PhysicsCBSE Class 12 · Current Electricity

EMF, Internal Resistance and Cells Class 12: Series and Parallel

Q: Why does terminal voltage drop when more current is drawn?

The current also flows through the cell's internal resistance, using up Ir volts inside the cell. The larger the current, the more of the emf is lost inside, so V = ε − Ir falls.

Q: Is emf the same as potential difference?

No. Emf is the terminal voltage when no current flows; it is a property of the cell. Potential difference between the terminals while current flows is the terminal voltage, which is less than the emf (more when charging).

Q: When should cells be connected in parallel?

When the external resistance is small compared with the internal resistance, since parallel connection lowers the internal resistance and lets more current flow.

Q: Can a cell be damaged by connecting it in parallel with a weaker cell?

The stronger cell drives current backwards through the weaker one, charging it and heating both. With very different emfs this wastes energy and can harm the cells.

Q: How do you find internal resistance experimentally?

Measure the terminal voltage for two different currents (or loads). The fall in voltage divided by the rise in current gives r, and the V-I line extended to I = 0 gives ε.

Updated 15 October 2026 · Fact-checked

The emf ε of a cell is its terminal voltage when no current flows. With current I, the terminal voltage is V = ε − Ir, r being the internal resistance, and I = ε/(R + r). Cells in series add emfs and resistances; in parallel, ε = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) and r = r₁r₂/(r₁ + r₂).

The cell's own resistance r uses up Ir volts, so V falls straight from ε as current rises; the line's slope is −r and it hits zero at Imax.

Illustration drawn for QuestionBuddy.

Understand EMF, Internal Resistance and Cells

A torch with old cells gives a dim beam even though a voltmeter across the cells, with the torch off, still reads close to 1.5 V each. The reason is the cell's own internal resistance, which grows as it ages.

EMF. Inside a cell, chemical reactions keep the positive terminal at a higher potential than the negative one. The emf ε is the potential difference between the terminals when no current is drawn (open circuit). It equals the work done by the cell per unit charge in moving charge round the circuit, so it is measured in volt. Despite its name, emf is not a force.

Internal resistance and terminal voltage. Current through a cell also passes through its electrolyte, which has resistance r, the internal resistance. With a load R, the current is $I = \frac{ε}{R + r}$ . Part of the emf is used inside the cell (the 'lost volts' Ir), so the terminal voltage is $V = I R = ε - I r$ . Consequences:

V < ε when the cell supplies current; V = ε only when I = 0.
The V versus I graph is a straight line with intercept ε on the V-axis and slope −r. It meets the I-axis at the short-circuit current $I_{ma x} = \frac{ε}{r}$ .
When a cell is charged (current forced into its positive terminal), $V = ε + I r$ , more than the emf.
A cell with large r cannot deliver a large current: that is why the old torch is dim.

Cells in series. Join the negative terminal of cell 1 to the positive terminal of cell 2. The terminal voltages add: $V = (ε_{1} + ε_{2}) - I (r_{1} + r_{2})$ . So the combination acts as one cell with $ε_{e q} = ε_{1} + ε_{2}$ and $r_{e q} = r_{1} + r_{2}$ . If one cell is connected the wrong way round, its emf subtracts, but its internal resistance still adds. For n identical cells: $I = \frac{n ε}{R + n r}$ .

Cells in parallel. Join positive to positive and negative to negative, so both cells have the same terminal voltage V. Currents $I_{1} = \frac{ε _{1} - V}{r _{1}}$ and $I_{2} = \frac{ε _{2} - V}{r _{2}}$ add to I: $I = \frac{ε _{1}}{r _{1}} + \frac{ε _{2}}{r _{2}} - V (\frac{1}{r _{1}} + \frac{1}{r _{2}})$ . Rearranging to the form $V = ε_{e q} - I r_{e q}$ gives $ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}$ and $r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$ , i.e. $\frac{ε _{e q}}{r _{e q}} = \frac{ε _{1}}{r _{1}} + \frac{ε _{2}}{r _{2}}$ . For m identical cells in parallel: emf ε and internal resistance $\frac{r}{m}$ , so $I = \frac{ε}{R + r / m}$ . If the emfs differ a lot, the weaker cell may be charged by the stronger one (its current comes out negative).

