PhysicsICSE Class 10 · Ohm's Law and Resistances

EMF, Terminal Voltage and Internal Resistance of a Cell (ICSE Class 10)

Q: What is the difference between emf and terminal voltage?

Emf is the pd across a cell's terminals when no current flows. Terminal voltage is the pd across them when current flows; it is less than the emf by Ir, the voltage lost across the internal resistance.

Q: What is the internal resistance of a cell?

It is the resistance offered by the electrolyte and electrodes inside the cell to the flow of current. It depends on the distance between and area of the electrodes, the electrolyte and the temperature.

Q: Why is terminal voltage less than emf?

When current flows, part of the energy supplied by the cell is used inside it to drive the current through its internal resistance, so less pd is available across its terminals.

Updated 15 October 2026 · Fact-checked

The emf ε of a cell is the pd across its terminals when no current is drawn. When current I flows, part of the emf is used inside the cell across its internal resistance r, so the terminal voltage is V = ε − Ir. Hence I = ε/(R + r), and V equals ε only in an open circuit.

Understand EMF, Terminal Voltage and Internal Resistance

Electromotive force (emf). A cell converts chemical energy into electrical energy. The emf $ε$ of a cell is the energy it supplies to each coulomb of charge passing through it; it equals the pd across its terminals when no current is drawn (open circuit). Unit: volt (V = J C⁻¹). Despite its name, emf is not a force.

Internal resistance. Current must pass through the electrolyte and electrodes inside the cell, which offer some resistance. This is the internal resistance $r$ of the cell. It depends on the cell itself, for example:

the distance between the electrodes (greater distance, greater $r$ );
the area of the electrodes dipping in the electrolyte (larger area, smaller $r$ );
the nature of the electrolyte;
the temperature (a warmer electrolyte usually has a smaller $r$ ).

As a cell ages, its internal resistance increases.

Terminal voltage. When the cell drives a current $I$ through an external resistance $R$ (closed circuit), the pd across its terminals is the terminal voltage $V$ . Some energy per coulomb is used up inside the cell, across $r$ ; this is the voltage drop or 'lost volts', $I r$ . So

$ε = V + I r, V = ε - I r, V = I R .$

Combining, the current in the circuit is

$I = \frac{ε}{R + r} .$

Open and closed circuits.

Situation	Current	Terminal voltage
Open circuit (no current)	$I = 0$	$V = ε$
Closed circuit	$I = \frac{ε}{R + r}$	$V = ε - I r < ε$
Short circuit ( $R \approx 0$ )	$I = \frac{ε}{r}$ (maximum)	$V \approx 0$

The larger the current drawn, the larger the lost volts and the smaller the terminal voltage. That is why car headlights dim for a moment when the starter motor draws a big current.

EMF and terminal voltage compared.

EMF ε	Terminal voltage V
pd across the terminals in open circuit	pd across the terminals in closed circuit
Does not depend on the external resistance	Depends on the current drawn (on R)
Always greater than V when current flows	Less than ε when current flows
Energy given per coulomb by the cell	Energy per coulomb delivered to the external circuit

Measuring them. A high-resistance voltmeter connected across a cell with the circuit open reads (practically) the emf. With the cell supplying current, the same voltmeter reads the terminal voltage. Two readings of $V$ and $I$ (with two different external resistances) give two equations $ε = V_{1} + I_{1} r = V_{2} + I_{2} r$ , from which both $ε$ and $r$ can be found.

Cells in series. If identical cells are joined in series (+ to −), their emfs add and so do their internal resistances: $n$ cells of emf $ε$ and internal resistance $r$ give $n ε$ and $n r$ .

EMF, Terminal Voltage and Internal Resistance: formulas and key facts

EMF and terminal voltage: $ε = V + I r$; Ir is the lost volts inside the cell.
Terminal voltage: $V = ε - I r = I R$; Minus, not plus, when the cell supplies current.
Current in the circuit: $I = \frac{ε}{R + r}$; Add the internal resistance to the external resistance.
Internal resistance from readings: $r = \frac{ε - V}{I}$; Or from two (V, I) pairs.
Three circuit states: Open circuit no current, V equals ε
Closed circuit V is less than ε by Ir
Short circuit largest current ε/r, V nearly zero; The closed-circuit pd across the terminals is the terminal voltage.

Open circuit	no current, V equals ε
Closed circuit	V is less than ε by Ir
Short circuit	largest current ε/r, V nearly zero

How to solve EMF, Terminal Voltage and Internal Resistance questions

Write the loop equation for the whole circuit, then use it with Ohm's law for the external part.

