PhysicsISC Class 12 · Current Electricity

Emf, Internal Resistance and the Combination of Cells

Updated 15 October 2026 · Fact-checked

Emf ε is the work a cell does per unit charge; internal resistance r opposes current inside it. With an external resistor R, $I = \frac{\varepsilon}{R + r}$ and the terminal p.d. is $V = \varepsilon - Ir$. n identical cells give emf nε and resistance nr in series, but ε and $\frac{r}{n}$ in parallel.

The cell's emf ε is shared between R and r: ε = IR + Ir, so the voltmeter at the terminals reads V = ε − Ir, less than ε.

Illustration drawn for QuestionBuddy.

Understand Emf, Internal Resistance and Combination of Cells

Emf. A cell uses chemical energy to push charge from its negative to its positive terminal inside the cell, against the electric field. Its emf is the work done per unit charge in doing this: $ε = \frac{d W}{d q}$ . Unit: volt. Emf is not a force; it is the energy given to each coulomb.

Internal resistance. The electrolyte and electrodes resist the current, so part of the energy is turned into heat inside the cell. This internal resistance r depends on the electrolyte, the distance between the plates, and their area; it grows as a cell ages.

**Deriving $I = \frac{ε}{R + r}$ .** In time dt, charge $d q = I d t$ passes. The cell does work $ε d q = ε I d t$ . This energy appears as heat in R and in r: $ε I d t = I^{2} R d t + I^{2} r d t$ . Dividing by $I d t$ :
$ε = I R + I r \Rightarrow I = \frac{ε}{R + r} .$

Terminal potential difference. The p.d. across the external resistor is $V = I R$ , so
$V = ε - I r .$
The term $I r$ is the "lost volts" inside the cell. Cases:

Open circuit (I = 0): $V = ε$ . A high-resistance voltmeter across a cell reads nearly its emf.
Short circuit (R = 0): $I = \frac{ε}{r}$ , the largest possible current; V = 0.
While charging (current forced into the positive terminal): $V = ε + I r$ .

Internal resistance from two readings: $r = (\frac{ε}{V} - 1) R$ .

V–I graph. Plotting the terminal p.d. V against the current I for different R gives a straight line $V = ε - r I$ : the V-intercept is ε (I = 0) and the magnitude of the slope is r. The 2025 ISC question asked for exactly this, and the examiners reported candidates who did not know that V = ε when I = 0. Measure I with an ammeter in series and V with a voltmeter in parallel.

Cells in series. n cells, all the same way: $ε_{e q} = ε_{1} + ε_{2} + \dots$ , $r_{e q} = r_{1} + r_{2} + \dots$ . A cell connected the wrong way round subtracts its emf but still adds its resistance. For identical cells, $I = \frac{n ε}{R + n r}$ ; series is best when $R ≫ r$ .

Cells in parallel. n identical cells side by side: the emf stays ε (not nε: the 2024 examiners saw "3e" for three cells in parallel) and $r_{e q} = \frac{r}{n}$ ; $I = \frac{ε}{R + \frac{r}{n}}$ . Parallel is best when $R ≪ r$ , and lets the battery deliver more current without draining one cell.
Two unequal cells in parallel (like terminals together):
$ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}, r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}} .$

Mixed grouping. N = mn identical cells: m rows in parallel, each with n cells in series. Each row has emf nε and resistance nr; the rows in parallel give emf nε and resistance $\frac{n r}{m}$ :
$I = \frac{n ε}{R + \frac{n r}{m}} = \frac{mn ε}{m R + n r} .$
For fixed N, the current is **maximum when $R = \frac{n r}{m}$ **: the external resistance equals the total internal resistance.

Power and maximum power transfer. Power delivered to R: $P = I^{2} R = \frac{ε ^{2} R}{( R + r ) ^{2}}$ . It is maximum when $R = r$ , with $P_{ma x} = \frac{ε ^{2}}{4 r}$ ; the efficiency is then only 50%, because as much power is wasted inside the cell.

