Current electricity

Class 10 Physics, unit 16, book pages 185–204

Electric current is the flow of charge, and it is the basis of every electrical device. This unit defines current and the ampere, explains conduction in metals and the difference between conventional current and the flow of electrons, and compares direct and alternating current. It then covers potential difference and emf, Ohm's law with ohmic and non-ohmic devices, resistance and the factors that change it, resistivity and its temperature dependence, the use of ammeters and voltmeters, and an experiment to find an unknown resistance.

Learning outcomes (15)
  • Define and calculate electric current.
  • Explain electrical conduction.
  • State that current is measured in amps (amperes) and that the amp is given by coulomb per second ().
  • Differentiate between direct current (d.c.) and alternating current (a.c.).
  • Differentiate between conventional and actual current.
  • Justify and illustrate the use of ammeters.
  • Define e.m.f.
  • Define p.d. (potential difference).
  • State that e.m.f. and p.d. are measured in volts and that the volt is given by joule per coulomb ().
  • Justify and illustrate the use of voltmeters.
  • Describe an experiment to determine resistance.
  • Define and calculate resistivity.
  • Define and apply Ohm's law.
  • Describe the effect of temperature increase on the resistance of a resistor.
  • Interpret current-voltage graphs.

16.1Electric currentOften examined

Electric current is produced by the flow of either positive or negative charges, or sometimes both. In metals the current is due to the flow of negatively charged electrons. The current in the beam of a particle accelerator is due to positively charged protons, while in gases and electrolytes it is due to the flow of both positive and negative charges.

Electric current
The time rate of flow of electric charges through a cross-sectional area.

Suppose a charge flows through any cross-sectional area of a conductor in a time (Figure 16.1). The electric current is then

16.1
Figure 16.1: Electric current in a simple circuit. The cell is the energy source, the bulb is the load and the coloured arrows show the flow of electrons in the connecting wires.
The ampere

The SI unit of current is the ampere (A). One ampere is the current when one coulomb of charge flows through any cross-section in one second.

The ampere is a large unit, so these sub-multiples are commonly used:

NameSymbolValue
milliampere
microampere

Conventional flow of current (16.1.1)

Before the discovery of the electron it was thought that electric current in conductors was due to the flow of positive charges. This assumed direction of current flow is now called conventional current. Current flowing from the positive terminal to the negative terminal of a battery, due to the flow of positive charges, is called conventional current.

For practical purposes, positive charges moving in one direction are equivalent to negative charges moving in the opposite direction. Conventional current produces the same effect as the current of negative charges flowing from the negative terminal to the positive terminal. That is why both are used in electrical circuit problems (Figure 16.2).

Figure 16.2: Conventional flow of current. Conventional current (dark arrows) leaves the positive terminal of the battery. The electrons, the actual moving charges in a metal, go the opposite way (coloured arrows).
Conventional currentElectron (actual) current in a metal
Charges assumed to movePositive chargesNegative electrons
Direction in the circuitPositive terminal to negative terminalNegative terminal to positive terminal
UseCircuit diagrams and calculationsThe real motion of charge in a metal wire

Electrical conduction in metals (16.1.2)

In metals the valence electrons are not tightly bound by the nucleus and can move around freely, like the movement of gas molecules in a container. When no battery or power supply is connected to a piece of copper wire (Figure 16.3 (a)), the free electrons move at random and their net motion is zero, so there is no electric current.

When a battery is connected across the ends of the wire (Figure 16.3 (b)), the free electrons start to move in one direction. The positive plate of the battery attracts the free electrons and the negative plate pushes them. The electrons drift through the metal, like leaves carried by a gentle breeze. On the way they collide with each other and with the nuclei (atoms) of the metal, which slows them down.

The net movement of these free electrons in one direction is an electric current. So electrical conduction in metals happens because of the movement of free electrons when an electric field is applied across the metal.

Figure 16.3: Electric current in metals. (a) The free electrons move at random, so the net motion is zero. (b) With a battery connected the electrons drift towards the positive end.
Example 16.1: Charge passing into a car battery
A battery of an electric car is charged at a charging station using a current of for two hours. How much charge passes to the battery?
Given
To find
  • Charge
  1. By the definition of electric current, . Make the subject.
  2. Put in the values.
  3. Charge in ampere-hours: divided by per hour is , a specification commonly used for batteries.
Worked example: charge, current and time from a graph
A torch battery delivers a steady current of for minutes. (a) How much charge flows round the circuit? (b) At the same current, how long does it take to move ? (c) Sketch the graph of charge against time and show how the current is read from it.
Given
To find
  • (a) charge
  • (b) time
  • (c) graph
  1. Change the units first: and minute is .
  2. (a) Use .
  3. (b) Rearrange to .
  4. (c) Because with constant, the graph is a straight line through the origin. Two points are and . The slope is the current.
Charge against time at a steady current
100
200
300
400
500
600
700
25
50
75
100
125
150
175
O
A steady current gives a straight line through the origin. The slope is the current: .

16.2Alternating and direct current

Electric current flows in two ways: in a fixed direction, called direct current (DC), or changing direction, called alternating current (AC). Current is nothing but the movement of charges (electrons here) through a conductor such as a wire. The difference between AC and DC lies in the direction in which the charges flow.

Direct current (DC)
Current that flows steadily in a single direction (forward). It may be steady (uniform) or varying.
Alternating current (AC)
Current in which the electrons keep switching direction, going forward half the time and backward the other half.

In AC the direction of the current depends on the changing polarity of the alternating voltage source. Electric generators can produce either DC or AC. Both AC and DC can be measured with a multimeter, as it has an option for both.

Figure 16.4: Circuit symbols. The DC symbol is a cell (long plate positive, short plate negative); the AC symbol is a circle with a wave.
(a) Direct current
2
4
6
8
10
12
1
2
3
4
O
Steady DC has a constant value. Varying DC changes in size but never changes direction (it stays on one side of the axis).
(b) Alternating current
2
4
6
8
10
12
−3
−2
−1
1
2
3
O
AC changes direction continuously: the current is positive for half of each cycle (one way) and negative for the other half (the other way).
Direct current (DC)Alternating current (AC)
Direction of flowOne fixed directionKeeps reversing (forward and backward)
Graph of against A horizontal line (steady) or a curve on one side of the axis (varying)A wave that goes above and below the axis
SourceCell, battery, DC generatorAlternating voltage source, AC generator

16.3Potential differenceOften examined

When one end of a conductor is connected to the positive terminal of a battery and the other end to the negative terminal, a potential difference is produced between the two ends of the conductor.

Potential difference
The difference of electric potential between two points.

Conventionally the positive terminal of a battery is at higher potential and the negative terminal is at lower potential. So the end of a circuit component connected to the positive terminal is at higher potential than the end connected to the negative terminal (Figure 16.5).

Figure 16.5: Potential difference. The end of the bulb joined to the positive terminal is at high potential and the other end is at low potential.

Current continues to flow as long as there is a potential difference. The device that provides the potential difference for a steady flow of current in a copper wire or any other conductor is the battery (or any other source, described under emf).

The charge leaving the positive terminal of the battery has potential energy, a part of which is lost in the device. So there is a difference of potential energy per coulomb of charge between the two ends of the device. This is the potential difference:

16.2

The SI unit of potential difference is the volt (V). One volt is the potential difference when joule of work (potential energy) is needed to move a charge of coulomb from one point to another inside the field.

Worked example: energy, charge and potential difference
A lamp transfers of electrical energy to heat and light as of charge passes through it. (a) What is the potential difference across the lamp? (b) A battery moves of charge through a circuit. How much energy does it give to the charge?
Given
To find
  • (a) across the lamp
  • (b) energy given by the battery
  1. (a) Potential difference is the energy transferred per coulomb, .
  2. (b) For the battery use the same relation written the other way, (energy supplied per coulomb is ).
  3. Check the units: , so and .

16.4Electromotive force (emf)Often examined

To maintain a steady current in a circuit there must be a source that continuously converts non-electrical energy into electrical energy.

Electromotive force (emf)
The energy supplied by a battery per unit positive charge when it flows through the closed circuit.
emf source
A device that converts non-electrical energy into electrical energy. It is needed in a circuit to maintain a constant potential difference across its components (devices).

Consider a battery connected through conducting wires to a device such as a light bulb (Figure 16.6). The positive charge leaves the positive terminal of the battery, passes through the conductor and reaches the negative terminal of the battery. As the positive charge enters the battery at its lower potential point (the negative terminal), the battery has to perform work on the positive charge to drive it to the point of higher potential, the positive terminal.

Figure 16.6: Electromotive force. Inside the source the charge is pushed from low to high potential (work is done on it); in the bulb it moves from high to low potential and loses energy.

If is the work done by the battery on a positive charge , the electromotive force (emf, symbol ) of the battery is

16.3

The SI unit of emf is the same as that of electric potential: . Examples of emf sources (with the energy conversion):

  • Batteries: chemical energy into electrical energy.
  • Solar panels: solar (light) energy into electrical energy.
  • Thermocouples: thermal energy into electrical energy.
  • Generators: mechanical energy into electrical energy.
emfPotential difference
MeaningEnergy supplied by the source to each coulomb of charge that flows round the closed circuitEnergy transferred per coulomb between two points (for example across a device)
Formula
Unitvolt ()volt ()
WhereProvided by the source (battery, generator, solar panel)Across any part of the circuit that uses energy (bulb, resistor)
EEG brain waves (Science tidbit)
Brain stateWave nameFrequency
Deep sleepDelta to
DrowsyTheta to
RelaxedAlpha to
FocusedBeta to

16.5Ohm's lawOften examined

Ohm's law states the relationship between the electric current flowing through a conductor and the potential difference applied across it.

Ohm's law
The current between two points in a conductor is directly proportional to the applied voltage across these points, as long as the temperature and the physical state of the conductor are kept constant.

If is the potential difference across the two ends of a conductor, then a current flows through it:

Replacing the sign of proportionality by an equals sign and putting in a constant gives

16.4

is the proportionality constant and is called the resistance.

Ohm's law is not applicable to all devices. It can be applied only to certain devices, mostly metals, which are called ohmic devices (examples: copper and silver wires and resistors). Devices that do not obey Ohm's law are called non-ohmic devices (examples: filament bulb, thermistor and semiconductor diode).

Limitations of Ohm's law (16.5.1)

Ohm's law is applicable only to materials for which the graph of against is a straight line. However, only metallic conductors show this behaviour, and that too only over a limited range of voltages. Many important devices do not obey Ohm's law and are called non-ohmic devices.

Ohmic and non-ohmic devices (16.5.2)

Figure 16.8 (a) shows that when is constant the graph of against is a straight line. All materials do not give a straight line, as the graphs (b), (c) and (d) show. The graphs of a filament bulb, a thermistor and a semiconductor diode are curved, so they are called non-ohmic.

