Like resistors, cells can be combined together in an electric circuit. And like resistors, one can, for calculating currents and voltages in a circuit, replace a combination of cells by an equivalent cell.
-- NCERT Class 12 Physics, Ch. 3, p. 95Cells Series Parallel
8 MCQs9-step worked example
Source: NCERT Current ElectricityOfficial key: NTA-verifiedLast updated: 7 Oct 2026
Try this first
Cells are joined in series so that the current leaves each cell from its positive electrode. What are the equivalent emf and equivalent internal resistance?
- A.The mean of the emfs and the mean of the internal resistances
- B.The emf of one cell and the sum of the internal resistances
- C.The sum of the emfs and the sum of the internal resistances
- D.The sum of the emfs and the parallel combination of the internal resistances
Tap to see the answer
Answer: C. C is correct: for cells in series, ε_eq is the sum of the individual emfs and r_eq is the sum of the internal resistances (NCERT Class 12 Physics, Chapter 3, pages 95–96).
A is wrong: A is wrong because nothing in the series rule averages; both quantities are sums.
B is wrong: B is wrong because the emfs of the cells in series do add; only the internal resistance part is correct.
D is wrong: D is wrong because the parallel formula for r belongs to the parallel arrangement; in series the internal resistances simply add.
Cells Series Parallel, explained for NEET
The question first: three identical cells, each 1.5 V with internal resistance 0.50 Ω, are joined first in series and then in parallel. What single cell replaces each arrangement? Series: 4.5 V and 1.5 Ω. Parallel: 1.5 V and 0.17 Ω. Adding the emfs in the parallel case is the trap.
The concept. Like resistors, cells can be combined, and for calculating currents and voltages a combination can be replaced by one equivalent cell of emf ε_eq and internal resistance r_eq (NCERT Class 12 Physics, Chapter 3, page 95).
- Series. ε_eq = ε₁ + ε₂ and r_eq = r₁ + r₂ (page 95). For n cells the emfs add and the internal resistances add, provided the current leaves each cell from its positive electrode. If it leaves a cell from the negative electrode, that cell's emf enters with a negative sign, so two cells joined negative to negative give ε₁ − ε₂ for ε₁ > ε₂ (page 96).
- Parallel. The currents leaving the positive electrodes add at the junction, I = I₁ + I₂ (page 96). The equivalent cell satisfies 1/r_eq = 1/r₁ + 1/r₂ and ε_eq/r_eq = ε₁/r₁ + ε₂/r₂ (Eqs. 3.56 and 3.57, page 97), and these extend to n cells (page 97). Equivalently, ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) and r_eq = r₁r₂/(r₁ + r₂) (page 96).
Bridge to NEET. Reduce the cells to ε_eq and r_eq, then I = ε_eq/(R + r_eq). The reduced circuit is the one question the paper asks.
Watch-out: parallel combination of identical cells leaves the emf unchanged and divides r by n. It helps with the current delivered to a low-resistance load, not with the voltage. No past-paper item in the corpus is tagged to this subtopic.
How do you solve a Cells Series Parallel question? A worked example
- 1
Given
Cell X has emf 9.0 V and internal resistance 3.0 Ω. Cell Y has emf 6.0 V and internal resistance 2.0 Ω. They are joined in parallel, positive to positive, across an external resistor of 4.8 Ω.
- 2
Required
The current through the 4.8 Ω resistor and the voltage across it.
- 3
Concept
Replace the two cells by one equivalent cell (ε_eq, r_eq) using the parallel rule on NCERT Class 12 Physics, Chapter 3, page 96, then treat it as a single cell driving the resistor.
- 4
Formula
ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂); r_eq = r₁r₂/(r₁ + r₂); I = ε_eq/(R + r_eq).
- 5
Substitution
ε_eq = (9.0 × 2.0 + 6.0 × 3.0) ÷ (3.0 + 2.0). r_eq = (3.0 × 2.0) ÷ (3.0 + 2.0). I = ε_eq ÷ (4.8 + r_eq).
- 6
Calculation
ε_eq = (18 + 18) ÷ 5.0 = 7.2 V. r_eq = 6.0 ÷ 5.0 = 1.2 Ω. I = 7.2 ÷ (4.8 + 1.2) = 7.2 ÷ 6.0 = 1.2 A. Voltage across R = 1.2 × 4.8 = 5.76 V.
