Definition
EMF and internal resistance
EMF (ε) is the voltage of a source on open circuit. Terminal voltage when current I flows: V = ε - Ir, where r is internal resistance.
-- NCERT Class 12 Physics, Ch. 3, p. 94Internal Resistance EMF
8 MCQs1 revision card9-step worked example
Source: NCERT Current ElectricityPYQ coverage: NEET 2021, 2024Official key: NTA-verifiedLast updated: 21 Sep 2026
Internal Resistance EMF, explained for NEET
The common confusion: treating the emf of a cell as the voltage across its terminals at all times. It is not. Emf is the potential difference between the two electrodes when no current is drawn (open circuit). NCERT is explicit that emf is "a potential difference and not a force", despite the name (NCERT Class 12 Physics Part I, Chapter 3, page 94). It is the sum of the potential jumps at the two electrode–electrolyte boundaries, ε = V₊ + V₋ (Eq 3.36).
Internal resistance. The electrolyte between the electrodes has a finite resistance r. When the cell drives a current I through an external resistor R, part of the emf is used up inside the cell. The terminal voltage is
V = ε − Ir (NCERT Eq 3.38, page 94)
Since the same V appears across R, V = IR, and combining the two gives the current in the circuit: I = ε/(R + r).
Three cases worth separating:
- Open circuit (I = 0): V = ε. A voltmeter drawing negligible current across an isolated cell reads the emf.
- Cell supplying current: V < ε, and V falls as I rises. Shorting the terminals (R = 0) gives the largest current, ε/r, with V = 0.
- Current forced through the cell from its positive to its negative terminal (another source pushing current against the cell): V = ε + Ir, so the terminal voltage exceeds the emf (NCERT Eq 3.60, in the Kirchhoff's rules section).
Bridge to NEET: questions give two of ε, r, R, I, V and ask for the rest, or give two different external resistors with the resulting currents and ask for ε and r. Each reading gives one equation ε = I(R + r); two readings solve for both unknowns.
Watch out: check the direction of current through the cell before choosing the sign in front of Ir, and do not drop r from the denominator of ε/(R + r) unless the question states the cell is ideal.
Can you answer these Internal Resistance EMF MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
According to NCERT, the emf of a cell is best described as:
Show answer and why every option is right or wrong
Answer: A. Emf is the potential difference between the positive and negative electrodes in open circuit; NCERT stresses it is a potential difference and not a force (NCERT Class 12 Physics Part I, Chapter 3, page 94).
Why B is wrong: B is wrong because the name 'electromotive force' misleads students; NCERT states emf is a potential difference, not a force, and it is measured in volts.
Why C is wrong: C is wrong because this is the terminal voltage V = ε − Ir, which is less than the emf whenever current flows; it equals the emf only when I = 0.
Why D is wrong: D is wrong because the drop across the internal resistance is Ir, the part of the emf lost inside the cell, not the emf itself.
MCQ 2Easy RecallPractice
A cell of emf ε and internal resistance r supplies a current I to an external resistor. The potential difference across its terminals is:
Show answer and why every option is right or wrong
Answer: B. While the cell drives current through the external circuit, the drop Ir inside the cell is subtracted: V = ε − Ir (NCERT Class 12 Physics Part I, Chapter 3, Eq 3.38, page 94).
Why A is wrong: A is wrong because ε + Ir applies only when current is forced through the cell from its positive to its negative terminal, not when the cell is supplying current.
Why C is wrong: C is wrong because it ignores the internal resistance; V = ε only in open circuit (I = 0) or for an ideal cell.
Why D is wrong: D is wrong because it reverses the sign of the whole expression, which would make the terminal voltage negative for any ordinary current.
MCQ 3Easy RecallPractice
An external source forces a current I through a cell (emf ε, internal resistance r) in the direction from the cell's positive terminal to its negative terminal. The potential difference across the cell's terminals is:
Show answer and why every option is right or wrong
Answer: B. When current flows through the cell from the positive to the negative electrode, the drop across r adds to the emf: V = ε + Ir (NCERT Class 12 Physics Part I, Chapter 3, Eq 3.60, Kirchhoff's rules section).