(a) Metal
1
2
3
1
2
3
O
A straight line through the origin: is constant, so the metal is ohmic.
(b) Filament bulb
1
2
3
1
2
3
O
The curve flattens: the current saturates, so a large change in gives only a small change in .
(c) Thermistor
1
2
3
1
2
3
O
The curve steepens: the resistance falls sharply, so even a small change in gives a large change in .
(d) Semiconductor diode
−1.5
−1
−0.5
0.5
1
2
4
6
8
O
Almost no current for reverse (negative) voltage and a very steep rise once the forward voltage is large: far from a straight line.
  • Filament bulb: the current saturates as it is increased, and at a large value even a large change in voltage gives only a small change in current .
  • Thermistor: the resistance decreases sharply, so at a large value even a small change in voltage gives a large change in current.
  • Semiconductor diode: the graph deviates very strongly from a straight line. In fact most modern electronic devices, such as transistors, have non-linear current-voltage relationships and are therefore non-ohmic.
Ohmic deviceNon-ohmic device
Graph of against Straight line through the originCurve
Resistance ConstantChanges with voltage (and temperature)
ExamplesCopper and silver wires, resistorsFilament bulb, thermistor, semiconductor diode
Example 16.2: Current through a torch bulb
A torch bulb uses two cells, each of , connected in series, to glow. What current will flow through the circuit of the torch when its resistance is ?
Given
To find
  • Current
  1. By Ohm's law , so .
  2. Put in the values.
Worked example: which device obeys Ohm's law?
A student measures the current through a resistor and through a torch bulb at four voltages. Resistor: , ; , ; , ; , . Bulb: , ; , ; , ; , . Find for each reading, plot against and decide which device is ohmic.
  1. Calculate for every reading and put the results in a table.
(V) resistor (A) resistor () bulb (A) bulb ()
1.50.10150.207.5
3.00.20150.2810.7
4.50.30150.3512.9
6.00.40150.4015
  1. Resistor: is every time, so . The graph is a straight line through the origin.
  2. Bulb: increases from to , so is not proportional to . The graph bends towards the axis, like the filament bulb of Figure 16.8 (b). The filament gets hotter as the current grows and its resistance increases.
  3. Conclusion: the resistor is ohmic and the bulb is non-ohmic.
I against V for a resistor and a bulb
1
2
3
4
5
6
0.1
0.2
0.3
0.4
0.5
O
The resistor gives a straight line through the origin (constant ). The bulb gives a curve: rises from to .

16.6ResistanceOften examined

Resistance
The opposition to the flow of charges.

From equation 16.4 the resistance can be written as

R=\dfrac{V}{I}
16.5

Its unit is volt per ampere, called the ohm (\Omega). The resistance of a wire is one ohm if a voltage of one volt causes a current of one ampere to flow through the wire.

1\,\Omega=\dfrac{1\,\mathrm{V}}{1\,\mathrm{A}}

The following multiples are used for a wide range of resistance values:

NameSymbolValue
kilo-ohm\mathrm{k}\Omega10^{3}\,\Omega
mega-ohm\mathrm{M}\Omega10^{6}\,\Omega

The value of resistance is independent of the voltage and current: if the voltage increases, the current increases too, and the ratio V/I remains constant. Resistance depends on the nature of the material, its geometrical properties (length and cross-sectional area) and its temperature.

There are many types of resistors (Figure 16.9): carbon composition, carbon film, metal film, metal oxide film, wire-wound resistor, thermistor, fusible resistor, carbon potentiometer, rheostat, trimmer, surface-mount resistor and wire-wound potentiometer.

Example 16.3: Resistance of a device
What will be the resistance of a device when connected to a battery of 12\,\mathrm{V} and it allows a current of 200\,\mathrm{mA}?
Given
  • V = 12.0\,\mathrm{V}
  • I = 200\,\mathrm{mA}=200\times10^{-3}\,\mathrm{A}
To find
  • Resistance R
  1. By the definition of resistance:
    R=\dfrac{V}{I}
  2. Put in the values (current first changed to amperes).
    R=\dfrac{12.0\,\mathrm{V}}{200\times10^{-3}\,\mathrm{A}}=60\,\Omega
R=60\,\Omega

16.6.1 to 16.6.4Factors affecting resistanceOften examined

The resistance is provided by the collisions of the free electrons with the lattice atoms. So any factor that affects the number of collisions will also affect the resistance of a material.

Length of wire (16.6.1)

The greater the length of a conductor, the greater its resistance: the resistance is directly proportional to the length.

R\propto L
1

Cross-sectional area of wire (16.6.2)

The greater the cross-sectional area of a conductor, the smaller its resistance: the resistance is inversely proportional to the cross-sectional area.

R\propto\dfrac{1}{A}
2

Temperature (16.6.3)

The resistance of a material changes as the temperature changes, depending on the nature of the material.

  • Conductors: as the temperature rises the resistance increases, because when the free electrons in a conductor get heated, the chances of collisions with each other increase.
  • Insulators and semiconductors: the resistance decreases as the temperature increases, because the thermal energy generates more charge carriers.

So the change in resistance is related to the change in temperature:

\Delta R\propto\Delta T
3

The filament of a bulb is made of tungsten, which is a conductor. Initially the temperature of the tungsten is low and so is its resistance, so a maximum current passes through it. This heats the tungsten and its temperature increases. As the temperature increases its resistance increases and the current drops, as shown in Figure 16.10.

Resistance of tungsten (Figure 16.10)
When the bulb is switched on, the cold tungsten filament has low resistance and the current is large. As the filament heats up its resistance increases and the current drops.

Material nature of wire (16.6.4)

The resistance of a body also depends upon the nature of the material, and its value changes from material to material. Usually insulators have greater resistivity while conductors have smaller. Different conductors have different resistivities, depending upon their internal structure.

FactorEffect on resistance of a metal wire
Length L increasesResistance increases (R\propto L)
Area A increasesResistance decreases (R\propto1/A)
Temperature risesResistance increases
MaterialDifferent for each material (resistivity)

16.7ResistivityOften examined

Resistivity (specific resistance)
The resistance \rho offered by a wire of length 1\,\mathrm{m} having cross-sectional area 1\,\mathrm{m^2}.

Keeping the temperature constant, the resistance of a conductor is directly proportional to its length L and inversely proportional to its cross-sectional area A:

R\propto\dfrac{L}{A}\qquad\text{or}\qquad R=\rho\,\dfrac{L}{A}
16.6

Here \rho is the proportionality constant and is called the resistivity of the material:

\rho=R\,\dfrac{A}{L}
16.7

The unit of resistivity is the ohm-metre (\Omega\,\mathrm{m}).

Figure 16.11: Resistivity. For a conductor with L=1\,\mathrm{m} and A=1\,\mathrm{m^2}, the resistance R equals the resistivity \rho.

Every material has a characteristic resistivity that depends on its nature and on temperature. Good electric conductors have very low resistivities and good insulators have very high resistivities. If L=1\,\mathrm{m} and A=1\,\mathrm{m^2}, then R=\rho. The resistivity of some materials is given in Table 16.1.

Table 16.1: Resistivities of some materials at 20\,^\circ\mathrm{C}
MaterialResistivity \rho (\Omega\,\mathrm{m})
Typical metals
Silver1.59\times10^{-8}
Copper1.72\times10^{-8}
Gold2.24\times10^{-8}
Aluminium2.56\times10^{-8}
Tin11.0\times10^{-8}
Tungsten5.52\times10^{-8}
Iron9.71\times10^{-8}
Lead20.6\times10^{-8}
Typical semiconductors (pure)
Carbon3.5\times10^{-5}
Germanium1 to 500\times10^{-3}
Silicon0.1 to 60
Typical insulators
Glass1 to 1000\times10^{9}
Fused quartz7.5\times10^{17}
Rubber1 to 100\times10^{13}

The resistivity of metals and alloys is very small, so they are very good conductors of electricity. The resistivity of insulators is very high, which is why they conduct current with difficulty. The resistivity of a semiconductor lies between those of conductors and insulators. By doping a semiconductor (introducing controlled amounts of impurities) its resistivity can be changed dramatically, which is one reason that semiconductors are used to make computer chips and other electronic devices.

Example 16.4: Resistance of a silver wire
A 1\,\mathrm{m} silver wire (resistivity =1.6\times10^{-8}\,\Omega\,\mathrm{m}) has a cross-sectional area of 0.4\,\mathrm{cm^2}. Calculate its resistance.
Given
  • L = 1\,\mathrm{m}
  • A = 0.4\,\mathrm{cm^2}=0.4\times10^{-4}\,\mathrm{m^2}
  • \rho = 1.6\times10^{-8}\,\Omega\,\mathrm{m}
To find
  • Resistance R
  1. Change the area to square metres: 1\,\mathrm{cm}=10^{-2}\,\mathrm{m}, so 1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2}.
    A=0.4\times10^{-4}\,\mathrm{m^2}
  2. The relation for resistance is R=\rho L/A. Put in the values.
    R=\dfrac{(1.6\times10^{-8}\,\Omega\,\mathrm{m})(1\,\mathrm{m})}{0.4\times10^{-4}\,\mathrm{m^2}}
  3. Divide: 1.6/0.4=4 and 10^{-8}/10^{-4}=10^{-4}.
    R=4\times10^{-4}\,\Omega=0.4\,\mathrm{m}\Omega
R=4\times10^{-4}\,\Omega=0.4\,\mathrm{m}\Omega
Worked example: length, area and resistivity
Wire X has a resistance of 6.0\,\Omega. Wire Y is made of the same material but is 3 times as long and has half the cross-sectional area. (a) Find the resistance of wire Y. (b) Find the resistance of a copper wire 50\,\mathrm{m} long with a cross-sectional area of 1.0\,\mathrm{mm^2}, and explain what the graph of R against L looks like for this wire.
Given
  • R_X = 6.0\,\Omega
  • L_Y/L_X = 3
  • A_Y/A_X = \tfrac12
  • L = 50\,\mathrm{m}
  • A = 1.0\,\mathrm{mm^2}=1.0\times10^{-6}\,\mathrm{m^2}
  • \rho_{Cu} = 1.72\times10^{-8}\,\Omega\,\mathrm{m}
To find
  • (a) R_Y
  • (b) R of the copper wire
  1. (a) Same material means the same \rho, so R=\rho L/A gives a ratio. A longer wire has more resistance and a thinner wire has more resistance, so both changes increase R.
    \dfrac{R_Y}{R_X}=\dfrac{L_Y}{L_X}\times\dfrac{A_X}{A_Y}=3\times2=6
  2. Multiply.
    R_Y=6\times6.0=36\,\Omega
  3. (b) Change the area to square metres: 1\,\mathrm{mm}=10^{-3}\,\mathrm{m}, so 1\,\mathrm{mm^2}=10^{-6}\,\mathrm{m^2}. Then use R=\rho L/A.
    R=\dfrac{(1.72\times10^{-8})(50)}{1.0\times10^{-6}}=\dfrac{8.6\times10^{-7}}{1.0\times10^{-6}}=0.86\,\Omega
  4. For one wire \rho and A are fixed, so R=(\rho/A)\,L is a straight line through the origin with slope \rho/A=1.72\times10^{-8}/10^{-6}=1.72\times10^{-2}\,\Omega\,\mathrm{m^{-1}}.
Resistance against length for two copper wires
Resistance is directly proportional to length (straight lines through the origin). Doubling the area halves the slope: at 100\,\mathrm{m} the 1.0\,\mathrm{mm^2} wire has 1.72\,\Omega and the 2.0\,\mathrm{mm^2} wire has 0.86\,\Omega.
R_Y=36\,\Omega,\qquad R=0.86\,\Omega

16.7.1Temperature dependence of resistivityOften examined

The resistance of every conductor varies with temperature. At higher temperatures the internal energy of the material increases, causing the positive ions to vibrate with greater amplitudes. This increased vibration leads to more frequent collisions between electrons and ions. Consequently the motion of electrons is hindered, reducing the current for a given electric field. This explains why the resistivity of metals typically increases with rising temperature.