Emf, Internal Resistance and Combination of Cells: formulas and key facts

Circuit current: $I = \frac{ε}{R + r}$; From $ε I d t = I^{2} R d t + I^{2} r d t$ .
Terminal p.d.: $V = ε - I r$; Charging: $V = ε + I r$ .
Internal resistance: $r = (\frac{ε}{V} - 1) R$; From open-circuit and loaded readings.
n identical in series: $ε_{e q} = n ε, r_{e q} = n r$; Reversed cell: subtract its emf only.
n identical in parallel: $ε_{e q} = ε, r_{e q} = \frac{r}{n}$; Emf does not add.
Unequal cells in parallel: $ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}$; $r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$ .
Mixed grouping: $I = \frac{mn ε}{m R + n r}$; Maximum when $R = \frac{n r}{m}$ .
Maximum power: $P_{ma x} = \frac{ε ^{2}}{4 r} at R = r$; Efficiency 50%.

How to solve Emf, Internal Resistance and Combination of Cells questions

Replace every battery by a single equivalent emf and internal resistance, then use Ohm's law for the whole circuit.

1Identify how the cells are grouped (series, parallel, mixed) and whether any is reversed.
2Find $ε_{e q}$ and $r_{e q}$ with the right formula.
3Find the current: $I = \frac{ε _{e q}}{R + r _{e q}}$ .
4Find the terminal p.d. $V = ε_{e q} - I r_{e q}$ (or $I R$ , which should agree).
5For a V–I graph, read ε from the V-intercept and r from the slope.
6For power, use $P = I^{2} R$ ; for the best arrangement, match R to the internal resistance.

Emf, Internal Resistance and Combination of Cells in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Lost volts in a power bank

Suppose a power bank has an emf of 5.1 V and an internal resistance of 0.15 Ω, and a phone draws 2 A from it while fast charging.

Find the p.d. at the power bank's terminals and the power wasted inside it.

Try it first, then show the solutionSolution

$V = ε - I r = 5.1 - 2 \times 0.15 = 4.8$ V.
Wasted power $I^{2} r = 4 \times 0.15 = 0.6$ W, which warms the power bank.

Answer: 4.8 V; 0.6 W lost as heat.

Memory hook: Lost volts = Ir.

Measuring internal resistance

Imagine a 9 V battery that reads 9.0 V on a voltmeter with nothing connected, but only 7.5 V when it drives a toy motor drawing 0.5 A.

Find its internal resistance. Why does the motor run slowly?

Try it first, then show the solutionSolution

Open-circuit reading gives ε = 9.0 V.
$I r = 9.0 - 7.5 = 1.5$ V, so $r = \frac{1.5}{0.5} = 3$ Ω.

Answer: 3 Ω: an old battery loses a sixth of its emf inside itself at this current.

Memory hook: Emf on open circuit, terminal p.d. under load; the gap is Ir.

Tricks to remember Emf, Internal Resistance and Combination of Cells

Short memory aids. Say each one aloud once, then use it on a question the same day.

Emf minus lost volts
$V = ε - I r$ : what you get = what the cell gives − what it keeps.
Use it when: Terminal p.d.
Intercept and slope
V–I graph: V-intercept = emf, |slope| = internal resistance.
Use it when: Graph questions.
Parallel cells: same push, less resistance
Emf stays ε, resistance becomes r/n.
Use it when: Cells in parallel.
Match for maximum
Maximum power when R = r; maximum current (mixed grouping) when R = nr/m.
Use it when: Optimisation questions.

Emf, Internal Resistance and Combination of Cells: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Charging a cell · 1 mark · MCQBoard level

A cell of emf ε and internal resistance r is being charged by a current I. The p.d. across its terminals is:
(a) ε
(b) ε − Ir
(c) ε + Ir
(d) Ir

Try it first, then show the solutionSolution

The charger must overcome the emf and also drive I through r.

Answer: (c).

Example 2: Current and terminal p.d. · 2 marks · Short answerBoard level

A cell of emf 9 V and internal resistance 1 Ω is connected to an 8 Ω resistor. Find the current, the terminal p.d. and the short-circuit current.

Try it first, then show the solutionSolution

$I = \frac{9}{8 + 1} = 1$ A; $V = 9 - 1 \times 1 = 8$ V. (1)
Short circuit: $I = \frac{ε}{r} = 9$ A. (1)

Answer: 1 A; 8 V; 9 A.

Example 3: Emf and r from a V–I graph · 3 marks · Short answerHarder board level

In an experiment with a cell, an ammeter and a voltmeter, the terminal p.d. is 1.8 V when the current is 0.2 A and 1.2 V when the current is 0.6 A. The V–I graph is a straight line. Find the emf and internal resistance, and state where the graph meets the V-axis.