Why A is wrong: A is wrong because ε − Ir is the terminal voltage when the cell itself supplies current (current leaves from the positive terminal); here the current direction through the cell is reversed.
Why C is wrong: C is wrong because Ir is only the potential drop across the internal resistance; the emf of the cell must still be included.
Why D is wrong: D is wrong because it ignores the internal resistance; with current flowing, the terminal voltage differs from ε by Ir.
MCQ 4Direct ApplicationPractice
A cell of emf 12 V and internal resistance 0.5 Ω supplies a current of 2 A to an external circuit. The terminal voltage of the cell is:
Show answer and why every option is right or wrong
Answer: D. V = ε − Ir = 12 − (2 × 0.5) = 12 − 1 = 11 V (NCERT Class 12 Physics Part I, Chapter 3, Eq 3.38, page 94).
Why A is wrong: A is wrong because 1 V is only the drop Ir across the internal resistance, not the terminal voltage.
Why B is wrong: B is wrong because 13 V comes from adding Ir (V = ε + Ir), the sign that applies only when current is forced through the cell from positive to negative terminal.
Why C is wrong: C is wrong because 12 V is the emf; it ignores the 1 V lost across the internal resistance while current flows.
MCQ 5Direct ApplicationPractice
A cell of emf 6 V and internal resistance 1 Ω is connected across a 5 Ω resistor. The current in the circuit is:
Show answer and why every option is right or wrong
Answer: D. I = ε/(R + r) = 6/(5 + 1) = 1 A. This follows from V = ε − Ir together with V = IR across the external resistor (NCERT Class 12 Physics Part I, Chapter 3, page 94).
Why A is wrong: A is wrong because 1.2 A = 6/5 uses only the external resistance and ignores the internal resistance of the cell.
Why B is wrong: B is wrong because 6 A = 6/1 is the short-circuit current ε/r, which uses only the internal resistance and ignores the 5 Ω load.
Why C is wrong: C is wrong because 1.5 A = 6/(5 − 1) subtracts the internal resistance from R instead of adding it; the two resistances are in the same current path.
MCQ 6Direct ApplicationPractice
A cell has emf 2 V and internal resistance 0.4 Ω. Its terminals are joined by a wire of negligible resistance. The current drawn from the cell is:
Show answer and why every option is right or wrong
Answer: C. With R = 0, I = ε/(R + r) = ε/r = 2/0.4 = 5 A, and the terminal voltage V = ε − Ir = 2 − 5 × 0.4 = 0 (NCERT Class 12 Physics Part I, Chapter 3, page 94).
Why A is wrong: A is wrong because 0.8 A = 2 × 0.4 multiplies emf by resistance instead of dividing emf by resistance.
Why B is wrong: B is wrong because 2 A treats the current as numerically equal to the emf, as if the resistance in the circuit were 1 Ω.
Why D is wrong: D is wrong because it treats the cell as ideal; the internal resistance limits the current to ε/r even when the external resistance is zero.
MCQ 7CalculationPractice
A cell of emf 1.5 V is connected across a 2.5 Ω resistor. A voltmeter across the cell's terminals reads 1.25 V. The internal resistance of the cell is:
Show answer and why every option is right or wrong
Answer: C. Step 1: current through the resistor, I = V/R = 1.25/2.5 = 0.5 A. Step 2: V = ε − Ir gives Ir = 1.5 − 1.25 = 0.25 V, so r = 0.25/0.5 = 0.5 Ω (NCERT Class 12 Physics Part I, Chapter 3, Eq 3.38, page 94).
Why A is wrong: A is wrong because 3.0 Ω = ε/I = 1.5/0.5 is the total resistance R + r; the 2.5 Ω external resistance was not subtracted.
Why B is wrong: B is wrong because 0.42 Ω comes from finding the current as ε/R = 1.5/2.5 = 0.6 A, using the emf instead of the terminal voltage across the resistor, then 0.25/0.6.
Why D is wrong: D is wrong because 12.5 Ω = V × R/(ε − V) = 1.25 × 2.5/0.25 inverts the ratio; the correct relation is r = R(ε − V)/V.