Consider a conductor having resistivity \rho_0 at 0\,^\circ\mathrm{C} and \rho_T at T\,^\circ\mathrm{C}. It is found that for normal temperatures the change in resistivity \Delta\rho=\rho_T-\rho_0 is directly proportional to:

  1. the initial resistivity \rho_0: \rho_T-\rho_0\propto\rho_0 (1)
  2. the rise in temperature \Delta T: \rho_T-\rho_0\propto\Delta T (2)

Combining the proportionalities 1 and 2 gives \rho_T-\rho_0\propto\rho_0\Delta T. Changing the sign of proportionality into equality:

\rho_T-\rho_0=\alpha\,\rho_0\,\Delta T

Here \alpha is the constant of proportionality and is called the temperature coefficient of resistivity. Its value depends only upon the nature of the material:

\alpha=\dfrac{\rho_T-\rho_0}{\rho_0\,\Delta T}\qquad\text{or}\qquad \rho_T=\rho_0(1+\alpha\Delta T)
16.8

The unit of the temperature coefficient of resistivity is ^\circ\mathrm{C^{-1}} or \mathrm{K^{-1}}. For common metals \alpha typically has a value of 0.003 to 0.005\,^\circ\mathrm{C^{-1}}. That is, an increase in temperature of 1\,\mathrm{C^\circ} increases the resistance by 0.3\% to 0.5\%.

The resistivity of a metallic conductor nearly always increases with increasing temperature. This is measured by the temperature coefficient of resistivity, which is positive for metals (Table 16.2). The temperature coefficient of resistivity is negative for semiconducting materials: unlike metals, their resistivity decreases with increasing temperature. The specific mechanisms by which this decrease happens in semiconductors are complex and beyond the scope of the book.

Table 16.2: Temperature coefficients of some materials at 20\,^\circ\mathrm{C}
Material\alpha (^\circ\mathrm{C^{-1}})
Typical metals
Silver3.8\times10^{-3}
Copper3.9\times10^{-3}
Gold3.4\times10^{-3}
Aluminium3.9\times10^{-3}
Manganin0.002\times10^{-3}
Tungsten4.5\times10^{-3}
Iron6.5\times10^{-3}
Lead3.9\times10^{-3}
Typical semiconductors (pure)
Carbon-0.5\times10^{-3}
Germanium-50\times10^{-3}
Silicon-75\times10^{-3}
Example 16.5: Resistivity after a rise in temperature
At some temperature the resistivity of a metallic conductor is 5.6\times10^{-8}\,\Omega\,\mathrm{m}. If the temperature is increased by 30\,\mathrm{K}, find the new resistivity. Take the temperature coefficient of resistivity as 0.0045\,\mathrm{K^{-1}}.
Given
  • \rho_0 = 5.6\times10^{-8}\,\Omega\,\mathrm{m}
  • \Delta T = 30\,\mathrm{K}
  • \alpha = 0.0045\,\mathrm{K^{-1}}
To find
  • Final resistivity \rho_T
  1. The temperature coefficient of resistivity is \alpha=(\rho_T-\rho_0)/(\rho_0\Delta T). Rearranged:
    \rho_T=\rho_0(1+\alpha\Delta T)
  2. Put in the values.
    \rho_T=(5.6\times10^{-8})\{1+(0.0045)(30)\}
  3. Work out the bracket: 0.0045\times30=0.135, so it is 1.135.
    \rho_T=5.6\times10^{-8}\times1.135=6.356\times10^{-8}\,\Omega\,\mathrm{m}\approx6.4\times10^{-8}\,\Omega\,\mathrm{m}
\rho_T\approx6.4\times10^{-8}\,\Omega\,\mathrm{m}\ (\text{book: }6.3\times10^{-8})
Worked example: copper wire heated by 50\,^\circ\mathrm{C}
The resistivity of copper at 20\,^\circ\mathrm{C} is 1.72\times10^{-8}\,\Omega\,\mathrm{m} and its temperature coefficient is 3.9\times10^{-3}\,^\circ\mathrm{C^{-1}} (Tables 16.1 and 16.2). (a) Find its resistivity at 70\,^\circ\mathrm{C}. (b) By what percentage has it increased? (c) Silicon has \alpha=-75\times10^{-3}\,^\circ\mathrm{C^{-1}}. What does the negative sign tell you?
Given
  • \rho_0 = 1.72\times10^{-8}\,\Omega\,\mathrm{m}
  • \alpha = 3.9\times10^{-3}\,^\circ\mathrm{C^{-1}}
  • \Delta T = 70-20=50\,^\circ\mathrm{C}
To find
  • (a) \rho_T
  • (b) percentage increase
  • (c) meaning of negative \alpha
  1. (a) Use \rho_T=\rho_0(1+\alpha\Delta T). First find \alpha\Delta T.
    \alpha\Delta T=3.9\times10^{-3}\times50=0.195
  2. Then the new resistivity.
    \rho_T=1.72\times10^{-8}\times(1+0.195)=1.72\times10^{-8}\times1.195=2.06\times10^{-8}\,\Omega\,\mathrm{m}
  3. (b) The increase is \alpha\Delta T as a fraction of \rho_0, so it is 0.195=19.5\%.
    \dfrac{\rho_T-\rho_0}{\rho_0}=\alpha\Delta T=0.195
  4. (c) A negative \alpha means that \rho_T-\rho_0 is negative when the temperature rises: the resistivity of silicon (a semiconductor) decreases with increasing temperature, unlike copper.
Resistivity of copper against temperature
Copper has a positive \alpha, so its resistivity rises in a straight line as the temperature rises: from 1.72\times10^{-8} at 20\,^\circ\mathrm{C} to 2.06\times10^{-8} at 70\,^\circ\mathrm{C}.
\rho_T=2.06\times10^{-8}\,\Omega\,\mathrm{m},\quad 19.5\%\ \text{increase}

16.8Electrical measuring instrumentsOften examined

Devices used for measuring electrical quantities such as current, voltage, resistance and power are called electrical measuring instruments.

Ammeter (16.8.1)

Ammeter
A device used to measure current.

To measure the current, an ammeter must be connected in series in the circuit, since it finds the current flowing through a component of the circuit. Figure 16.12 shows an analogue and a digital ammeter along with the circuit diagram. The ammeter must disturb the current in the circuit as little as possible, so ammeters are designed to have as low a resistance as possible (because they are connected in series), usually of the order of 1\,\Omega. Then they do not have an appreciable effect on the currents they measure.

Figure 16.12: Ammeter connected in series with the load in a circuit (the book also shows an analogue and a digital ammeter).

Range of ammeter: ammeters of different ranges are used, based on the amount of current in the circuit. To measure the current in home appliances an ammeter with values from 1\,\mathrm{A} to 100\,\mathrm{A} is usually used. For small currents a milliammeter (\mathrm{mA}) or a microammeter (\mu\mathrm{A}) is used (Figure 16.13).

Voltmeter (16.8.2)

Voltmeter
A device used to measure potential difference.

To measure the potential difference, a voltmeter must be connected in parallel with the component across which the potential difference is to be measured. Figure 16.14 shows a voltmeter placed in the circuit to measure the potential drop across a resistor. The voltmeter must make measurements while disturbing the circuit as little as possible, so voltmeters are designed to have as high a resistance as possible, usually of the order of 10\,\mathrm{M}\Omega (10^{7}\,\Omega). Then they have a negligible effect on the potential differences they measure.

Figure 16.14: Voltmeter connected in parallel with the load, across which it measures the potential difference.

Range of voltmeter: like ammeters, voltmeters have different ranges based on their use. For measuring home voltage a voltmeter with values from 1\,\mathrm{V} to 500\,\mathrm{V} is used. A millivoltmeter (\mathrm{mV}) is used for small voltage measurements (Figure 16.15).

AmmeterVoltmeter
MeasuresElectric currentPotential difference
ConnectedIn series with the componentIn parallel with the component
ResistanceAs low as possible (about 1\,\Omega)As high as possible (about 10\,\mathrm{M}\Omega)
ReasonSo that it does not reduce the current it measuresSo that it does not draw current from the circuit
Ranges\mathrm{A}, \mathrm{mA}, \mu\mathrm{A}\mathrm{V}, \mathrm{mV}
Worked example: connecting the meters correctly
A 6\,\mathrm{V} cell is joined to a bulb of resistance 20\,\Omega. (a) Describe how an ammeter and a voltmeter must be connected to measure the current in the bulb and the potential difference across it. (b) An ammeter of resistance 1\,\Omega is wrongly connected across the bulb. Compare the current in the two branches.
Given
  • V = 6\,\mathrm{V}
  • R_{bulb} = 20\,\Omega
  • R_A = 1\,\Omega
To find
  • (a) correct connections
  • (b) currents in the two branches
  1. (a) The ammeter measures the current through a component, so it is connected in series with the bulb (Figure 16.12). The voltmeter measures the potential difference across the bulb, so it is connected in parallel with the bulb (Figure 16.14).
  2. (b) Both branches are joined across the same two points, so each has the potential difference 6\,\mathrm{V}. Use I=V/R for each branch.
    I_{bulb}=\dfrac{6}{20}=0.3\,\mathrm{A},\qquad I_{ammeter}=\dfrac{6}{1}=6\,\mathrm{A}
  3. The ammeter gives the current an easy path, taking 20 times as much current as the bulb: the bulb nearly goes out and the ammeter (and cell) may be damaged. The ammeter is not meant to be across a component.
    \dfrac{6}{0.3}=20
The ammeter connected across the bulb. Its very low resistance makes the coloured branch an easy path, so most of the current bypasses the bulb. The correct connections are in Figures 16.12 and 16.14.
I_{bulb}=0.3\,\mathrm{A},\ I_{ammeter}=6\,\mathrm{A}

16.9Experiment for determination of resistance

Aim: to find an unknown resistance by measuring the current through it and the potential difference across it, using R=V/I.