Try it first, then show the solutionSolution

$∣ s l o p e ∣ = r = \frac{1.8 - 1.2}{0.6 - 0.2} = 1.5$ Ω. (1)
$ε = V + I r = 1.8 + 0.2 \times 1.5 = 2.1$ V. (1)
The line meets the V-axis (I = 0) at 2.1 V, the emf. (1)

Answer: ε = 2.1 V, r = 1.5 Ω; V-intercept 2.1 V.

Example 4: Unequal cells in parallel · 3 marks · Short answerHarder board level

A 6 V cell of internal resistance 1 Ω and a 3 V cell of internal resistance 2 Ω are connected in parallel (positive to positive) across a 4 Ω resistor. Find the current in the resistor.

Try it first, then show the solutionSolution

$ε_{e q} = \frac{6 \times 2 + 3 \times 1}{1 + 2} = 5$ V. (1)
$r_{e q} = \frac{1 \times 2}{3} = \frac{2}{3}$ Ω. (1)
$I = \frac{5}{4 + \frac{2}{3}} = \frac{15}{14} \approx 1.07$ A. (1)

Answer: About 1.07 A.

Example 5: Best grouping of cells · 4 marks · Long answerHOTS / competency

Sixteen identical cells, each of emf 2 V and internal resistance 1 Ω, are to be arranged in m parallel rows of n cells in series to send the largest possible current through a 4 Ω resistor. Find m, n and the current.

Try it first, then show the solutionSolution

Maximum current when $R = \frac{n r}{m}$ , i.e. $\frac{n}{m} = 4$ . (1)
With $mn = 16$ : $n = 8$ , $m = 2$ . (1)
$I = \frac{n ε}{R + \frac{n r}{m}} = \frac{16}{4 + 4} = 2$ A. (1)
Check: all 16 in series gives $\frac{32}{20} = 1.6$ A; all in parallel gives $\frac{2}{4.0625} \approx 0.49$ A. Both smaller. (1)

Answer: 2 rows of 8 cells; 2 A.

Example 6: Maximum power from a battery · 4 marks · Case-basedHOTS / competency

Suppose a battery of emf 12 V and internal resistance 2 Ω powers a heater whose resistance R can be chosen.
(i) Find the power in R for R = 2 Ω and R = 4 Ω. (2 marks)
(ii) For what R is the power greatest, and what is it? (1 mark)
(iii) Find the efficiency at that R. (1 mark)

Try it first, then show the solutionSolution

(i) R = 2 Ω: $I = 3$ A, $P = 9 \times 2 = 18$ W. R = 4 Ω: $I = 2$ A, $P = 4 \times 4 = 16$ W.
(ii) $P = \frac{ε ^{2} R}{( R + r ) ^{2}}$ is greatest when R = r = 2 Ω: $P_{ma x} = \frac{144}{8} = 18$ W.
(iii) Efficiency $= \frac{R}{R + r} = 50%$ : another 18 W is wasted in the battery.

Answer: (i) 18 W, 16 W; (ii) R = 2 Ω, 18 W; (iii) 50%.

Where students lose marks in Emf, Internal Resistance and Combination of Cells

The slips examiners see most often, and what to write instead.

Mistake: Adding emfs of parallel cells
In ISC 2024 candidates used series ideas and answered 3e for three cells in parallel.
Do this: Parallel identical cells: emf ε, resistance r/n.
Usually costs: 1 mark.
Mistake: Missing V = ε at I = 0
The 2025 examiners saw candidates who did not know the V-intercept of a V–I graph is the emf, and confused emf with voltage.
Do this: Write the line $V = ε - r I$ and read ε at I = 0, r from the slope.
Usually costs: 1 to 2 marks.
Mistake: Wrong instruments
Also in 2025: X and Y were called galvanometer or potentiometer instead of ammeter and voltmeter.
Do this: Ammeter in series measures I; voltmeter in parallel measures V.
Usually costs: 1 mark.
Mistake: Unclear polarity in combinations
In 2025 candidates could not read whether cells in a diagram were in series or parallel.
Wrong working
Taking two cells joined + to − as parallel.
Right working
+ to − in a chain is series: emfs add.
Do this: Trace the path from one terminal of the battery to the other.
Usually costs: 1 mark.

Exam tips

In the derivation of $I = \frac{ε}{R + r}$ , use the energy argument the syllabus describes.
Draw the V–I graph with labelled axes, intercept and slope when asked to explain.
Use the rubric's symbol for emf (ε or E as given) consistently; the 2024 examiners penalised odd symbols.
For grouping questions, check your answer by comparing with all-series and all-parallel.