Check: cell X delivers (9.0 − 5.76) ÷ 3.0 = 1.08 A and cell Y delivers (6.0 − 5.76) ÷ 2.0 = 0.12 A, and 1.08 + 0.12 = 1.20 A. All data are measured values; there are no exact constants to adjust. - 7
Final answer
I = 1.2 A and the voltage across the 4.8 Ω resistor is 5.8 V.
- 8
Common trap
Adding the emfs in parallel (9.0 + 6.0 = 15 V) or averaging them (7.5 V). The weighted rule gives 7.2 V because the cell with the larger internal resistance counts for less.
- 9
Similar NEET-style question
Two identical cells, each 4.0 V with internal resistance 1.0 Ω, are joined in parallel across a 1.5 Ω resistor. Find the current. (Answer: ε_eq = 4.0 V, r_eq = 1.0 ÷ 2 = 0.50 Ω, so I = 4.0 ÷ (1.5 + 0.50) = 4.0 ÷ 2.0 = 2.0 A.)
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Can you answer these Cells Series Parallel 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
Cells are joined in series so that the current leaves each cell from its positive electrode. What are the equivalent emf and equivalent internal resistance?
Show answer and why every option is right or wrong
Answer: C. C is correct: for cells in series, ε_eq is the sum of the individual emfs and r_eq is the sum of the internal resistances (NCERT Class 12 Physics, Chapter 3, pages 95–96).
Why A is wrong: A is wrong because nothing in the series rule averages; both quantities are sums.
Why B is wrong: B is wrong because the emfs of the cells in series do add; only the internal resistance part is correct.
Why D is wrong: D is wrong because the parallel formula for r belongs to the parallel arrangement; in series the internal resistances simply add.
MCQ 2Easy RecallPractice
Two cells of emf ε₁ and ε₂ (ε₁ > ε₂) are joined in series with their negative terminals connected together. What is the equivalent emf?
Show answer and why every option is right or wrong
Answer: A. A is correct: when the current leaves a cell from its negative electrode, that cell's emf enters the sum with a negative sign, giving ε_eq = ε₁ − ε₂ for ε₁ > ε₂ (NCERT Class 12 Physics, Chapter 3, page 96).
Why B is wrong: B is wrong because ε₁ + ε₂ is the result when the cells aid each other, positive to negative.
Why C is wrong: C is wrong because that is a product over sum, the shape of the parallel resistance formula, not an emf rule.
Why D is wrong: D is wrong because the emf would cancel only if ε₁ = ε₂; for ε₁ > ε₂ there is a net emf ε₁ − ε₂.
MCQ 3Easy RecallPractice
Two cells are connected in parallel with their positive terminals joined together. How is the equivalent internal resistance r_eq related to r₁ and r₂?
Show answer and why every option is right or wrong
Answer: D. D is correct: for cells in parallel, 1/r_eq = 1/r₁ + 1/r₂ (Eq. 3.56), the same as r_eq = r₁r₂/(r₁ + r₂) (NCERT Class 12 Physics, Chapter 3, pages 96–97).
Why A is wrong: A is wrong because adding internal resistances is the series rule.
Why B is wrong: B is wrong because an average lies between r₁ and r₂, whereas for parallel cells r_eq is smaller than either.
Why C is wrong: C is wrong because the internal resistances of cells never subtract in either arrangement.
MCQ 4Direct ApplicationPractice
Four identical cells, each of emf 2.0 V and internal resistance 0.40 Ω, are joined in series across an external resistor of 2.4 Ω. What is the current?
Show answer and why every option is right or wrong
Answer: B. B is correct: ε_eq = 4 × 2.0 = 8.0 V and r_eq = 4 × 0.40 = 1.6 Ω, so I = 8.0 ÷ (1.6 + 2.4) = 2.0 A (NCERT Class 12 Physics, Chapter 3, pages 95–96).
Why A is wrong: A is wrong because 2.0 ÷ (0.40 + 2.4) = 0.71 A is the current for a single cell; the other three cells were ignored.
Why C is wrong: C is wrong because 8.0 ÷ 2.4 = 3.3 A leaves out the internal resistance 1.6 Ω of the combination.
Why D is wrong: D is wrong because 8.0 ÷ 1.6 = 5.0 A leaves out the external resistor of 2.4 Ω.
MCQ 5Direct ApplicationPractice
Three identical cells, each of emf 1.5 V and internal resistance 0.60 Ω, are joined in parallel. What are ε_eq and r_eq of the combination?