Setup (Figure 16.16): the unknown resistance is connected in series with the ammeter, while the voltmeter is connected in parallel with the unknown resistance. A variable resistance called a rheostat is also connected in series with the unknown resistance and the ammeter.

Figure 16.16: Resistance measurement. The ammeter is in series with the unknown resistance and the rheostat; the voltmeter is in parallel with the unknown resistance.

Procedure

  1. Connect the circuit as in Figure 16.16: the unknown resistance in series with the ammeter and a rheostat, and the voltmeter in parallel with the unknown resistance.
  2. Make sure the indicators of both the voltmeter and the ammeter are at the zero mark of their scales.
  3. Record the full-scale value and the smallest division value (least count) for both the voltmeter and the ammeter.
  4. Close the switch (key K) and adjust the rheostat towards the position that gives the lowest current. Note the readings of both the voltmeter and the ammeter.
  5. Open the switch (remove the key K) and allow the wire to return to ambient temperature. Then close the circuit again and gradually raise the voltage by changing the position of the rheostat.
  6. Record the readings of the ammeter and the voltmeter and put them in the relation R=V/I (Table 16.3). Continue, adjusting the rheostat step by step, until you have four different voltage levels.
Table 16.3: Resistance calculation (to be filled in from your readings)
Sr. No.Ammeter reading (A)Voltmeter reading (V)Resistance R=V/I
1
2
3
4

Formula sheet

Electric current
I=\dfrac{\Delta Q}{\Delta t}
I: current; \Delta Q: charge passing a cross-section; \Delta t: time (16.1). Also \Delta Q=I\Delta t, and 1\,\mathrm{A}=1\,\mathrm{C\,s^{-1}}
Unit: \mathrm{A} (\mathrm{C\,s^{-1}})
Potential difference
\Delta V=\dfrac{\Delta U}{\Delta q}=\dfrac{\Delta W}{\Delta q}
\Delta V: potential difference; \Delta W: work done or energy transferred; \Delta q: charge moved (16.2)
Unit: \mathrm{V} (\mathrm{J\,C^{-1}})
Electromotive force
\varepsilon=\dfrac{W}{q}
\varepsilon: emf; W: work done by the source on the charge q (16.3)
Unit: \mathrm{V} (\mathrm{J\,C^{-1}})
Ohm's law
V=IR\quad\text{or}\quad I=\dfrac{V}{R}
V: potential difference across the conductor; I: current; R: resistance; temperature constant (16.4)
Units: \mathrm{V}, \mathrm{A}, \Omega
Resistance
R=\dfrac{V}{I}
R: resistance (16.5)
Unit: \Omega (\mathrm{V\,A^{-1}})
Factors of resistance
R\propto L,\qquad R\propto\dfrac{1}{A}
L: length of the wire; A: cross-sectional area (16.6.1, 16.6.2)
Resistance and resistivity
R=\rho\dfrac{L}{A}
\rho: resistivity of the material; L: length; A: cross-sectional area (16.6)
Unit: \Omega
Resistivity
\rho=R\dfrac{A}{L}
Resistance of a wire of unit length and unit cross-sectional area (16.7)
Unit: \Omega\,\mathrm{m}
Temperature coefficient of resistivity
\alpha=\dfrac{\rho_T-\rho_0}{\rho_0\Delta T}
\rho_0: initial resistivity; \rho_T: resistivity after a temperature change \Delta T (16.8)
Units: ^\circ\mathrm{C^{-1}} or \mathrm{K^{-1}}
Resistivity at a new temperature
\rho_T=\rho_0(1+\alpha\Delta T)
The same relation rearranged (Example 16.5)
Unit: \Omega\,\mathrm{m}
Conductance and conductivity
G=\dfrac{1}{R},\qquad \sigma=\dfrac{1}{\rho}
Reciprocals of resistance and resistivity (science fact)
Units: \mathrm{S} (siemens); \mathrm{S\,m^{-1}}

Key terms

Electric current
The rate of flow of charge: the charge passing through a cross-section per unit time, I=\Delta Q/\Delta t.
Ampere
The SI unit of electric current; 1\,\mathrm{A} is one coulomb of charge flowing per second.
Conventional current
The assumed direction of flow of positive charges, from the positive terminal to the negative terminal of a battery.
Direct current (DC)
Electric current that is unidirectional and does not change its direction.
Alternating current (AC)
Electric current that changes its direction continuously.
Potential difference
The difference of electric potential between two points; the work done in moving a unit positive charge from one point to another inside an electric field.
Volt
The SI unit of potential difference and emf; one joule per coulomb.
emf source
Any device that converts non-electrical energy into electrical energy (battery, solar panel, thermocouple, generator).
Electromotive force (emf)
The energy supplied by a battery per unit positive charge when it flows through the closed circuit, \varepsilon=W/q.
Ohm's law
The current through a conductor is directly proportional to the potential difference across it, if temperature and physical state are constant.
Ohmic and non-ohmic devices
Ohmic devices (metal wires, resistors) obey Ohm's law and give a straight-line I-V graph; non-ohmic devices (filament bulb, thermistor, semiconductor diode) do not.
Resistance
The opposition to the flow of charges; R=V/I, measured in ohms.
Ohm
The SI unit of resistance; the resistance of a wire in which a voltage of one volt causes a current of one ampere.
Resistivity
The resistance of a wire of unit length and unit cross-sectional area; \rho=RA/L, unit \Omega\,\mathrm{m}.
Conductance
The reciprocal of resistance, G=1/R: how easily charge can flow through a material.
Conductivity
The reciprocal of resistivity, \sigma=1/\rho.
Temperature coefficient of resistivity
A measure of how a material's resistivity (resistance) changes with temperature; positive for metals and negative for semiconductors.
Ammeter
A device used to measure electric current; connected in series and of very low resistance.
Voltmeter
A device used to measure potential difference; connected in parallel and of very high resistance.
Rheostat
A variable resistance connected in series to change the current in a circuit.

Multiple choice questionsp. 201–202

1
Ampere-hour (Ah) is the unit of:
  1. Aelectric current
  2. BchargeAnswer
  3. Cenergy
  4. Dresistance
Why

An ampere-hour is current multiplied by time: 1\,\mathrm{Ah}=1\,\mathrm{A}\times3600\,\mathrm{s}=3600\,\mathrm{C}. Since \Delta Q=I\,\Delta t, the product of current and time is charge, so the unit is a unit of charge.

Current is measured in amperes (A), energy in joules (J) and resistance in ohms (\Omega), so options A, C and D are wrong.

2
Which of the following is a non-ohmic device?
  1. Acopper wire
  2. Bcarbon resistor
  3. CdiodeAnswer
  4. Dall of these
Why

A diode is a non-ohmic device. Its current does not rise in direct proportion to the voltage, so its V–I graph is curved (Figure 16.8 (d)). Non-ohmic devices break Ohm's law.

A copper wire and a carbon resistor are ohmic devices. The book lists both as examples of ohmic devices, so option D is wrong.

4
Electric current is measured using:
  1. AammeterAnswer
  2. Bvoltmeter
  3. Cohmmeter
  4. Dmeter rod
Why

Electric current is measured with an ammeter, which is connected in series with the component (Section 16.8.1). A voltmeter measures potential difference (B), an ohmmeter measures resistance (C) and a meter rod measures length (D).

Note. The printed exercise has no question 3. The numbering runs 1, 2, 4 to 13 as printed.

5
Voltmeter measures
  1. Apotential differenceAnswer
  2. Belectric current
  3. Cresistance
  4. Dresistivity
Why

A voltmeter measures potential difference. It is connected in parallel with the component across which the potential difference is to be measured (Section 16.8.2).

Current is measured by an ammeter (B). Resistance and resistivity are calculated from measured values of V and I or from the material's dimensions, so they are not what a voltmeter reads.

6
If 3 A of current flows for 2 minutes, the amount of charge crosses the cross-section of a conductor will be:
  1. A3 C
  2. B6 C
  3. C60 C
  4. D360 CAnswer
Why

The time must be in seconds: \Delta t=2\,\mathrm{min}=2\times60\,\mathrm{s}=120\,\mathrm{s}.

\Delta Q=I\,\Delta t=3\,\mathrm{A}\times120\,\mathrm{s}=360\,\mathrm{C}

Option B comes from using 2 s instead of 120 s, and option A from leaving out the time altogether.

7
A current of 1 mA is passing through a wire. Number of electrons passing through a point in the wire in 10 seconds is:
  1. A6.25\times10^{19}
  2. B6.25\times10^{18}
  3. C6.25\times10^{17}
  4. D6.25\times10^{16}Answer
Why

First find the charge that passes in 10 s, converting 1 mA to 1\times10^{-3}\,\mathrm{A}:

\Delta Q=I\,\Delta t=1\times10^{-3}\,\mathrm{A}\times10\,\mathrm{s}=1\times10^{-2}\,\mathrm{C}

Each electron carries e=1.6\times10^{-19}\,\mathrm{C}, so the number of electrons is:

n=\dfrac{\Delta Q}{e}=\dfrac{1\times10^{-2}}{1.6\times10^{-19}}=6.25\times10^{16}

The other options are wrong by factors of 10, 100 and 1000, which come from misplacing the decimal point in the milliampere conversion.

8
Resistance of a material always decreases when:
  1. Aits temperature increases
  2. Bits temperature decreases
  3. Cnumber of free electrons increasesAnswer
  4. Dnumber of free electrons decreases
Why

Resistance depends on how many free charge carriers there are to carry the current. With more free electrons, the same voltage drives a larger current, so the resistance falls whatever the material. Option D is the reverse of this.

Temperature does not act the same way in every material. The resistance of a metal rises when it is heated, so option A is not always true. The resistance of a semiconductor or insulator falls when it is heated (Section 16.6.3), so option B is not always true either.

9
The resistance of a metallic conductor varies inversely as:
  1. Aarea of cross sectionAnswer
  2. Blength
  3. Ctemperature
  4. Dall of these
Why

For a conductor, R=\rho\dfrac{L}{A}, so the resistance is inversely proportional to the cross-sectional area A (Section 16.6.2).

Length varies directly with resistance, not inversely (B is wrong). Temperature raises the resistance of a metal, so it is not an inverse relation (C is wrong), which makes D wrong too.

10
Directions of actual current and conventional current are:
  1. Asame
  2. Bopposite to each otherAnswer
  3. Cperpendicular to each other
  4. Dnone of these
Why

Conventional current is taken to flow from the positive to the negative terminal of the battery (Section 16.1.1). In a metal the charges that really move are electrons, which are negative, so they travel from the negative terminal to the positive terminal.

The two directions are therefore opposite, which makes B correct and D wrong.

11
As temperature of a semiconductor increases, its resistance:
  1. Aincreases
  2. BdecreasesAnswer
  3. Cremains constant
  4. Dbecomes zero
Why

In a semiconductor, rising temperature releases more charge carriers, so more current can flow and the resistance decreases (Section 16.6.3). Option A describes a metal, not a semiconductor.