Series or parallel? Series raises the emf and suits a large external resistance ( $R ≫ r$ ). Parallel lowers the internal resistance and suits a small external resistance ( $R ≪ r$ ). With a fixed number of cells in rows, the current is largest when the total internal resistance roughly equals R.

EMF, Internal Resistance and Cells: formulas and key facts

Current in a circuit with one cell: $I = \frac{ε}{R + r}$; r is the internal resistance.
Terminal voltage (discharging): $V = ε - I r = I R$; V = ε only when I = 0.
Terminal voltage (charging): $V = ε + I r$; Current enters at the positive terminal.
Internal resistance from a measurement: $r = R (\frac{ε}{V} - 1)$; From V = εR/(R + r).
Cells in series: $ε_{e q} = ε_{1} + ε_{2}, r_{e q} = r_{1} + r_{2}$; A reversed cell's emf subtracts; its r still adds.
Cells in parallel: $ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}, r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$; Equivalently $\frac{ε _{e q}}{r _{e q}} = \frac{ε _{1}}{r _{1}} + \frac{ε _{2}}{r _{2}}$ .
Mixed grouping (n per row, m rows): $I = \frac{n ε}{R + \frac{n r}{m}}$; Current is largest when nr/m is close to R.

How to solve EMF, Internal Resistance and Cells questions

Use this method for emf, terminal voltage and cell combination questions.

1Draw each cell with its emf and its internal resistance r shown as a small resistor beside it.
2Replace a group of cells by one equivalent cell: add emfs (with signs) and resistances in series; use the parallel formula for cells side by side.
3Find the current with $I = \frac{ε _{e q}}{R + r _{e q}}$ .
4Find the terminal voltage with $V = ε - I r$ (or $ε + I r$ if the cell is being charged).
5For a V-I graph, read ε as the intercept on the V-axis and r as minus the slope.
6For cells in parallel with different emfs, find each cell's current from $I_{k} = \frac{ε _{k} - V}{r _{k}}$ ; a negative value means that cell is being charged.
7Check: the sum of lost volts and terminal voltage equals the total emf.

EMF, Internal Resistance and 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.

Terminal voltage drops under heavy load

Suppose a car battery has an emf of 12 V and an internal resistance of 0.025 Ω. When the starter motor turns, it draws 120 A, and the headlights are connected across the same battery terminals.

(a) Find the terminal voltage while the starter runs. (b) Explain why the headlights dim at that moment. (c) What would the terminal voltage be with only the headlights on, drawing 8 A?

Try it first, then show the solutionSolution

(a) $V = ε - I r = 12 - 120 \times 0.025 = 9$ V.
(b) The headlights now get only 9 V instead of nearly 12 V, so they draw less current and give less light.
(c) $V = 12 - 8 \times 0.025 = 11.8$ V, close to the emf.

Answer: (a) 9 V; (b) the lost volts Ir reduce the voltage across the lights; (c) 11.8 V.

Memory hook: Big current, big lost volts: the starter steals voltage from the lights.

Internal resistance grows as cells age

Imagine a torch bulb of resistance 6 Ω run by two 1.5 V cells in series. New cells have 0.25 Ω internal resistance each; old ones have 2 Ω each, with the emf still close to 1.5 V.

(a) Find the current with new and with old cells. (b) Find the power used by the bulb in each case. (c) Why does a voltmeter across old cells (torch off) still read about 3 V?

Try it first, then show the solutionSolution

New: $I = \frac{3}{6 + 0.5} \approx 0.46$ A. Old: $I = \frac{3}{6 + 4} = 0.30$ A.
Bulb power $= I^{2} R$ : new $\approx 0.4 6^{2} \times 6 \approx 1.28$ W; old $= 0.09 \times 6 = 0.54$ W.
With the torch off almost no current flows, so Ir ≈ 0 and the voltmeter reads nearly the emf.

Answer: (a) about 0.46 A and 0.30 A; (b) about 1.28 W and 0.54 W; (c) on open circuit the terminal voltage equals the emf.

Memory hook: An old cell keeps its emf but loses its strength: r grows.