1Draw the cell as an ideal emf ε in series with a small resistor r.
2Find the total resistance $R + r$ (reduce the external network to one R first).
3Current: $I = \frac{ε}{R + r}$ .
4Terminal voltage: $V = I R$ , and check with $V = ε - I r$ ; lost volts $= I r$ .
5With two sets of readings, write $ε = V_{1} + I_{1} r$ and $ε = V_{2} + I_{2} r$ and solve the pair for $r$ and $ε$ .

EMF, Terminal Voltage and Internal Resistance 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.

A torch with old cells

Suppose a torch uses two 1.5 V cells in series, each with internal resistance 0.5 Ω, and a bulb of resistance 5 Ω. After months of use, each cell's internal resistance has risen to 1.5 Ω while its emf is still about 1.5 V.

Compare the current and the pd across the bulb when the cells are new and when they are old.

Try it first, then show the solutionSolution

New: $ε = 3$ V, $r = 1$ Ω; $I = \frac{3}{5 + 1} = 0.5$ A; $V = 0.5 \times 5 = 2.5$ V.
Old: $r = 3$ Ω; $I = \frac{3}{5 + 3} = 0.375$ A; $V = 0.375 \times 5 = 1.875$ V.

Answer: Current falls from 0.5 A to 0.375 A and the bulb's pd from 2.5 V to 1.875 V, so the torch is dimmer.

Memory hook: Old cells hide their losses inside.

Dimming headlights

Imagine a car's headlights are on, and the driver turns the key to start the engine. The headlights dim for a moment.

Explain using terminal voltage.

Try it first, then show the solutionSolution

The starter motor draws a very large current from the battery.
The lost volts $I r$ inside the battery become large, so the terminal voltage $V = ε - I r$ falls, and the headlights get a smaller pd.

Answer: The large current raises the lost volts, lowering the terminal voltage across the headlights.

Memory hook: Big current, big lost volts, dim lights.

Tricks to remember EMF, Terminal Voltage and Internal Resistance

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

EMF = Volts out + Volts lost
ε = V + Ir: 'Everything Made = Value delivered + Inside rent'.
Use it when: Writing the relation with the right sign.
Open means equal
Open circuit, no current, nothing lost: V = ε.
Use it when: Fill-in and MCQ items.
Add r to R
The cell's own resistance always joins the external resistance in series.
Use it when: Finding the current.
Two readings, two unknowns
Two (V, I) pairs give two equations for ε and r.
Use it when: Finding emf and internal resistance.

EMF, Terminal Voltage and Internal Resistance: solved examples

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

Example 1: A voltmeter across an idle cell · 1 mark · MCQBoard level

A voltmeter of very high resistance is connected directly across a cell that is not supplying any other circuit. It reads:
(a) zero
(b) the emf of the cell
(c) less than the emf
(d) more than the emf

Try it first, then show the solutionSolution

Practically no current flows, so there are no lost volts and the reading is the emf.

Answer: (b) the emf of the cell

Example 2: Defining the three terms · 3 marks · Short answerBoard level

Define (a) emf of a cell, (b) terminal voltage, (c) internal resistance. (d) Write the relation between emf and terminal voltage.

Try it first, then show the solutionSolution

(a) The pd across the terminals of a cell when no current is drawn from it (the energy supplied per unit charge). (1)
(b) The pd across the terminals when the cell supplies current in a closed circuit; (c) the resistance offered by the electrolyte and electrodes inside the cell. (1)
(d) $ε = V + I r$ , i.e. $V = ε - I r$ . (1)

Answer: (a) Open-circuit pd (b) Closed-circuit pd (c) Resistance inside the cell (d) $ε = V + I r$

Example 3: Current, terminal voltage and lost volts · 3 marks · Short answerHarder board level

A dry cell with an emf of 2 V and an internal resistance of 0.4 Ω lights a small lamp of resistance 4.6 Ω. Find (a) the current, (b) the terminal voltage, (c) the lost volts.

Try it first, then show the solutionSolution

(a) $I = \frac{2}{4.6 + 0.4} = 0.4$ A. (1)
(b) $V = I R = 0.4 \times 4.6 = 1.84$ V. (1)
(c) Lost volts $= I r = 0.4 \times 0.4 = 0.16$ V (check: $1.84 + 0.16 = 2$ V). (1)

Answer: (a) 0.4 A (b) 1.84 V (c) 0.16 V

Example 4: Finding internal resistance · 2 marks · Short answerHarder board level

A cell of emf 1.5 V delivers a current of 0.3 A to an external resistor, and its terminal voltage is then 1.35 V. Find (a) the internal resistance, (b) the external resistance.