The resistance does not stay constant (C) or become zero (D) when a semiconductor is heated.

12
The SI unit of temperature coefficient of resistance (\alpha) is:
  1. AK
  2. B\mathrm{K}^{-1}Answer
  3. C\Omega\,\mathrm{K}
  4. D\Omega/\mathrm{K}
Why

The temperature coefficient is defined as:

\alpha=\dfrac{\rho_T-\rho_o}{\rho_o\,\Delta T}

The numerator and the denominator in this ratio both carry the unit of resistivity, so they cancel. What remains is one over the unit of temperature, \mathrm{K}^{-1} (or ^\circ\mathrm{C}^{-1}, Section 16.7.1). Option A is the unit of temperature itself, and options C and D keep an unwanted ohm.

13
By increasing temperature, the resistivity of semiconductors:
  1. Aincreases
  2. BdecreasesAnswer
  3. Cremains constant
  4. Dfirst increases then decreases
Why

The temperature coefficient of resistivity is negative for semiconductors (carbon, germanium and silicon in Table 16.2). Their resistivity therefore decreases steadily as the temperature rises.

Option A is the behaviour of metals, and options C and D do not describe semiconductors.

Constructed response questionsp. 202

1
Draw connections so that the emf sources given in figure become in series and parallel.
Solution

In series

In a series connection the sources are joined end to end. The positive terminal of one source is connected to the negative terminal of the next, so the emfs add up and the same current flows through every source.

Four emf sources in series. The total emf is the sum of the four emfs.
\varepsilon_{\text{total}}=\varepsilon_1+\varepsilon_2+\varepsilon_3+\varepsilon_4

In parallel

In a parallel connection all the positive terminals are joined to one wire and all the negative terminals to the other. The sources then share the load, and the current is divided between them. The emf of the group is the same as the emf of one source, provided that the four sources have equal emfs.

Four emf sources in parallel. With four equal emfs \varepsilon the combined emf is \varepsilon.
\varepsilon_{\text{total}}=\varepsilon\quad(\text{all four equal})

Book answer. Series and parallel connection diagrams (no numerical answer).

Note. The emf values of the four sources are not printed. The series total is the sum of all four. A parallel group gives a useful emf only when the sources are equal, because unequal sources would drive currents round the group against each other.

2
Compare and contrast the roles of ammeters and voltmeters in electrical circuits. Include a discussion on how each instrument measures different electrical properties.
Solution

Both instruments are electrical measuring instruments used to find properties of a circuit. They measure different quantities, they are connected in different ways, and they are designed with opposite resistances so that they do not disturb the circuit.

What each instrument measures

  • An ammeter measures the electric current, in amperes (A). It reads the current flowing through the component it is connected to.
  • A voltmeter measures the potential difference (p.d.), in volts (V). It reads the energy transferred per coulomb across the component it is connected across.

How each is connected

An ammeter is connected in series with the component, so that the whole current of the component passes through it. A voltmeter is connected in parallel with the component, so that it reads the p.d. across that component.

Design and effect on the circuit

An ammeter must not change the current it measures, so its resistance is made as low as possible (of the order of 1\,\Omega). A voltmeter must not change the p.d. it measures, so its resistance is made as high as possible (of the order of 10\,\mathrm{M}\Omega).

Ammeters and voltmeters compared.
FeatureAmmeterVoltmeter
Quantity measuredElectric current (I), unit APotential difference (V), unit V
ConnectionIn series with the componentIn parallel with the component
ResistanceVery low (about 1\,\Omega)Very high (about 10\,\mathrm{M}\Omega)
Effect on circuitNegligible, because its resistance is lowNegligible, because little current goes through it
Typical rangesmA, \muA and A, from 1 A to 100 A for home appliancesmV for small voltages; 1 V to 500 V for home use

Why the connection matters

If an ammeter were connected in parallel, its very low resistance would divert most of the current away from the component, so the reading would be wrong and the circuit altered. If a voltmeter were connected in series, its very high resistance would cut the current almost to zero, so the component would hardly work at all.

In short, the ammeter tells you how much charge flows each second through a component, and the voltmeter tells you how much energy each coulomb gives up across it.

Book answer. Comparison only; the book gives the properties of both instruments in Sections 16.8.1 and 16.8.2.

Note. The ratings are the book's typical values: an ammeter resistance of about 1 ohm and a voltmeter resistance of about 10 megohm (10^7 ohm).

Short response questionsp. 203

1
Why is copper preferred over silver for household wiring, considering cost, conductivity, and durability?
Solution

Silver is a slightly better conductor than copper. From Table 16.1, the resistivity of silver is 1.59\times10^{-8}\,\Omega\,\mathrm{m} and that of copper is 1.72\times10^{-8}\,\Omega\,\mathrm{m}, and the lower value means less resistance for the same wire. The difference is only about 8\%, so conductivity is not the deciding factor.

Copper is far cheaper than silver and is much more widely available, which keeps the cost of long lengths of household wiring low. It is also strong and ductile, so it can be drawn into long wires without breaking. Together these make copper the practical choice for wiring, even though silver conducts a little better.

2
Why free electrons in metals don't create a current without a potential difference?
Solution

Free electrons in a metal are in constant, random thermal motion, moving in every direction. Because the motion is random, the electrons on average cross any cross-section of the wire equally in both directions, so the net movement of charge is zero and there is no current (Section 16.1.2).

When a potential difference is applied, the battery sets up an electric field in the wire. The electrons then drift slowly in one direction, and the net flow of charge in that direction is the electric current. Without a potential difference there is no field, so the drift does not start.

3
How conventional and actual current directions affect circuit behaviour, and why conventional current is important in diagrams and calculations.
Solution

In a metal, the actual current is a flow of electrons from the negative terminal of the battery through the external circuit to the positive terminal. The conventional current is taken in the opposite sense, from the positive terminal through the circuit back to the negative terminal (Section 16.1.1).

The two directions describe the same flow. Both give the same effects in the circuit, such as heating a filament, so the behaviour of the circuit does not depend on which direction is used. What matters is that each quantity is used consistently.

Conventional current is used in diagrams and calculations because it was adopted before the electron was discovered, and it is used in every circuit rule, such as Ohm's law and the arrangement of series and parallel components. Using it keeps the signs of the potential differences and emfs correct, so the calculations give the right answers.

4
Compare emf and potential difference, and why this distinction matters in circuit analysis.
Solution
emf and potential difference compared.
Propertyemf (\varepsilon)Potential difference (V)
Where it occursAcross the source (battery, cell)Across a component (bulb, resistor)
MeaningEnergy given to each coulomb of charge by the source, \varepsilon=\dfrac{W}{q}Energy transferred to each coulomb in the component, V=\dfrac{\Delta U}{\Delta q}
RoleCauses the current and keeps a steady p.d. across the devicesDrives the current through the component, according to V=IR
Unitvolt (V), =\mathrm{J\,C^{-1}}volt (V), =\mathrm{J\,C^{-1}}

The distinction matters because the source supplies the energy and the components use it. When the current in the circuit is worked out with Ohm's law, the potential difference across each component is what enters the calculation, while the emf describes the source that maintains it.

5
How is a current-carrying wire electrically neutral?
Solution

A metal wire has equal numbers of positive protons in its nuclei and negative electrons around them, so each atom is neutral. The free electrons that carry the current are only moved along the wire, not created or destroyed.

When current flows, electrons enter one end of the wire at the same rate as they leave the other end. The wire therefore never builds up a net charge, and the whole wire stays neutral while the current flows through it.

6
Would the lights still work if a car battery's positive and negative terminals were interchanged?
Solution

Yes, a filament bulb would still light. Reversing the battery reverses the potential difference across the bulb, and by Ohm's law the current reverses with it. The resistance of the filament is the same, and the heating effect of the current, which makes the filament glow, does not depend on the direction of the current.

The current would flow in the opposite direction through the circuit. Devices that only work one way round, such as a diode, would stop conducting in the reversed connection, so the circuit would not behave in the same way.

7
Evaluate the limits of Ohm's law for different materials, especially under extreme conditions.
Solution

Ohm's law, V=IR, holds only for ohmic devices, such as metals at constant temperature. For these the V–I graph is a straight line through the origin (Figure 16.8 (a)).

  • Non-ohmic materials. A filament bulb, a thermistor and a semiconductor diode do not obey Ohm's law. Their V–I graphs are curved (Figure 16.8 (b) to (d)).
  • Temperature. The law assumes that the temperature is constant. When the current heats a conductor, its resistance changes, so the ratio V/I changes too.
  • Range of voltage. A metal obeys the law only over a limited range of voltages. At high voltages the heating is large and the straight line no longer holds.
  • Extreme conditions. High temperatures or very large voltages make the material depart from the straight-line graph, so the law can be used only for the range over which it is tested.

In practice, Ohm's law is a useful model for metals in ordinary conditions, and it must be checked before it is used for other materials or extreme conditions.

8
How does a wire's shape affect its resistance, whether bending changes it?
Solution

The resistance of a wire depends on its length and cross-sectional area, as R=\rho\dfrac{L}{A}. A longer wire has a larger resistance, because the electrons collide more often along a longer path. A wire with a smaller cross-sectional area has a larger resistance, because there is less room for the current to pass through.

Bending a wire does not change its resistance on its own. The wire keeps the same length, cross-section and material, so the value of R is unchanged. Only if the bend stretches or squeezes the wire, so that L or A changes, is the resistance changed.

9
The temperature coefficient of copper is 0.004\,^\circ\mathrm{C}^{-1}. How does this affect copper's performance in electrical systems under varying temperatures?
Solution

The temperature coefficient of resistivity is positive for copper, so its resistivity increases as the temperature rises. The relation is:

\rho_T=\rho_o\,(1+\alpha\,\Delta T)

With \alpha=0.004\,^\circ\mathrm{C}^{-1}, a rise of 25\,^\circ\mathrm{C} raises the resistivity by 0.004\times25=0.10, or 10\%. The resistance of a copper wire rises by the same fraction, so for a fixed voltage the current falls.

In electrical systems, copper wires in a warm environment carry less current for the same voltage and lose more energy as heat in the wire. The designer must allow for this when a system operates over a wide range of temperatures. (The book's own value for copper is \alpha=3.9\times10^{-3}\,\mathrm{K}^{-1}, which the rounded value 0.004 approximates.)

Note. The book's Table 16.2 gives the temperature coefficient of copper as 3.9 × 10^-3 per degree C. The question rounds it to 0.004 per degree C.

10
How would you define fluid resistance for a water-carrying pipe?
Solution

Fluid resistance is the opposition that a pipe offers to the flow of a fluid through it. It is defined in the same way as electrical resistance, which the book describes as the opposition to the flow of charges.

By analogy with the electrical case, the fluid resistance of a pipe rises with its length and falls with its cross-sectional area, as R\propto\dfrac{L}{A}. A long, narrow pipe therefore opposes the flow of water more than a short, wide pipe does.