Tricks to remember EMF, Internal Resistance and Cells

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

The cell pays a toll
Every coulomb pays Ir of 'toll' inside the cell, so only ε − Ir reaches the outside.
Use it when: Remembering V = ε − Ir.
Graph: intercept is emf, slope is r
On a V-I graph, read ε where the line meets the V-axis and r from how steeply it falls.
Use it when: Graph-based questions.
Series adds everything, parallel averages
Series: add emfs and add r. Parallel: emf is a weighted average, r drops like resistors in parallel.
Use it when: Combination questions.
Charging flips the sign
Discharging: minus Ir. Charging: plus Ir.
Use it when: Battery charger questions.

EMF, Internal Resistance and Cells: solved examples

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

Example 1: Current and terminal voltage · 2 marks · Very short answerBoard level

A cell of emf 6 V and internal resistance 0.4 Ω is connected to a 2.6 Ω resistor. Find (a) the current, (b) the terminal voltage and (c) the potential drop inside the cell.

Try it first, then show the solutionSolution

(a) $I = \frac{6}{2.6 + 0.4} = 2$ A. (1)
(b) $V = ε - I r = 6 - 0.8 = 5.2$ V (check: $I R = 2 \times 2.6 = 5.2$ V). (½)
(c) Lost volts $= I r = 0.8$ V. (½)

Answer: (a) 2 A; (b) 5.2 V; (c) 0.8 V.

Example 2: Derivation: two cells in parallel · 3 marks · Short answerBoard level

Two cells of emfs ε₁ and ε₂ and internal resistances r₁ and r₂ are connected in parallel. Derive the emf and internal resistance of the equivalent cell.

Try it first, then show the solutionSolution

Both cells have the same terminal voltage V. Currents through them: $I_{1} = \frac{ε _{1} - V}{r _{1}}$ , $I_{2} = \frac{ε _{2} - V}{r _{2}}$ . (1)
Total current $I = I_{1} + I_{2} = (\frac{ε _{1}}{r _{1}} + \frac{ε _{2}}{r _{2}}) - V (\frac{1}{r _{1}} + \frac{1}{r _{2}})$ . (1)
Solving for V: $V = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}} - I \frac{r _{1} r _{2}}{r _{1} + r _{2}}$ . Comparing with $V = ε_{e q} - I r_{e q}$ : $ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}$ , $r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$ . (1)

Answer: $ε_{e q} = \frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}$ and $r_{e q} = \frac{r _{1} r _{2}}{r _{1} + r _{2}}$ .

Example 3: EMF and internal resistance from readings · 3 marks · Short answerHarder board level

When a cell supplies 0.2 A its terminal voltage is 1.4 V; when it supplies 0.8 A the terminal voltage is 1.1 V. Find (a) the internal resistance, (b) the emf and (c) the largest current it can supply. Sketch the V-I graph.

Try it first, then show the solutionSolution

(a) $1.4 = ε - 0.2 r$ and $1.1 = ε - 0.8 r$ . Subtracting: $0.3 = 0.6 r$ , r = 0.5 Ω. (1)
(b) $ε = 1.4 + 0.2 \times 0.5 = 1.5$ V. (1)
(c) Short circuit: $I = \frac{ε}{r} = 3$ A. The graph is a straight line from (0, 1.5 V) to (3 A, 0), slope −0.5 Ω. (1)

Answer: (a) 0.5 Ω; (b) 1.5 V; (c) 3 A.

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

A 6 V cell (internal resistance 1 Ω) and a 3 V cell (internal resistance 2 Ω) are connected in parallel, positive to positive, across a 4 Ω resistor. Find (a) the equivalent emf and internal resistance, (b) the current in the 4 Ω resistor, and (c) the current through each cell. Comment on the 3 V cell.

Try it first, then show the solutionSolution

(a) $ε_{e q} = \frac{6 \times 2 + 3 \times 1}{1 + 2} = 5$ V; $r_{e q} = \frac{2}{3}$ Ω. (1)
(b) $I = \frac{5}{4 + 2/3} \approx 1.07$ A; terminal voltage $V = 4 \times 1.07 \approx 4.29$ V. (1)
(c) $I_{1} = \frac{6 - 4.29}{1} \approx 1.71$ A; $I_{2} = \frac{3 - 4.29}{2} \approx - 0.64$ A. The negative sign means current enters the 3 V cell at its positive terminal: it is being charged by the 6 V cell. (1)

Answer: (a) 5 V and 2/3 Ω; (b) about 1.07 A; (c) about 1.71 A from the 6 V cell and 0.64 A into the 3 V cell, which is being charged.