Practice questions on Emf, Internal Resistance and Combination of Cells

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Four identical cells, each of emf 1.5 V, are connected in parallel. The emf of the battery is:
(a) 6 V
(b) 1.5 V
(c) 0.375 V
(d) zero
(A variation on ISC 2024.)
Hint
In parallel the emf is that of one cell.
Answer
(b) 1.5 V.
Q2 · 1 mark · Very short answerBoard level
When can the terminal p.d. of a cell be greater than its emf?
Hint
Think about the direction of current through the cell.
Answer
When the cell is being charged: $V = ε + I r$ .
Q3 · 2 marks · Short answerHarder board level
A battery drives 1.5 A through a 4 Ω resistor, and its terminal p.d. is 0.75 V less than its open-circuit value. Find its emf and internal resistance.
Hint
V = IR, lost volts = Ir.
Answer
V = 6 V; $r = \frac{0.75}{1.5} = 0.5$ Ω; ε = 6.75 V.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Identical cells in parallel give the same emf as one cell but can supply a larger current through a small resistance.
Reason (R): The internal resistance of n identical cells in parallel is $\frac{r}{n}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$I = \frac{ε}{R + r / n}$ .
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
A battery of 5 cells, each of emf 2 V and internal resistance 0.4 Ω, is charged from a 15 V d.c. supply through a 3 Ω series resistor. Find the charging current and the p.d. across the battery terminals.
Hint
The battery emf opposes the supply.
Answer
$I = \frac{15 - 10}{3 + 2} = 1$ A; $V = 10 + 1 \times 2 = 12$ V.
Q6 · 3 marks · Short answerHOTS / competency
A cell delivers the same power to an external resistor of 2 Ω as to one of 8 Ω. Find its internal resistance.
Hint
Equate $\frac{ε ^{2} R}{( R + r ) ^{2}}$ for both.
Answer
$\frac{2}{( 2 + r ) ^{2}} = \frac{8}{( 8 + r ) ^{2}}$ gives $r^{2} = 16$ , r = 4 Ω.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a torch uses two cells in series, each of emf 1.5 V and internal resistance 0.3 Ω, with a bulb of resistance 2.4 Ω.
(i) Find the current and the bulb's power. (1 mark)
(ii) Find the terminal p.d. of the battery. (1 mark)
(iii) What happens if one cell is put in the wrong way round? (1 mark)
(iv) As the cells age, each internal resistance rises to 1.2 Ω. Find the new bulb power. (1 mark)
Hint
Total r = 0.6 Ω, then 2.4 Ω.
Answer
(i) $I = \frac{3}{3.0} = 1$ A; 2.4 W. (ii) 2.4 V. (iii) Net emf 0: no current, the bulb stays off. (iv) $I = \frac{3}{4.8} = 0.625$ A; $P \approx 0.94$ W.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define emf.Show answer
The work done per unit charge by the source in moving charge round the complete circuit (from − to + inside the cell).
2.Terminal p.d. on open circuit?Show answer
Equal to the emf.
3.Cell 2 V, r = 0.5 Ω, R = 3.5 Ω: current?Show answer
0.5 A.
4.Condition for maximum power in the external resistor?Show answer
R = r.
5.Equivalent internal resistance of 4 identical cells (r each) in parallel?Show answer
$\frac{r}{4}$ .

Emf, Internal Resistance and Combination of Cells: frequently asked questions

What is the difference between emf and terminal p.d.?

Emf is the energy per coulomb the cell provides; terminal p.d. is the energy per coulomb delivered to the external circuit. They differ by the lost volts Ir inside the cell, and are equal only when no current flows.

Why does a car's light dim when the engine starts?

The starter motor draws a very large current, so the lost volts Ir inside the battery become large and the terminal p.d. falls for a moment.

Is maximum power the same as maximum efficiency?

No. Maximum power occurs at R = r, where the efficiency is only 50%. Efficiency \frac{R}{R + r} approaches 100% when R is much larger than r, but the power is then small.

Which combinations of cells are in the ISC 2027 syllabus?

Identical cells in series, parallel and mixed grouping, two cells of unequal emf in parallel, and n cells of unequal emf in series.

How was this topic tested recently?

In 2025 a 3-mark question asked for emf and internal resistance from a V–I graph and a 2-mark question compared series and parallel pairs of cells; in 2024 a 1-mark item asked for the emf of cells in parallel.