Note. The book gives no formal definition of fluid resistance. This answer uses the book's definition of resistance and the analogy with the electrical case.

Long response questionsp. 203–204

1
Explain electric current and its unit (ampere) by exploring charge flow. Discuss the difference between conventional and actual current and when this distinction matters.
Solution

Electric current

Electric current is the time rate of flow of electric charge through a cross-sectional area of a conductor. If a charge \Delta Q flows through a cross-section in time \Delta t, the current is:

I=\dfrac{\Delta Q}{\Delta t}
16.1

The SI unit of current is the ampere (A). One ampere is the current when one coulomb of charge flows through a cross-section in one second, so 1\,\mathrm{A}=1\,\mathrm{C\,s^{-1}}. Smaller units are the milliampere, 1\,\mathrm{mA}=10^{-3}\,\mathrm{A}, and the microampere, 1\,\mu\mathrm{A}=10^{-6}\,\mathrm{A}.

Charge flow

Current can be produced by positive charges moving in one direction, by negative charges moving in the opposite direction, or by both at once. In a metal the charges that move are electrons, which are negative. In a gas or an electrolyte, positive and negative charges both move.

Conventional and actual current

The actual current is the flow of the charges that really move. In a metal this is a flow of electrons from the negative terminal to the positive terminal of the battery. The conventional current is the flow of positive charge, taken from the positive terminal through the circuit to the negative terminal. The two currents are equal in size and opposite in direction (Section 16.1.1).

FeatureActual currentConventional current
CarriersElectrons in a metalA positive charge in the same sense
DirectionNegative to positive terminalPositive to negative terminal
UseDescribes the physical motion of chargesUsed in diagrams and all circuit calculations

When the distinction matters

The distinction does not change the effects of the current, such as heating or the light from a bulb. It matters when the physical motion of the charges is being described, for example when the electrons in a metal are considered. For diagrams, for Ohm's law and for every calculation, the conventional direction is used, so that the signs of the currents and potential differences are consistent.

Book answer. Descriptive question; the answer uses Sections 16.1 and 16.1.1.

2
Analyse electrical conduction in metals and how external factors like temperature and magnetic fields affect electron movement and conduction efficiency.
Solution

Conduction in metals

In a metal the valence electrons are not tightly bound to the nuclei, so they move freely through the metal like gas molecules in a container. With no battery connected, their net motion is zero and there is no current. When a battery is connected, the electric field makes the electrons drift in one direction, and the net drift of electrons is the current. The electrons keep colliding with the ions of the metal, which slows them down (Section 16.1.2).

Effect of temperature

When the temperature rises, the positive ions vibrate with greater amplitude. The electrons then collide with the ions more often, so their motion is hindered and the current for a given electric field is reduced (Section 16.7.1). The resistivity of a metal therefore rises with temperature, and the conduction efficiency falls.

Magnetic fields

This unit does not treat the effect of magnetic fields on moving charges. That topic belongs to the unit on Electromagnetism (Unit 19), so the answer for this unit is limited to the electric field and the temperature. The electric field supplies the drift of the electrons, while the temperature decides how often they collide and so how efficiently the metal conducts.

Book answer. Descriptive question. The magnetic-field part is outside this unit's content.

3
Compare DC and AC in terms of their behaviour in circuit elements. Discuss the pros and cons of each in applications like power transmission and electronics.
Solution

Direct and alternating current

In direct current (DC) the electrons flow in one fixed direction. The current may be steady or may vary in size, as shown in Figure 16.4 (a). In alternating current (AC) the electrons go forward for half the time and backward for the other half, so the direction of the current changes continuously (Section 16.2).

Behaviour in circuit elements

At any instant, a resistor obeys Ohm's law for both kinds of current, so V=IR holds whether the current is DC or AC. The current through a resistor follows the polarity of the voltage. A diode, which is non-ohmic, conducts only in one direction, so it does not behave the same way for AC, where the current reverses.

Pros and cons

DC and AC compared.
DCAC
BehaviourSteady flow in one directionCurrent reverses direction continuously
SourcesBatteries, solar cells and some generatorsGenerators (the book notes that generators can produce either form)
ElectronicsSuited to electronic circuits and to storage in batteriesUsed directly in most mains supplies; electronic circuits usually convert it to DC
Power transmissionDifficult to change its voltage easily over long distancesIts voltage can be changed easily, which is why it is used for long distance transmission
SafetyAt the same voltage, a DC shock usually causes a single muscle contractionAt high voltage (above 500 V) AC is more dangerous, because it can keep the muscles contracting

In summary, DC is simple and steady, and it suits batteries and electronics. AC is easy to generate and to change in voltage, so it is preferred for transmission, but at high voltage it is more hazardous to people.

Book answer. Descriptive question. The transmission point uses the change of voltage by transformers, which is outside this unit.

4
Examine electromotive force (emf) and its role in maintaining current.
Solution

Definition

The emf of a source is the energy supplied by the source per unit positive charge when the charge flows through the complete circuit. If W is the work done by the battery on a charge q, then:

\varepsilon=\dfrac{W}{q}
16.3

The SI unit of emf is the same as that of electric potential, \mathrm{J\,C^{-1}}=\mathrm{V}. A source of emf converts non-electrical energy into electrical energy. Examples are batteries (chemical to electrical), solar panels (light to electrical), thermocouples (heat to electrical) and generators (mechanical to electrical).

Role in maintaining current

Inside the battery, positive charge moves from the low potential terminal to the high potential terminal. Work must be done on the charge to drive it uphill in potential, and the battery does this using its stored energy. Outside the battery, the charge gives up its energy in the devices in the circuit.

The circuit cycle
  1. The battery raises each coulomb of charge to a higher potential, so it has more energy.
  2. The charge passes through the devices and loses energy in them, which keeps a potential difference across each device.
  3. The charge then returns to the battery and the cycle repeats. The emf is what keeps the charge moving round the circuit.

Without an emf source, the charges would lose their energy and stop moving once the potential difference had been used up. The emf is therefore the cause of the steady current in a closed circuit.

Book answer. Descriptive question; the definition is in Section 16.4 and Equation 16.3.

5
Distinguish between emf and potential difference, using real-world examples like batteries.
Solution

Definitions

The emf of a battery is the energy it gives to each coulomb of charge when the charge goes round the closed circuit, \varepsilon=W/q. The potential difference across a component is the energy each coulomb gives up in that component, V=\Delta U/\Delta q (Equation 16.2).

emf and potential difference.
PropertyemfPotential difference
Belongs toA source, such as a batteryA component, such as a bulb
What it measuresEnergy supplied by the source per coulombEnergy used by the component per coulomb
Symbol\varepsilonV (or \Delta V)
Unitvoltvolt
Present whenThere is a source in the circuit, even on an open circuitA current is flowing or the component is connected across the source

Real-world example: a torch

Two cells of 1.5\,\mathrm{V} each, connected in series, have a total emf of 2\times1.5\,\mathrm{V}=3.0\,\mathrm{V} (the same arrangement as Example 16.2). This is the energy that the cells supply to each coulomb of charge. When the bulb is lit, the potential difference across the bulb is 3.0\,\mathrm{V} and the current through it is 0.15\,\mathrm{A}, so the bulb uses 3.0\,\mathrm{J} of energy from each coulomb that passes through it.

The battery's emf and the bulb's potential difference have the same value here because the wires have negligible resistance, so the whole emf appears across the bulb. The two quantities are different in concept: the emf is what the battery provides, and the potential difference is what the bulb receives.

Why the distinction matters

The emf is a property of the source. The potential difference is a property of a component, and it changes with the component. Ohm's law and all circuit calculations use the potential difference across each component, while the emf describes the source that maintains it.

Book answer. Descriptive question; Section 16.3 and Section 16.4.

6
Critically analyse Ohm's law and its limitations for non-ohmic devices. Discuss the importance of I-V graphs and propose experiments to explore non-linear materials.
Solution

Ohm's law and its statement

Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, as long as the temperature and the physical state of the conductor are kept constant (Section 16.5). It is written V=IR, where R is a constant for the conductor.

Limitations

  • Ohm's law is applicable only to materials whose V–I graph is a straight line through the origin.
  • Only metallic conductors show this behaviour, and only over a limited range of voltages. The temperature must also stay constant.
  • Many important devices, such as a filament bulb, a thermistor and a semiconductor diode, are non-ohmic. Their graphs are curved, so V/I is not constant (Figure 16.8).

The importance of I-V graphs

A plot of current against voltage shows directly whether a device is ohmic. A straight line through the origin means that R=V/I is the same at every voltage. A curve means that the resistance changes with voltage or current, and the shape of the curve shows how it changes. The graph below sketches the three shapes described in the book, on the same axes.

I-V curves of ohmic and non-ohmic devices
Schematic I-V curves. The straight line is ohmic. The filament bulb's current saturates at high voltage, and the thermistor's current rises faster than the voltage.

Experiments to explore non-linear materials

Method
  1. Build a circuit with a cell, a rheostat, an ammeter in series with the device and a voltmeter in parallel across it.
  2. Close the switch, set the rheostat to give the lowest current, and record V and I for the device.
  3. Raise the voltage in several steps with the rheostat, and record V and I at each step, letting the device cool between readings.
  4. Plot I against V. A straight line means the device is ohmic, and a curve means it is not. Reverse the connections of a diode to see its behaviour in the other direction.

Book answer. Descriptive question; Sections 16.5, 16.5.1 and 16.5.2 and Figure 16.8.

7
Investigate the relationship between resistance and resistivity and design an experiment to measure resistance. Discuss factors affecting accuracy, such as temperature.
Solution

Relationship between resistance and resistivity

The resistance of a uniform wire depends on its length and cross-sectional area. Keeping the temperature constant, the resistance is directly proportional to the length and inversely proportional to the cross-sectional area (Section 16.7):

R=\rho\dfrac{L}{A}
16.6

The resistivity \rho is a property of the material only. It is the resistance of a wire of length 1\,\mathrm{m} and cross-sectional area 1\,\mathrm{m}^2, so the resistance of a given wire is its resistivity multiplied by a factor that depends on its shape.

Experiment design

Circuit for measuring resistance. The ammeter is in series with the unknown resistance and the voltmeter is in parallel across it. The rheostat controls the current.
Method
  1. Set the ammeter and voltmeter needles to zero. Record the full-scale reading and the least count of each instrument.
  2. Close the switch and move the rheostat until the current is at its lowest value. Record the ammeter reading I and the voltmeter reading V.
  3. Open the switch, let the wire return to room temperature, then close it again at a higher voltage. Repeat for four different voltage levels.
  4. For each reading, calculate R=V/I and enter it in a table. Find the average of the four values.
  5. Measure the length L and the cross-sectional area A of the wire, and calculate the resistivity from \rho=RA/L.