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

Eight identical cells, each of emf 2 V and internal resistance 0.5 Ω, are to drive current through a 1 Ω resistor. Compare the current when they are arranged (a) all in series, (b) all in parallel, (c) in 2 rows of 4, and (d) in 4 rows of 2. Which arrangement gives the largest current, and why?

Try it first, then show the solutionSolution

General: n cells per row, m rows: $I = \frac{n ε}{R + n r / m}$ . (1)
(a) n = 8, m = 1: $I = \frac{16}{1 + 4} = 3.2$ A. (1)
(b) n = 1, m = 8: $I = \frac{2}{1 + 0.0625} \approx 1.88$ A. (1)
(c) n = 4, m = 2: $I = \frac{8}{1 + 1} = 4$ A. (d) n = 2, m = 4: $I = \frac{4}{1 + 0.25} = 3.2$ A. (1)
Largest in (c), where the total internal resistance (1 Ω) equals the external resistance. (1)

Answer: (a) 3.2 A; (b) about 1.88 A; (c) 4 A; (d) 3.2 A. The 2 × 4 grouping wins, as its internal resistance equals R.

Example 6: Charging a battery · 4 marks · Case-basedHarder board level

Read the passage and answer.

A 12 V battery of internal resistance 0.5 Ω is charged from a 15 V supply (negligible internal resistance) through a 2.5 Ω resistor, with the positive terminals connected together.
(a) Find the charging current. (1)
(b) Find the terminal voltage of the battery during charging. (1)
(c) Find the rate at which energy is stored in the battery as chemical energy and the rate of heating inside it. (1)
(d) Why is a series resistor needed? (1)

Try it first, then show the solutionSolution

(a) The emfs oppose: $I = \frac{15 - 12}{2.5 + 0.5} = 1$ A. (1)
(b) $V = ε + I r = 12 + 0.5 = 12.5$ V (also $15 - 1 \times 2.5 = 12.5$ V). (1)
(c) Chemical energy: $ε I = 12$ W; heat in the battery: $I^{2} r = 0.5$ W. (Supply gives 15 W; the other 2.5 W heats the resistor.) (1)
(d) Without it, $I = \frac{3}{0.5} = 6$ A, which could overheat and damage the battery; the resistor limits the current. (1)

Answer: (a) 1 A; (b) 12.5 V; (c) 12 W stored, 0.5 W as heat in the battery; (d) to limit the charging current.

Where students lose marks in EMF, Internal Resistance and Cells

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

Mistake: Calling emf a force.
The name 'electromotive force' misleads.
Do this: Define emf as the potential difference across the terminals when no current is drawn (work done per unit charge by the cell).
Usually costs: 1 mark in a definition.
Mistake: Forgetting internal resistance when finding current.
Using I = ε/R by habit.
Wrong working
I = 6/2.6 ≈ 2.31 A
Right working
I = 6/(2.6 + 0.4) = 2 A
Do this: Draw r next to every cell in the circuit diagram.
Usually costs: The final 1 to 2 marks.
Mistake: Adding emfs of cells in parallel.
Confusing parallel with series.
Wrong working
Two 2 V cells in parallel give 4 V.
Right working
Two identical 2 V cells in parallel give 2 V with half the internal resistance.
Do this: In parallel, the emf is a weighted average, never a sum.
Usually costs: Most of the question.
Mistake: Subtracting the internal resistance of a reversed cell.
Treating r like ε.
Do this: Reversing a cell flips the sign of its emf only; resistances always add in series.
Usually costs: 1 mark.
Mistake: Reading the slope of a V-I graph as +r.
The line slopes downward.
Do this: r = −(slope) = (fall in V)/(rise in I).
Usually costs: ½ to 1 mark.

Exam tips

Write 'V = ε − Ir' as the first line of any terminal-voltage answer, and say which way the current flows in the cell.
For the parallel-cell derivation, a labelled diagram with I₁, I₂ and V earns a step mark.
In graph questions, mark the intercept (ε) and the short-circuit current (ε/r) on the axes.
Board papers have set internal resistance and combinations of cells as 2-, 3- and 5-mark items (for example 2023 55/5/1 Q31(b), 2025 55/1/1 Q17 and 2026 55/1/1 Q31(b)).
When cells of different emfs are in parallel, check whether the weaker one is being charged.

