (110 W, marks 1+2), and the energy in 5 s in Set M. MCQ 4 and Numerical 3 of the book use the same formulas.
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Explain the kilowatt-hour as a unit of energy, convert it to joules (1\,\mathrm{kWh}=3.6\times10^{6}\,\mathrm{J}) and use it for a bill. (short)
The cost and energy questions of Q.2 (iv) in all three 2026 sets depend on it, and short question 11 of the book asks for the conversion.
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Compare series and parallel circuits: current, voltage, equivalent resistance and what happens when one device fails. (long)
Not asked directly in 2026, but the parallel and series derivations of Q.5 OR rest on this comparison, and long question 3 of the book is exactly this; it is also why homes are wired in parallel (short question 2).
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