Factors affecting accuracy

  • Temperature. The current heats the wire, so its resistance rises during the reading. Keep the current small and let the wire cool between readings.
  • Zero error. If the needles do not rest on zero, every reading is shifted. Correct the zero before starting.
  • Least count. The smallest division limits the precision of each reading, so use the most sensitive range that does not overload the instrument.
  • Contact and connections. Poor joints add a resistance that is not part of the wire, so the connections must be clean and tight.

With these steps and precautions, the average of several readings gives a reliable value of the resistance of the wire.

Book answer. Descriptive question; Sections 16.7 and 16.9 and Figure 16.16 of the book describe the method.

8
Analyse how factors like temperature, material, length, and area affect resistance. Discuss practical applications, such as wiring or heating elements, and ways to optimize resistance.
Solution

Factors affecting resistance

Factors that affect the resistance of a wire.
FactorEffect on resistanceReason
Length LDirectly proportionalLonger path, more collisions (Equation 1)
Cross-sectional area AInversely proportionalLarger area gives more room for the current (Equation 2)
MaterialDepends on resistivity \rhoInternal structure of the material (Section 16.6.4)
TemperatureRises for metals, falls for semiconductors and insulatorsCollisions increase in metals; charge carriers increase in semiconductors (Section 16.6.3)
R=\rho\dfrac{L}{A}

Practical applications

  • Wiring. Connecting wires must have very low resistance, so that little energy is wasted as heat. Copper, with a low resistivity, is the usual choice, and the wire is made thick (large A) and kept short (small L).
  • Heating elements. A heater needs a large resistance so that the current produces a lot of heat. A material of high resistivity is chosen, and the wire is made long and thin.
  • Filaments. A tungsten filament is a conductor with a fairly high resistivity. When it is cold its resistance is low, so a large current flows at switch-on, which heats the filament to a high temperature.

Ways to optimise resistance

To reduce resistance, use a material with a low resistivity, a short length and a large cross-sectional area. To increase it, do the reverse. The temperature of the material must also be considered, because a warm metal has a higher resistance than the same metal when cold.

Book answer. Descriptive question; Sections 16.6 and 16.7.

9
How does temperature influence the resistance of materials, and what implications does this have for electrical components in different environments?
Solution

Metals

For a metal, the resistance increases as the temperature rises, because the ions vibrate more and collide with the electrons more often (Section 16.6.3). The relation between the resistivity and the temperature is:

\rho_T=\rho_o\,(1+\alpha\,\Delta T)

The graph below shows this straight-line increase for copper, starting from \rho_o=1.7\times10^{-8}\,\Omega\,\mathrm{m} at 0\,^\circ\mathrm{C}, with \alpha=3.9\times10^{-3}\,\mathrm{K}^{-1}.

Resistivity of copper against temperature
Resistivity of copper rises linearly with temperature. The vertical axis is in units of 10^{-8}\,\Omega\,\mathrm{m}.

Semiconductors and insulators

For insulators and semiconductors the resistance decreases as the temperature rises, because the thermal energy releases more charge carriers (Section 16.6.3). The temperature coefficient of these materials is negative (Table 16.2).

Implications for components

  • Hot environments. A copper or tungsten wire has a higher resistance when it is hot, so for a fixed voltage it carries less current. A circuit designed in a cold place may therefore work differently in a hot one.
  • Cold start. A tungsten filament is cold and has a low resistance when first switched on, so a large current flows and the filament heats up quickly. This is why a bulb often fails when it is switched on.
  • Sensors. A thermistor is made of a semiconductor, and its resistance falls as it gets hot. It can therefore be used to measure temperature, because its resistance shows how hot its surroundings are.

Book answer. Descriptive question; Sections 16.6.3 and 16.7.1 and Table 16.2.

10
Examine the placement of ammeters and voltmeters in circuits and how misconnection affects measurement. Propose strategies to avoid errors.
Solution

Correct placement

An ammeter is placed in series with the component, so that all the current passes through it (Section 16.8.1). A voltmeter is placed in parallel with the component, so that it reads the potential difference across it (Section 16.8.2).

Correct connection. The ammeter is in series with R and the voltmeter is in parallel across R.

Effect of misconnection

  • Ammeter in parallel. The ammeter has a very low resistance, of about 1\,\Omega. Connected in parallel with a component, it takes most of the current and gives a reading that is not the current through the component. The large current can also damage the meter.
  • Voltmeter in series. The voltmeter has a very high resistance, of about 10\,\mathrm{M}\Omega. Connected in series, it limits the current to a tiny value, so the circuit hardly works and the reading is not the potential difference across the component.
  • Wrong range. Choosing a range that is too small can overload the meter. Choosing one that is too large gives a reading with little precision.

Strategies to avoid errors

  1. Draw the circuit first and check that the ammeter is in series and the voltmeter is in parallel.
  2. Connect the positive terminal of each meter towards the positive side of the circuit, and the negative terminal towards the negative side.
  3. Set the needles to zero before taking readings, and note the full-scale value and least count.
  4. Start with the highest range and reduce it until the needle is in the upper half of the scale.
  5. Disconnect the supply before changing the connections, and check the circuit once more before closing the switch.

Book answer. Descriptive question; Sections 16.8.1 and 16.8.2.

Numerical response questionsp. 204

1
How much time will it take to charge a 200 Ah battery when it is connected to a current of 25 A?
Solution
Given
  • Q = 200\,\mathrm{Ah} charge to be delivered
  • I = 25\,\mathrm{A} charging current
To find
  • t
  1. Charge and current are related by Q=I\,t, so the time is:
    t=\dfrac{Q}{I}
  2. Substitute the values. The ampere-hours divided by amperes give hours.
    t=\dfrac{200\,\mathrm{A\,h}}{25\,\mathrm{A}}=8\,\mathrm{h}
  3. Check in seconds: 8\,\mathrm{h}\times3600\,\mathrm{s\,h^{-1}}=2.88\times10^{4}\,\mathrm{s}, and 25\,\mathrm{A}\times2.88\times10^{4}\,\mathrm{s}=7.2\times10^{5}\,\mathrm{C}, which is 200\,\mathrm{A\,h} of charge.
t=8\,\mathrm{h}

Book answer. 8 hr

2
A 12.0 V emf source is connected to a purely resistive electrical appliance (a light bulb). An electric current of 2.0 A flows through it. Consider the conducting wires to be resistance-free, calculate the resistance offered by the electrical appliance.
Solution
Given
  • V = 12.0\,\mathrm{V} potential difference across the appliance
  • I = 2.0\,\mathrm{A} current through it
To find
  • R
  1. Ohm's law gives the resistance from V and I, because the wires have no resistance and the whole emf appears across the appliance.
    R=\dfrac{V}{I}
  2. Substitute.
    R=\dfrac{12.0\,\mathrm{V}}{2.0\,\mathrm{A}}=6.0\,\Omega
R=6.0\,\Omega

Book answer. 6 Ω

3
The potential difference across the two ends of the component decreases to one fourth of its initial value while its resistance remains constant. What change will occur in the current through it?
Solution
Given
  • V_2 = \dfrac{1}{4}V_1 new potential difference
  • R = \text{constant} resistance
To find
  • I_2/I_1
  1. Write Ohm's law for the initial and final states.
    I_1=\dfrac{V_1}{R},\qquad I_2=\dfrac{V_2}{R}
  2. Divide the second equation by the first. The resistance cancels because it is the same in both states.
    \dfrac{I_2}{I_1}=\dfrac{V_2}{V_1}
  3. Substitute the new p.d.
    \dfrac{I_2}{I_1}=\dfrac{V_1/4}{V_1}=\dfrac{1}{4}
I_2=\dfrac{1}{4}\,I_1

Book answer. 1/4

4
What voltage is necessary to pass a current of 50 amperes through a resistance of 100 Ω?
Solution
Given
  • I = 50\,\mathrm{A} current
  • R = 100\,\Omega resistance
To find
  • V
  1. Ohm's law, V=IR, gives the voltage needed.
    V=IR
  2. Substitute.
    V=50\,\mathrm{A}\times100\,\Omega=5000\,\mathrm{V}
V=5.0\times10^{3}\,\mathrm{V}

Book answer. 5,000 V

5
The current passing through a resistor in a circuit is 0.5 A when the voltage across it is 6.0 V. What current will pass through the same resistor when the voltage across it is 9.0 V?
Solution
Given
  • I_1 = 0.5\,\mathrm{A} at V_1
  • V_1 = 6.0\,\mathrm{V}
  • V_2 = 9.0\,\mathrm{V}
To find
  • I_2
  1. Find the resistance from the first state, since it does not change.
    R=\dfrac{V_1}{I_1}=\dfrac{6.0\,\mathrm{V}}{0.5\,\mathrm{A}}=12\,\Omega
  2. Use the same resistance with the second voltage.
    I_2=\dfrac{V_2}{R}=\dfrac{9.0\,\mathrm{V}}{12\,\Omega}=0.75\,\mathrm{A}
I_2=0.75\,\mathrm{A}

Book answer. 0.75 A

6
The length and area of wire are given as 2 m and 0.4 mm^2 respectively. The resistance of that wire is 0.085 Ω, calculate the resistivity.
Solution
Given
  • L = 2\,\mathrm{m} length
  • A = 0.4\,\mathrm{mm^2}=0.4\times10^{-6}\,\mathrm{m^2} cross-sectional area
  • R = 0.085\,\Omega resistance
To find
  • \rho
  1. Resistivity follows from R=\rho\dfrac{L}{A}, so \rho=\dfrac{RA}{L} (Equation 16.7).
    \rho=\dfrac{RA}{L}
  2. Convert the area to square metres: 1\,\mathrm{mm^2}=10^{-6}\,\mathrm{m^2}.
    A=0.4\times10^{-6}\,\mathrm{m^2}
  3. Substitute.
    \rho=\dfrac{0.085\,\Omega\times0.4\times10^{-6}\,\mathrm{m^2}}{2\,\mathrm{m}}=1.7\times10^{-8}\,\Omega\,\mathrm{m}
\rho=1.7\times10^{-8}\,\Omega\,\mathrm{m}

Book answer. 1.7\times10^{8}\,\Omega\,\mathrm{m} as printed (the minus sign of the exponent is missing)

Note. The printed answer shows 10^{8}, but the minus sign of the exponent has dropped out. The data give 1.7\times10^{-8}\,\Omega\,\mathrm{m}, which matches copper in Table 16.1 (1.72\times10^{-8}\,\Omega\,\mathrm{m}). The same missing minus sign appears in the printed answers to question 7 and question 8, and in the printed value of the resistivity in question 7.

