E.m.f., potential difference and resistance

A torch cell is marked 1.5 V and a phone charger 5 V. A voltage tells you how much energy each coulomb of charge carries: given out by the source, and transferred in each component. Resistance links the voltage across a component to the current through it, and it depends on the length and thickness of a wire. The calculations use current and Q = It.

E.m.f., potential difference and energy

Current tells you how much charge passes each second. It does not tell you how much energy that charge carries. A voltage does: it counts joules per coulomb.

A source’s e.m.f. is measured in volts

The electromotive force (e.m.f.) of a source, such as a cell, a battery or a power supply, is measured in volts (V). A source labelled 6.0 V gives each coulomb that passes through it 6.0 J of energy. (Treat every source here as ideal: none of that energy is transferred inside the source itself.)

Despite its name, e.m.f. is not a force. A force is measured in newtons; e.m.f. is measured in volts. A cell transfers energy from its chemical store to the charge. It does not make the charge: the same charge keeps circulating.

A component’s p.d. is work done per unit charge

The potential difference (p.d.) across a component is the work done per unit charge in driving charge through that component. It is also measured in volts:

W is the work done in joules, which is the energy transferred in the component. Rearranged, the energy transferred when a charge Q passes through a component with p.d. V is

A p.d. of 4.0 V across a motor means each coulomb that passes through the motor transfers 4.0 J there.

The same energy account as a table, for 3.0 C passing round the circuit:

Charge goes all the way round and is not used up. The source gives energy to the charge, and the charge transfers it in the components: 18 J = 12 J + 6.0 J.

Check your understanding 1

A component transfers 27 J when 6.0 C passes through it. Find the p.d. across it and say what the number means.

p.d. across the component

VUnit: V

What the p.d. tells you

Coulombs passing through the component per second.CorrectIncorrectCoulombs used up as charge passes through the component.CorrectIncorrectJoules transferred in the component per coulomb.CorrectIncorrect

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V = W/Q = 27/6.0 = 4.5 V. Each coulomb transfers 4.5 J in the component.

Match the voltage to the part of the circuit

E.m.f. and p.d. are both measured in volts, so it is easy to use the wrong one. Ask one question: does the energy come from the source (use its e.m.f.) or is it transferred in one component (use the p.d. across that component)?

Worked example 1

E.m.f. and p.d. in an energy account

Problem

A 12 V battery drives charge through a heater and a lamp in series. The p.d. across the heater is 9.0 V and the p.d. across the lamp is 3.0 V. 5.0 C passes. Find the energy the battery supplies and the energy transferred in the heater.

Show full solution

1. Energy from the battery

   Method

   Use the battery’s e.m.f.

   Reason

   The battery supplies the energy, and e.m.f. is the energy it gives each coulomb.

   Working

   Answer

   E = QV = 5.0 × 12 = 60 J

2. Energy in the heater

   Method

   Use the p.d. across the heater.

   Reason

   The heater is one component, and p.d. is the energy each coulomb transfers in it.

   Working

   Answer

   E = QV = 5.0 × 9.0 = 45 J

3. Check the energy account

   Working

   Answer

   The lamp transfers 5.0 × 3.0 = 15 J, and 45 + 15 = 60 J: all the energy from the source is transferred.

Check your understanding 2

A 12 V supply drives a motor and a lamp in series. The p.d. across the motor is 8.0 V. 3.0 C passes. How much energy does the supply give the charge, and how much does the motor transfer? The rest is transferred in the lamp: how much?

Energy the supply gives the charge

JUnit: J

Energy transferred in the motor

JUnit: J

Show the steps Optional working

Step 1: Energy the supply gives the charge

JUnit: J

Step 2: Energy transferred in the motor

JUnit: J

Energy transferred in the lamp

JUnit: J

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Supply: E = QV = 3.0 × 12 = 36 J. Motor: E = QV = 3.0 × 8.0 = 24 J. The remaining 36 - 24 = 12 J is transferred in the lamp.

Check your understanding 3

A 9.0 V source drives a circuit in which the p.d. across a heater is 6.0 V. 8.0 C passes. A learner calculates the heater’s energy as 8.0 × 9.0 = 72 J. Find the mistake and the correct energy.

The mistake

It ignored charge used up before reaching the heater.CorrectIncorrectIt multiplied voltage by charge instead of dividing.CorrectIncorrectIt used the source voltage instead of the heater voltage.CorrectIncorrect

Correct energy transferred in the heater

JUnit: J

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The learner used the source’s e.m.f. for one component. Use the p.d. across the heater: E = QV = 8.0 × 6.0 = 48 J. The other 24 J of the 72 J from the source is transferred elsewhere in the circuit.

Resistance and wire dimensions

Resistance is p.d. divided by current

Resistance, R, is measured in ohms (Ω). V is the p.d. across the component and I is the current through that same component at the measured operating condition. Reusing that resistance for a different condition assumes it stays unchanged.

What one ohm means. 1 Ω = 1 V/A: a 1 Ω component needs a p.d. of 1 V for each ampere through it. A 30 Ω component needs 30 V for each ampere. The larger the resistance, the larger the p.d. needed to drive the same current, and the smaller the current that the same p.d. drives.

Worked example 2

Resistance from p.d. and current

Problem

A component has a p.d. of 9.0 V across it and a current of 0.30 A through it. Find its resistance.

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1. Apply the definition

   Method

   Divide the p.d. across the component by the current through it.

   Reason

   Resistance is the p.d. needed for each ampere.

   Working

   Answer

   R = V/I = 9.0/0.30 = 30 Ω

2. Interpret

   Working

   Answer

   The component needs 30 V for each ampere of current.

