12 min readConcept sheetSection A objective parts (primary or secondary cell, the kilowatt hour); Section B very short answer (2 marks) on EMF and terminal voltage or showing 1 kWh = 3.6 × 10⁶ J; short answer (3 marks) numericals on E = V + Ir

EMF, internal resistance, cells, and electric power

A cell's EMF is its voltage with no current drawn; with current, E = V + Ir. The power used is P = VI = I²R = V²/R.

EMF EThe potential difference between the terminals of a cell when no current is drawn (open circuit); unit volt

Terminal voltageV = E − Ir while the cell gives a current I; r = internal resistance; I = E/(R + r)

Electric powerP = VI = I²R = V²/R; unit watt (W), 1 W = 1 J s⁻¹

Joule heating and energyHeat H = I²Rt (Joule's law); 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J, one unit on the bill

EMF, terminal voltage and internal resistance

Join a high-resistance voltmeter across a cell. With no current drawn, it reads the cell's EMF (electromotive force) E: the potential difference between its terminals on open circuit. Despite its name, EMF is a potential difference, not a force. Now let the cell drive a current I through a resistor R. The current also passes through the electrolyte inside the cell, which has a resistance: the internal resistance r. Part of the EMF is used up inside the cell, as a drop Ir. So the voltmeter now reads less: the terminal voltage V = E − Ir. The EMF depends on the electrolyte, the electrode materials and the temperature, not on the size of the cell. The internal resistance depends on the plate area, the distance between the plates and the strength of the electrolyte.

Derivation: current from a cell

A cell of EMF E and internal resistance r drives a current I through an external resistor R.

1
Potential difference across R: V = IR.
2
Inside the cell, the current passes through r, so the potential falls by Ir there.
3
The EMF supplies both drops: E = V + Ir = IR + Ir.
4★
So, I = E/(R + r) and V = E − Ir. V equals E only when no current flows.
5
The larger the current drawn, the lower the terminal voltage. That is why a car's headlights dim while its starter motor runs.

★ key step

Primary cells vs secondary cells

Primary cellsChemical energy changes directly into electrical energy. The materials are used up and the reaction cannot be reversed, so the cell cannot be recharged. Examples: the dry cell (torch cell), the Daniell cell and the voltaic cell.

Secondary cellsElectrical energy is stored as a reversible chemical change. Drawing current runs the reaction one way; charging runs it back. So the cell can be recharged again and again. Example: the lead-acid accumulator in cars and home inverters.

Quick check: a cell you throw away when it runs down is primary; a battery you put on charge is secondary.

Worked example: finding E and r from two readings

Problem

A cell gives a terminal voltage of 1.4 V when the current is 0.20 A, and 1.2 V when the current is 0.60 A. Find its EMF and internal resistance.

Working

Use V = E − Ir for each reading.
1.4 = E − 0.20r ... (i)
1.2 = E − 0.60r ... (ii)
(i) − (ii): 0.20 = 0.40r, so r = 0.50 Ω.
From (i): E = 1.4 + 0.20 × 0.50 = 1.4 + 0.10 = 1.5 V.

Answer

E = 1.5 V, r = 0.50 Ω

Derivation: power dissipated in a conductor

A conductor AB carries a steady current I. Point A is at a higher potential than B, and V = V(A) − V(B).

1
In a time Δt, a charge ΔQ = IΔt moves from A to B.
2
Its potential energy falls by ΔU = ΔQ × V = IVΔt.
3
The moving charges collide with the ions and hand over this energy. The ions vibrate more, so the conductor heats up.
4
Energy given out in time Δt: ΔW = IVΔt.
5
Power = energy per unit time: P = ΔW/Δt = VI.
6★
Using V = IR: P = I²R = V²/R. So, the power dissipated in a resistor is P = VI = I²R = V²/R. In a time t the heat produced is H = I²Rt: this is Joule's law of heating.

★ key step

Worked example: power, energy and units

Problem

(a) Show that 1 kWh = 3.6 × 10⁶ J. (b) A 100 W, 220 V bulb is used for 5 hours a day. Find its current, its resistance and the energy it uses in 30 days.

Working

(a) 1 kWh = 1000 W × 1 h = 1000 J s⁻¹ × 3600 s = 3.6 × 10⁶ J.
(b) Current: I = P/V = 100 W ÷ 220 V ≈ 0.45 A.
Resistance: R = V²/P = (220)² ÷ 100 = 48 400 ÷ 100 = 484 Ω.
Energy per day: 100 W × 5 h = 500 Wh = 0.5 kWh.
In 30 days: 0.5 × 30 = 15 kWh, that is 15 units. The bill is 15 × the rate per unit.

Answer

(a) 3.6 × 10⁶ J; (b) about 0.45 A, 484 Ω and 15 kWh (15 units)

In real life

Your electricity bill counts energy in units, where 1 unit = 1 kWh. A 2000 W geyser running for half an hour uses 1 unit, as much as a 10 W LED bulb burning for 100 hours.

Exam angle

1 marks: “Is a car battery made of primary or secondary cells?” Secondary cells: a lead-acid accumulator can be recharged.

2 marks: “Why is the terminal voltage of a cell less than its EMF when it gives a current?” Part of the EMF is used to drive the current through the cell's own internal resistance r. So V = E − Ir, which is less than E whenever I is not zero.

3 marks: “A battery of EMF 6.0 V gives a current of 0.50 A through an external resistance of 11 Ω. Find its internal resistance and its terminal voltage.” I = E/(R + r): 0.50 = 6.0/(11 + r), so 11 + r = 12 and r = 1.0 Ω. V = IR = 0.50 × 11 = 5.5 V (check: E − Ir = 6.0 − 0.5 = 5.5 V).

Remember it

EMF is the full price, terminal voltage is what reaches the shop: V = E − Ir, the Ir lost inside the cell. And 1 unit = 1 kWh = 3.6 million joules.