7
Resistivity of a given copper wire of length 2 m is 1.7\times10^{-8}\,\Omega\,\mathrm{m}. (a) If the wire is stretched to double of its length, what will be the new resistivity? (b) If its temperature is increased by 20 K, find the new resistivity. (Take \alpha=3.9\times10^{-3}\,\mathrm{K^{-1}})
Solution
Given
  • \rho_o = 1.7\times10^{-8}\,\Omega\,\mathrm{m} resistivity of the wire
  • \alpha = 3.9\times10^{-3}\,\mathrm{K^{-1}} temperature coefficient of copper
To find
  • (a) \rho_{\text{new}}
  • (b) \rho_T
(a) If the wire is stretched to double of its length, what will be the new resistivity?
  1. Resistivity is a property of the material and does not depend on the length or the shape of the wire. Stretching changes R but not \rho.
  2. The new resistivity is therefore the same as before.
    \rho=1.7\times10^{-8}\,\Omega\,\mathrm{m}
\rho=1.7\times10^{-8}\,\Omega\,\mathrm{m}
(b) If its temperature is increased by 20 K, find the new resistivity.
  1. Use the temperature relation for resistivity, with \Delta T=20\,\mathrm{K}.
    \rho_T=\rho_o\,(1+\alpha\,\Delta T)
  2. Substitute.
    \rho_T=1.7\times10^{-8}\,\Omega\,\mathrm{m}\,(1+3.9\times10^{-3}\times20)
  3. Evaluate the bracket, then multiply.
    \rho_T=1.7\times10^{-8}\times1.078=1.83\times10^{-8}\,\Omega\,\mathrm{m}
\rho_T\approx1.83\times10^{-8}\,\Omega\,\mathrm{m}

Book answer. (a) 1.7\times10^{8}\,\Omega\,\mathrm{m}; (b) 1.83\times10^{8}\,\Omega\,\mathrm{m} (both as printed)

Note. As in question 6, the printed answers lose the minus sign of the exponent. The values are 1.7\times10^{-8} and 1.83\times10^{-8}\,\Omega\,\mathrm{m}. The printed resistivity in the question text also shows 10^{8} and is read here as 10^{-8}.

8
At some temperature T_o, the resistivity of a metallic conductor is 1.7\times10^{-8}\,\Omega\,\mathrm{m}. If the temperature is increased by 50 K, find the new resistivity by taking temperature coefficient of resistivity as 0.004.
Solution
Given
  • \rho_o = 1.7\times10^{-8}\,\Omega\,\mathrm{m} resistivity at T_o
  • \Delta T = 50\,\mathrm{K} rise in temperature
  • \alpha = 0.004\,\mathrm{K^{-1}}
To find
  • \rho_T
  1. Use the temperature relation for resistivity.
    \rho_T=\rho_o\,(1+\alpha\,\Delta T)
  2. Substitute.
    \rho_T=1.7\times10^{-8}\,\Omega\,\mathrm{m}\,(1+0.004\times50)
  3. Evaluate.
    \rho_T=1.7\times10^{-8}\times1.2=2.04\times10^{-8}\,\Omega\,\mathrm{m}
\rho_T=2.04\times10^{-8}\,\Omega\,\mathrm{m}

Book answer. 2.04\times10^{8}\,\Omega\,\mathrm{m} as printed (the minus sign of the exponent is missing)

Note. The printed answer has the same missing minus sign as questions 6 and 7. The value is 2.04\times10^{-8}\,\Omega\,\mathrm{m}.

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, MCQ 71 mark
A charge of 10 C flows through a conductor in 5 s. The value of current is:
  1. A10 A
  2. B0.5 A
  3. C2 AAnswer
  4. D5 A
Why

Current is the charge passing per second, I=\dfrac{Q}{t}=\dfrac{10\,\mathrm{C}}{5\,\mathrm{s}}=2\,\mathrm{A}. Option D (5 A) comes from dividing by the wrong number, and option A (10 A) is just the charge, not the charge per second.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (vi)1.5+1.5 = 3 marks
State two differences between d.c. and a.c.
Solution

Direct current (d.c.) flows steadily in one fixed direction. Alternating current (a.c.) keeps changing direction, going forward for half the time and backward for the other half.

Direct current (d.c.)Alternating current (a.c.)
DirectionFlows in one fixed direction onlyReverses its direction continuously (forward for half a cycle, backward for the other half)
Graph of I against tA horizontal line (steady) or a curve that stays on one side of the time axisA wave (sine curve) that goes above and below the time axis
SourceCell, battery, d.c. generatorAlternating voltage source, a.c. generator
  • Direction: d.c. does not change its direction, while a.c. reverses its direction continuously.
  • Graph: the I-t graph of d.c. is a horizontal line (or a curve on one side of the time axis), while that of a.c. is a wave that goes above and below the time axis.
FBISE SSC-II 2026 Annual I, Set B, Q.2 (viii) OR1+2 = 3 marks
State Ohm's law. If a resistor has resistance 10\,\Omega and current 2 A flows through it, find the potential difference.
Solution
Ohm's law
The current through a conductor is directly proportional to the potential difference across its ends, as long as the temperature and physical state of the conductor stay constant. That is, I\propto V, or V=IR.
Given
  • R = 10\,\Omega
  • I = 2\,\mathrm{A}
To find
  • Potential difference V
  1. Ohm's law gives V=IR.
    V=IR
  2. Put in the values.
    V=2\,\mathrm{A}\times10\,\Omega=20\,\mathrm{V}
V=20\,\mathrm{V}
FBISE SSC-II 2026 Annual I, Set D, MCQ 71 mark
If the charge passing through a wire in a given time is doubled, the current will:
  1. ABecome zero
  2. BBecome half
  3. CDoubledAnswer
  4. DRemain same
Why

Since I=\dfrac{Q}{t}, with t fixed the current is directly proportional to the charge. Doubling Q gives I'=\dfrac{2Q}{t}=2I. Option B (becomes half) would be true if the time were doubled instead.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (vi)1x3 = 3 marks
Why is alternating current preferred for domestic use? Give three reasons.
Solution

Alternating current is preferred for domestic use because of the following reasons.

  1. Its voltage can be changed (stepped up or down) easily, so it can be sent over long distances at high voltage and brought down to a safe value (about 220 V) for homes.
  2. Generators produce a.c. directly, so it is easy to generate and supply to homes.
  3. Most household appliances such as lights, heaters, fans and motors work directly on a.c. mains, and where d.c. is needed (electronic circuits) a.c. can be converted into it.
FBISE SSC-II 2026 Annual I, Set D, Q.2 (viii) OR1+2 = 3 marks
State Ohm's law. A potential difference of 12 V produces a current of 3 A in a conductor. Calculate its resistance.
Solution
Ohm's law
The current through a conductor is directly proportional to the potential difference across its ends, as long as the temperature and physical state of the conductor stay constant. That is, I\propto V, or V=IR.
Given
  • V = 12\,\mathrm{V}
  • I = 3\,\mathrm{A}
To find
  • Resistance R
  1. From V=IR, the resistance is R=V/I.
    R=\dfrac{V}{I}
  2. Put in the values.
    R=\dfrac{12\,\mathrm{V}}{3\,\mathrm{A}}=4\,\Omega
R=4\,\Omega
FBISE SSC-II 2026 Annual I, Set M, MCQ 71 mark
Correct formula of electric current is:
  1. AI = Q / tAnswer
  2. BI = t / Q
  3. CI = Q \times t
  4. DQ = I / t
Why

Current is the rate of flow of charge: charge divided by time, I=\dfrac{Q}{t} (coulomb per second, or ampere). Option D is the same equation turned upside down and wrongly written, since Q=It, not I/t.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (vi)1.5+1.5 = 3 marks
Briefly describe direct current (d.c) and alternating current (a.c). Give one advantage of each.
Solution
Direct current (d.c.)
Current that flows steadily in a single direction. It may be steady or may vary in size, but it never reverses.
Alternating current (a.c.)
Current in which the charges keep switching direction, going forward half the time and backward the other half.
  • Advantage of d.c.: it is steady and in one direction, so it is suited to electronic circuits and can be stored in batteries.
  • Advantage of a.c.: its voltage can be changed easily, so it can be transmitted over long distances.
FBISE SSC-II 2026 Annual I, Set M, Q.2 (viii) OR1+2 = 3 marks
State Ohm's law. Calculate the current through a resistor of 6\,\Omega when a voltage of 12 V is applied.
Solution
Ohm's law
The current through a conductor is directly proportional to the potential difference across its ends, as long as the temperature and physical state of the conductor stay constant. That is, I\propto V, or V=IR.
Given
  • R = 6\,\Omega
  • V = 12\,\mathrm{V}
To find
  • Current I
  1. From V=IR, the current is I=V/R.
    I=\dfrac{V}{R}
  2. Put in the values.
    I=\dfrac{12\,\mathrm{V}}{6\,\Omega}=2\,\mathrm{A}
I=2\,\mathrm{A}

Exam focus

  1. State Ohm's law, write V=IR and solve a numerical for V, I or R (for example 10\,\Omega and 2 A, or 12 V and 3 A). (short)
    Q.2 (viii) OR in all three 2026 sets (B, D and M), 3 marks (1+2), always the statement plus one V=IR calculation. Numericals 4 and 5 of the book practise the same.
    Go to answer
  2. Use I=Q/t: find the current, the charge or the time (for example 10 C in 5 s, or what happens to I when Q doubles). (mcq)
    MCQ 7 in Set B, Set D and Set M, all on the definition I=Q/t; also book exercise MCQs 1, 6 and 7 and Numerical 1.
    Go to answer
  3. Distinguish direct current from alternating current (two differences, one advantage of each) and say why a.c. is used in homes. (short)
    Q.2 (vi) in all three 2026 sets: two differences (Set B), three reasons for a.c. at home (Set D), describe both with an advantage of each (Set M). Long question 3 of the book.
    Go to answer
  4. Draw the I-t graphs of d.c. and a.c. and compare how they behave in a circuit. (long)
    The two graphs of Figure 16.4 back up every d.c./a.c. answer in the 2026 papers (Q.2 (vi) in Sets B, D and M); asked in full as Long question 3.
    Go to answer
  5. Explain ohmic and non-ohmic devices with I-V graphs and the limits of Ohm's law. (long)
    Follows from the Ohm's law question that appeared in all three 2026 sets; Short question 7 and Long question 6 of the book ask for the graphs of a metal, filament bulb, thermistor and diode.
    Go to answer
  6. Distinguish emf from potential difference and give the unit of each. (short)
    Not asked in the 2026 papers, but it is a standard conceptual pair (Short question 4, Long question 5) and a likely next question from the SLOs on emf and p.d.
    Go to answer
  7. Define resistivity and use R=\rho L/A; explain how resistance depends on length, area, material and temperature. (long)
    The formula R=\rho L/A is printed on the 2026 formula lists of Sets B and D, though no unit 16 question used it this year; Numericals 6 to 8 and Long questions 7 to 9 of the book.
    Go to answer
  8. Describe the experiment to find an unknown resistance with an ammeter and a voltmeter, and draw the circuit. (long)
    An SLO of the unit ('describe an experiment to determine resistance'), not asked in 2026; Long questions 7 and 10 of the book.
    Go to answer

Mark this unit as done to fill the Class 10 Physics ring.

Next, unit 17
Electric circuits

© 2026 FBISE Fold. Independent study notes, not affiliated with the Federal Board of Intermediate and Secondary Education.