Try it first, then show the solutionSolution

(a) $r = \frac{ε - V}{I} = \frac{1.5 - 1.35}{0.3} = 0.5$ Ω. (1)
(b) $R = \frac{V}{I} = \frac{1.35}{0.3} = 4.5$ Ω. (1)

Answer: (a) 0.5 Ω (b) 4.5 Ω

Example 5: EMF and r from two readings · 3 marks · Short answerHOTS / competency

Across an 8 Ω resistor a cell maintains a pd of 2.4 V; across a 4 Ω resistor the pd drops to 2.0 V. Find its emf and internal resistance.

Try it first, then show the solutionSolution

Currents: $I_{1} = \frac{2.4}{8} = 0.3$ A; $I_{2} = \frac{2.0}{4} = 0.5$ A. (1)
$ε = 2.4 + 0.3 r$ and $ε = 2.0 + 0.5 r$ ; subtracting, $0.4 = 0.2 r$ , so $r = 2$ Ω. (1)
$ε = 2.4 + 0.3 \times 2 = 3.0$ V. (1)

Answer: ε = 3.0 V, r = 2 Ω

Example 6: Short-circuiting a cell · 3 marks · Short answerHarder board level

A cell of emf 1.5 V has an internal resistance of 0.25 Ω. (a) Find the current if its terminals are joined by a thick copper wire of negligible resistance. (b) What is the terminal voltage then? (c) Why should a cell never be short-circuited?

Try it first, then show the solutionSolution

(a) $I = \frac{ε}{r} = \frac{1.5}{0.25} = 6$ A. (1)
(b) Nearly zero: all the emf is lost inside the cell. (1)
(c) The very large current heats the cell and wire and drains the cell quickly; it can damage the cell. (1)

Answer: (a) 6 A (b) About 0 V (c) Huge current heats and damages the cell

Example 7: Case study: a cell and a parallel load · 4 marks · Case-basedHOTS / competency

A 6 V battery whose internal resistance is 0.5 Ω lights two 5 Ω lamps joined in parallel.
(a) Find the external resistance.
(b) Find the current from the battery.
(c) Find the terminal voltage and the current in each lamp.
(d) One lamp fuses. Without calculating, say whether the terminal voltage rises or falls, and why.

Try it first, then show the solutionSolution

(a) $R = \frac{5}{2} = 2.5$ Ω. (1)
(b) $I = \frac{6}{2.5 + 0.5} = 2$ A. (1)
(c) $V = 2 \times 2.5 = 5$ V (check: $6 - 2 \times 0.5 = 5$ V); each lamp $\frac{5}{5} = 1$ A. (1)
(d) It rises: the external resistance increases to 5 Ω, the current falls, so the lost volts $I r$ fall and $V = ε - I r$ rises. (1)

Answer: (a) 2.5 Ω (b) 2 A (c) 5 V; 1 A each (d) Rises, since less current means fewer lost volts

Where students lose marks in EMF, Terminal Voltage and Internal Resistance

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

Mistake: Answering 'emf' for the pd across a cell in a closed circuit
In a closed circuit current flows, so the pd across the terminals is the terminal voltage.
Wrong working
'The pd between the terminals of a cell in a closed circuit is its emf.'
Right working
'It is the terminal voltage.'
Do this: Underline 'open' or 'closed' in the question.
Usually costs: Reported by CISCE examiners for 2024 Q1(x).
Mistake: Using V = ε + Ir
A cell supplying current loses Ir inside, so the terminal voltage is smaller.
Wrong working
$V = ε + I r$ .
Right working
$V = ε - I r$ .
Do this: Derive it: ε is shared between the outside (V) and the inside (Ir).
Usually costs: Reported by CISCE examiners for 2025 Q1(ix).
Mistake: Thinking emf and terminal voltage differ in open circuit
With no current there are no lost volts.
Wrong working
'Emf is greater than the terminal voltage in open circuit.'
Right working
'In open circuit they are equal.'
Do this: Put I = 0 in V = ε − Ir.
Usually costs: Reported by CISCE examiners for 2025 Q2(i)(d).
Mistake: Leaving out r when finding the current
The internal resistance is in series with the external circuit.
Wrong working
$I = \frac{ε}{R}$ for a cell with given r.
Right working
$I = \frac{ε}{R + r}$ .
Do this: Draw r inside the cell symbol.
Usually costs: The method mark.