Both readings must belong to the component: the p.d. across it and the current through it. When a circuit has more than one component, dividing the supply voltage by the current does not give the resistance of any one of them.

The same relationship, arranged for each unknown:

Check your understanding 4

A component has 4.5 V across it and 180 mA through it. Write the current in amperes, then find its resistance.

Current in amperes

AUnit: A

Show the steps Optional working

Step 1: Current in amperes

AUnit: A

Resistance of the stated component

ΩUnit: Ω

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I = 0.180 A, so R = V/I = 4.5/0.180 = 25 Ω.

Compare wires one dimension at a time

For wires of the same material at the same temperature, resistance depends on the wire’s length and its cross-sectional area:

• Length: resistance is proportional to length. Double the length and the resistance doubles.
• Cross-sectional area: resistance is inversely proportional to area. Double the area and the resistance halves.

A longer wire gives the charge a longer path through the material. A wire with a larger cross-sectional area gives it more paths side by side.

To find out how one dimension affects resistance, change only that one. When a question changes two dimensions at once, apply each factor in turn.

Diameter is not area. The cross-section of a wire is a circle, so its area is proportional to the diameter squared. Doubling the diameter makes the area 2² = 4 times as large, so the resistance becomes one quarter.

Check your understanding 5

Wires A and B are the same material, temperature and length, but B has twice A’s cross-sectional area. Compare their resistances. Then compare them again if B were also twice as long.

Resistance of B compared with A (same length)

A quarter of A’s resistanceCorrectIncorrectHalf A’s resistanceCorrectIncorrectThe same as A’s resistanceCorrectIncorrectTwice A’s resistanceCorrectIncorrect

If B were also twice as long

Twice A’s resistanceCorrectIncorrectThe same as A’s resistanceCorrectIncorrectHalf A’s resistanceCorrectIncorrect

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With the length the same, B has half A’s resistance. If B were also twice as long, that would double its resistance again, so B would have the same resistance as A. To test one factor, keep the other the same.

Check your understanding 6

A wire carries 0.20 A when the p.d. across it is 7.2 V. Find its resistance. A second wire of the same material and temperature is twice as long and has twice the diameter. Find its resistance.

Resistance of the first wire

ΩUnit: Ω

Show the steps Optional working

Step 1: Resistance of the first wire

ΩUnit: Ω

Step 2: Cross-sectional area factor

Resistance of the second wire

ΩUnit: Ω

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R = V/I = 7.2/0.20 = 36 Ω. Twice the length multiplies R by 2. Twice the diameter makes the area 4 times as large, which divides R by 4. Together: 36 × 2 ÷ 4 = 18 Ω.

Put current, p.d. and resistance together

Worked example 3

Current, p.d. and resistance from charge and energy

Problem

A steady current passes 24 C through a component in 8.0 s. Its p.d. stays constant and it transfers 48 J during that interval. Find the current, the p.d. and the resistance.

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1. Current

   Method

   Divide charge by time.

   Reason

   Current is the rate of flow of charge.

   Working

   Answer

   I = Q/t = 24/8.0 = 3.0 A

2. p.d.

   Method

   Divide energy by charge.

   Reason

   p.d. is the energy transferred per coulomb.

   Working

   Answer

   V = W/Q = 48/24 = 2.0 V: each coulomb transfers 2.0 J

3. Resistance

   Working

   Answer

   R = V/I = 2.0/3.0 = 0.67 Ω to 2 significant figures

Every quantity here belongs to the same component over the same 8.0 s.

Check your understanding 7 Choose which component the data describe

An ideal 9.0 V source drives two components in series. During 2.0 s, a steady current carries 4.0 C through each. Component X transfers 12 J, and its p.d. is constant. Find the current, X’s p.d. and X’s resistance, and the energy transferred in the other component in those 2.0 s.

Current

AUnit: A

p.d. across X

VUnit: V

Show the steps Optional working

Step 1: Current

AUnit: A

Step 2: p.d. across X

VUnit: V

Resistance of X

ΩUnit: Ω

Show the steps Optional working

Step 1: Energy supplied by the source

JUnit: J

Energy transferred in the other component

JUnit: J

Check answer

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Current = 4.0/2.0 = 2.0 A. X’s p.d. = 12/4.0 = 3.0 V, so X’s resistance = 3.0/2.0 = 1.5 Ω. The source supplies 9.0 × 4.0 = 36 J; after X transfers 12 J, the other component transfers 24 J.

Check your understanding 8 Explain the wrong voltage choice

A learner finds X’s resistance by dividing the source’s 9.0 V by the 2.0 A current. Explain why this does not give X’s resistance, say what 9.0 V ÷ 2.0 A does give, and use the p.d.s to check the energy account.

Your explanation and p.d. check

Show answer

The source’s 9.0 V is shared by the whole external circuit, not X alone, so it cannot give X’s resistance. Dividing it by 2.0 A gives 4.5 Ω for the complete circuit. The other component’s p.d. is 24 J/4.0 C = 6.0 V, and 3.0 V + 6.0 V = 9.0 V, so the p.d.s account for the whole e.m.f.

Common mistakes

• Calling e.m.f. a force. It is measured in volts: the energy per coulomb a source gives.
• Using the supply voltage for one component. Use the p.d. across that component.
• Saying a thicker wire has more resistance. A larger cross-sectional area gives a smaller resistance.
• Saying doubling the diameter halves the resistance. The area becomes 4 times as large, so the resistance is divided by 4.
• Using mA in R = V/I. For 6.0 V and 250 mA, R = 6.0/0.250 = 24 Ω, not 6.0/250.