Practice questions on EMF, Internal Resistance and 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
The terminal voltage of a cell equals its emf when
(a) the cell is short-circuited (b) no current is drawn from it (c) the external resistance equals the internal resistance (d) the cell is being charged
Hint
V = ε − Ir.
Answer
(b) no current is drawn from it.
Q2 · 2 marks · Very short answerBoard level
A cell of emf 2 V and internal resistance 0.1 Ω is connected to a 3.9 Ω resistor. Find the current and the terminal voltage.
Hint
I = ε/(R + r).
Answer
$I = \frac{2}{4} = 0.5$ A; $V = 2 - 0.5 \times 0.1 = 1.95$ V.
Q3 · 2 marks · Very short answerHarder board level
A cell sends 0.5 A through a 4 Ω resistor and 0.25 A through a 9 Ω resistor. Find its emf and internal resistance.
Hint
Write ε = I(R + r) twice.
Answer
$0.5 (4 + r) = 0.25 (9 + r)$ gives r = 1 Ω; $ε = 0.5 \times 5 = 2.5$ V.
Q4 · 1 mark · Very short answerHarder board level
Six identical cells, each of emf ε and internal resistance r, are joined in series, but one cell is connected the wrong way round. Find the emf and internal resistance of the battery.
Hint
A reversed cell cancels one correct cell's emf.
Answer
emf = 5ε − ε = 4ε; internal resistance = 6r (all resistances still add).
Q5 · 2 marks · Very short answerHarder board level
Two cells of emf 2 V and 1.5 V, each with internal resistance 0.5 Ω, are connected in parallel (positive to positive). Find the emf and internal resistance of the combination.
Hint
Use the parallel-cell formula.
Answer
$ε_{e q} = \frac{2 \times 0.5 + 1.5 \times 0.5}{1} = 1.75$ V; $r_{e q} = \frac{0.25}{1} = 0.25$ Ω.
Q6 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): The terminal voltage of a cell can be greater than its emf.
Reason (R): When a cell is being charged, current enters its positive terminal and V = ε + Ir.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Think of a battery charger.
Answer
(a).
Q7 · 3 marks · Short answerHOTS / competency
Two identical cells (emf ε, internal resistance r) are connected to an external resistance R, first in series and then in parallel. Show that the current through R is the same in both cases only if R = r.
Hint
Write both currents and equate.
Answer
Series: $I_{s} = \frac{2 ε}{R + 2 r}$ . Parallel: $I_{p} = \frac{ε}{R + r /2} = \frac{2 ε}{2 R + r}$ . Equal when $R + 2 r = 2 R + r$ , i.e. R = r. For R > r, series gives more current; for R < r, parallel does.

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 of a cell.Show answer
The potential difference between its terminals when no current is drawn from it.
2.Terminal voltage of a discharging cell?Show answer
V = ε − Ir.
3.What does the slope of a V-I graph of a cell give?Show answer
Minus the internal resistance.
4.Equivalent emf of two cells in parallel?Show answer
$\frac{ε _{1} r _{2} + ε _{2} r _{1}}{r _{1} + r _{2}}$ .
5.Internal resistance of m identical cells in parallel?Show answer
r/m.
6.Terminal voltage of a cell being charged?Show answer
V = ε + Ir.

EMF, Internal Resistance and Cells: frequently asked questions

Why does terminal voltage drop when more current is drawn?

The current also flows through the cell's internal resistance, using up Ir volts inside the cell. The larger the current, the more of the emf is lost inside, so V = ε − Ir falls.

Is emf the same as potential difference?

No. Emf is the terminal voltage when no current flows; it is a property of the cell. Potential difference between the terminals while current flows is the terminal voltage, which is less than the emf (more when charging).

When should cells be connected in parallel?

When the external resistance is small compared with the internal resistance, since parallel connection lowers the internal resistance and lets more current flow.

Can a cell be damaged by connecting it in parallel with a weaker cell?

The stronger cell drives current backwards through the weaker one, charging it and heating both. With very different emfs this wastes energy and can harm the cells.

How do you find internal resistance experimentally?

Measure the terminal voltage for two different currents (or loads). The fall in voltage divided by the rise in current gives r, and the V-I line extended to I = 0 gives ε.