Exam tips

Use the exact terms emf, terminal voltage and voltage drop (lost volts); examiners treat them as three different pds.
When both V and I are known, $r = \frac{ε - V}{I}$ is the quickest route.
Check every answer with $ε = V + I r$ .
In 'what happens if' questions, reason through the current: more current means more lost volts and a smaller terminal voltage.

Practice questions on EMF, Terminal Voltage and Internal Resistance

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 'lost volts' inside a cell supplying current I is equal to:
(a) $ε r$
(b) $I r$
(c) $I R$
(d) $\frac{ε}{R}$
Hint
Ohm's law for the inside of the cell.
Answer
(b) $I r$
Q2 · 1 mark · Very short answerBoard level
A 12 V battery of internal resistance 0.2 Ω supplies 2 A. Find its terminal voltage.
Hint
$V = ε - I r$ .
Answer
$V = 12 - 2 \times 0.2 = 11.6$ V.
Q3 · 2 marks · Short answerHarder board level
A 9 V battery drives 0.5 A through a 17 Ω resistor. Find the internal resistance of the battery.
Hint
$R + r = \frac{ε}{I}$ .
Answer
$R + r = \frac{9}{0.5} = 18$ Ω, so $r = 1$ Ω.
Q4 · 2 marks · Short answerBoard level
State two factors on which the internal resistance of a cell depends, and how.
Hint
Think of the path through the electrolyte.
Answer
Any two: distance between the electrodes (greater distance, greater $r$ ); area of the electrodes in the electrolyte (larger area, smaller $r$ ); nature of the electrolyte; temperature (higher temperature, usually smaller $r$ ).
Q5 · 2 marks · Short answerHarder board level
A cell of emf 2 V gives a pd of 1.6 V across a 4 Ω resistor. Find the current and the internal resistance.
Hint
Current from the external part first.
Answer
$I = \frac{1.6}{4} = 0.4$ A; $r = \frac{2 - 1.6}{0.4} = 1$ Ω.
Q6 · 2 marks · Short answerHarder board level
Give two differences between emf and terminal voltage.
Hint
Open vs closed; dependence on R.
Answer
(1) Emf is the pd across the terminals in open circuit; terminal voltage is the pd in closed circuit. (2) Emf does not depend on the external resistance; terminal voltage depends on the current drawn and is less than the emf when current flows.
Q7 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The terminal voltage of a cell in a closed circuit is less than its emf.
Reason (R): Part of the emf is used to drive the current through the internal resistance of the cell.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Lost volts.
Answer
(c) Both are true, and (R) explains (A).
Q8 · 3 marks · Short answerHOTS / competency
A battery has emf 4.5 V and internal resistance 0.3 Ω. (a) What is the largest current it can supply? (b) What external resistance makes its terminal voltage 4.2 V?
Hint
Largest current: R = 0.
Answer
(a) $I_{m a x} = \frac{4.5}{0.3} = 15$ A. (b) Lost volts $= 0.3$ V, so $I = \frac{0.3}{0.3} = 1$ A; $R = \frac{4.2}{1} = 4.2$ Ω.
Q9 · 2 marks · Short answerHarder board level
Why must a car battery, which supplies a large current to the starter motor, have a very small internal resistance?
Hint
Lost volts at large current.
Answer
With a large current, even a moderate $r$ would make the lost volts $I r$ large, leaving little terminal voltage for the motor and wasting energy as heat inside the battery. A very small $r$ keeps the terminal voltage close to the emf.

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.Relation between ε, V, I and r?Show answer
$ε = V + I r$ .
2.Terminal voltage in open circuit?Show answer
Equal to the emf.
3.Current in a circuit with internal resistance?Show answer
$I = \frac{ε}{R + r}$ .
4.What are lost volts?Show answer
$I r$ , the pd across the internal resistance.
5.Effect of a larger electrode area on r?Show answer
It decreases r.

EMF, Terminal Voltage and Internal Resistance: frequently asked questions

What is the difference between emf and terminal voltage?

Emf is the pd across a cell's terminals when no current flows. Terminal voltage is the pd across them when current flows; it is less than the emf by Ir, the voltage lost across the internal resistance.

What is the internal resistance of a cell?

It is the resistance offered by the electrolyte and electrodes inside the cell to the flow of current. It depends on the distance between and area of the electrodes, the electrolyte and the temperature.

Why is terminal voltage less than emf?

When current flows, part of the energy supplied by the cell is used inside it to drive the current through its internal resistance, so less pd is available across its terminals.