MCQ 8CalculationPractice
When a cell is connected across a 2 Ω resistor the current is 1.0 A; when the same cell is connected across a 5 Ω resistor the current is 0.50 A. The emf and internal resistance of the cell are:
Show answer and why every option is right or wrong
Answer: A. Each reading gives ε = I(R + r): ε = 1.0(2 + r) and ε = 0.50(5 + r). Equating, 2 + r = 2.5 + 0.5r, so 0.5r = 0.5 and r = 1 Ω; then ε = 1.0 × (2 + 1) = 3 V. Check: 0.50 × (5 + 1) = 3 V (NCERT Class 12 Physics Part I, Chapter 3, page 94).
Why B is wrong: B is wrong because it takes the terminal voltage in the first reading (1.0 A × 2 Ω = 2 V) as the emf, ignoring internal resistance; it is inconsistent with the second reading, which gives 2.5 V.
Why C is wrong: C is wrong because it takes the terminal voltage in the second reading (0.50 A × 5 Ω = 2.5 V) as the emf, ignoring internal resistance; it is inconsistent with the first reading.
Why D is wrong: D is wrong because r = 6 Ω comes from dividing the change in external resistance (3 Ω) by the change in current (0.50 A), which is not a resistance; with r = 6 Ω the two readings give 8 V and 5.5 V, which do not agree.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Internal Resistance EMF: quick recall before you leave
How do you solve a Internal Resistance EMF question? A worked example
- 1
Given
A cell of emf ε = 6.0 V and internal resistance r = 0.50 Ω is connected across an external resistor R = 2.5 Ω.
- 2
Required
The current in the circuit and the terminal voltage of the cell.
- 3
Concept
The internal resistance and the external resistor carry the same current. Part of the emf (Ir) is lost inside the cell; the remainder appears across the terminals and across R.
- 4
Formula
V = ε − Ir and V = IR, which combine to I = ε/(R + r).
- 5
Substitution
I = 6.0/(2.5 + 0.50); V = 6.0 − I × 0.50.
- 6
Calculation
R + r = 3.0 Ω, so I = 6.0/3.0 = 2.0 A.
Ir = 2.0 × 0.50 = 1.0 V, so V = 6.0 − 1.0 = 5.0 V.
Check: IR = 2.0 × 2.5 = 5.0 V, matching the terminal voltage. No exact constants enter this calculation; every value is a given measurement, so the answers keep two significant figures. - 7
Final answer
I = 2.0 A; terminal voltage V = 5.0 V (less than the emf of 6.0 V).
- 8
Common trap
Writing I = ε/R = 6.0/2.5 = 2.4 A drops the internal resistance, and quoting V = 6.0 V treats the emf as the terminal voltage. Both errors come from treating the cell as ideal.
- 9
Similar NEET-style question
"A cell of emf 9.0 V and internal resistance 1.0 Ω is connected across an 8.0 Ω resistor. Find the terminal voltage."
Strategy: I = 9.0/(8.0 + 1.0) = 1.0 A; V = 9.0 − 1.0 × 1.0 = 8.0 V.
---
What to remember before solving Internal Resistance EMF questions
Which Internal Resistance EMF formulas do you need for NEET?
1 formula — click to collapse
EMF and terminal voltage
Terminal voltage of cell with internal resistance r when current I flows.
| Symbol | Quantity | SI Unit |
|---|---|---|
| epsilon | EMF | V |
| r | internal resistance | Ω |
| I | current | A |
| V_t | terminal voltage | V |
Valid when
- Steady current
- Cell with constant EMF
More in Current Electricity: 3 exam traps and mistakes · 6 formulas · 6 question patterns from its other lessons.
Internal Resistance EMF questions from past NEET papers
2 questions from NEET 2021, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
14 V
26 V
38 V
410 V
NTA Answer: Option 3(revised_final)
NEET 2021
162 cm
260 cm
321.6 cm
464 cm
NTA Answer: Option 2(final)
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →Prefer to watch this?
8 real NEET Current Electricity questions (2022–2025) on animated circuits: watch the current flow (15 min)
Watch on YouTube