10 min readConcept sheet1-mark MCQ and assertion-reason, 2-mark definitions and graphs, and 3-mark numericals; also a NEET favourite

EMF, terminal voltage and internal resistance

A cell's emf ε is its terminal voltage on open circuit. When it drives a current I, the terminal voltage drops to V = ε − Ir.

EMF εPotential difference between the terminals of a cell when no current flows (open circuit); unit volt

Internal resistance rResistance of the electrolyte between the electrodes; it causes a drop Ir inside the cell

Terminal voltageGiving current: V = ε − Ir. Being charged: V = ε + Ir. Current in a load R: I = ε/(R + r)

Largest currentI_max = ε/r, when the terminals are joined by a wire of nearly zero resistance (R = 0)

Inside a cell

A cell has two electrodes, positive (P) and negative (N), dipped in an electrolyte. Chemical action keeps P at a higher potential than N. With no current, the potential difference between P and N is the emf ε of the cell. The name 'electromotive force' is historical: emf is a potential difference, not a force, and it is measured in volts. When the cell drives a current I, the current also flows through the electrolyte, from N to P. The electrolyte has a resistance, the internal resistance r. Part of the emf is used up inside the cell as a drop Ir. So the terminal voltage V is less than ε while the cell gives current.

Derivation: current and terminal voltage

A cell of emf ε and internal resistance r is joined to an external resistor R. A current I flows.

1
Potential difference across the external resistor: V = IR.
2
Inside the cell the current passes through r, so the potential falls by Ir there.
3
Terminal voltage: V = ε − Ir.
4
Equate the two: IR = ε − Ir, so I(R + r) = ε.
5★
So, I = ε/(R + r) and V = εR/(R + r). V equals ε only when I = 0. The largest current, with R = 0, is ε/r.

★ key step

EMF vs terminal voltage

EMF εPD between the terminals when no current flows. It is set by the chemistry of the cell and does not depend on the external resistance.

Terminal voltage VPD between the terminals when current flows. V = ε − Ir while the cell gives current, so V < ε. V = ε + Ir while it is being charged, so V > ε.

A voltmeter that draws almost no current, joined straight across a cell, reads very nearly its emf.

How to draw the V against I graph of a cell

Draw current I (A) on the x-axis and terminal voltage V (V) on the y-axis.
Draw a straight line sloping downwards from the V-axis to the I-axis.
Mark the V-intercept as ε (where I = 0, open circuit).
Mark the I-intercept as ε/r (where V = 0, terminals shorted).
Write 'slope = −r' beside the line. A steeper line means a larger internal resistance.

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

Answer

ε = 1.5 V, r = 0.50 Ω

Worked example: giving current and being charged

Problem

A 12 V battery has an internal resistance of 0.50 Ω. Find its terminal voltage (a) when it supplies 4.0 A to a load, and (b) when it is being charged with a current of 4.0 A. (c) What is the largest current it can give?

Working

(a) Giving current: V = ε − Ir = 12 V − 4.0 A × 0.50 Ω = 12 V − 2.0 V = 10 V.
(b) Being charged: V = ε + Ir = 12 V + 2.0 V = 14 V.
(c) I_max = ε/r = 12 V ÷ 0.50 Ω = 24 A.

Answer

(a) 10 V; (b) 14 V; (c) 24 A

In real life

When a two-wheeler or car is started, the headlights dim for a moment. The starter motor draws a huge current, so the battery's terminal voltage V = ε − Ir drops. An old inverter battery has a larger internal resistance, so its voltage drops even more under load.

Exam angle

1 marks: “When is the terminal voltage of a cell equal to its emf?” When no current is drawn from it (open circuit).

2 marks: “A cell of emf 2.0 V and internal resistance 0.50 Ω is connected across a 3.5 Ω resistor. Find the current and the terminal voltage.” I = ε/(R + r) = 2.0 ÷ 4.0 = 0.50 A. V = ε − Ir = 2.0 − 0.25 = 1.75 V (also IR = 0.50 × 3.5 = 1.75 V).

3 marks: “Draw the graph of terminal voltage V against current I for a cell. Show how its emf and internal resistance can be found from it.” A straight line with a negative slope. The V-intercept is ε. The slope is −r, so r = (fall in V) ÷ (rise in I). The I-intercept is ε/r.

Remember it

Open: V = ε. Giving current: V = ε − Ir. Being charged: V = ε + Ir.