Show answer and why every option is right or wrong
Answer: D. D is correct: ε_eq/r_eq = 3 × (1.5 ÷ 0.60) = 7.5 and 1/r_eq = 3 ÷ 0.60 = 5.0 per ohm, so r_eq = 0.20 Ω and ε_eq = 7.5 × 0.20 = 1.5 V (NCERT Class 12 Physics, Chapter 3, page 97).
Why A is wrong: A is wrong because it adds the emfs, which is the series rule; parallel identical cells keep the emf of one cell.
Why B is wrong: B is wrong because the internal resistance 1.8 Ω is the series sum 3 × 0.60; in parallel it falls to 0.20 Ω.
Why C is wrong: C is wrong because both quantities are taken from the series rule.
MCQ 6Direct ApplicationPractice
Cell 1 has ε₁ = 6.0 V and r₁ = 2.0 Ω. Cell 2 has ε₂ = 3.0 V and r₂ = 1.0 Ω. They are joined in parallel, positive to positive. What are ε_eq and r_eq?
Show answer and why every option is right or wrong
Answer: B. B is correct: ε_eq = (ε₁r₂ + ε₂r₁)/(r₁ + r₂) = (6.0 × 1.0 + 3.0 × 2.0) ÷ 3.0 = 4.0 V and r_eq = r₁r₂/(r₁ + r₂) = 2.0 ÷ 3.0 = 0.67 Ω (NCERT Class 12 Physics, Chapter 3, page 96).
Why A is wrong: A is wrong because 9.0 V is the sum of the emfs, which applies in series, not in parallel.
Why C is wrong: C is wrong because 4.5 V is the plain mean of 6.0 and 3.0, and 3.0 Ω is the series sum of the resistances; the parallel rule weights each emf by the other cell's resistance.
Why D is wrong: D is wrong because the emf is right but 3.0 Ω is r₁ + r₂; in parallel r_eq = r₁r₂/(r₁ + r₂) = 0.67 Ω.
MCQ 7CalculationPractice
Three cells of emf 3.0 V, 3.0 V and 2.0 V have internal resistances 0.30 Ω, 0.30 Ω and 0.40 Ω. They are joined in series, but the 2.0 V cell is connected the wrong way round (its current leaves from the negative electrode). The combination drives a 3.0 Ω resistor. What is the current?
Show answer and why every option is right or wrong
Answer: A. A is correct: ε_eq = 3.0 + 3.0 − 2.0 = 4.0 V and r_eq = 0.30 + 0.30 + 0.40 = 1.0 Ω, so I = 4.0 ÷ (1.0 + 3.0) = 1.0 A. The reversed cell's emf is negative but its internal resistance still adds (NCERT Class 12 Physics, Chapter 3, page 96).
Why B is wrong: B is wrong because 8.0 ÷ (1.0 + 3.0) = 2.0 A counts the reversed cell as aiding the others.
Why C is wrong: C is wrong because 4.0 ÷ 1.0 = 4.0 A leaves out the external resistor of 3.0 Ω.
Why D is wrong: D is wrong because 4.0 ÷ 3.0 = 1.3 A leaves out the internal resistance of 1.0 Ω.
MCQ 8CalculationPractice
Six identical cells, each of emf 2.0 V and internal resistance 1.2 Ω, are arranged as two parallel rows of three cells in series. The arrangement drives an external resistor of 4.2 Ω. What is the current?
Show answer and why every option is right or wrong
Answer: C. C is correct: each row has ε = 3 × 2.0 = 6.0 V and r = 3 × 1.2 = 3.6 Ω. Two identical rows in parallel give ε_eq = 6.0 V and r_eq = 3.6 ÷ 2 = 1.8 Ω, so I = 6.0 ÷ (1.8 + 4.2) = 1.0 A (NCERT Class 12 Physics, Chapter 3, pages 95–97).
Why A is wrong: A is wrong because 6.0 ÷ (3.6 + 4.2) = 0.77 A is the current from a single row; the second row was ignored.
Why B is wrong: B is wrong because 12.0 ÷ (1.8 + 4.2) = 2.0 A adds the emfs of the two rows, which parallel connection does not do.
Why D is wrong: D is wrong because 6.0 ÷ 4.2 = 1.4 A leaves out the equivalent internal resistance 1.8 Ω.
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What to remember before solving Cells Series Parallel questions
5 NCERT lines
Comparing the last two equations, we get eeq = e1 + e2 (3.45) and req = r1 + r2 (3.46)
-- NCERT Class 12 Physics, Ch. 3, p. 95The rule for series combination clearly can be extended to any number of cells: (i) The equivalent emf of a series combination of n cells is just the sum of their individual emf’s, and (ii) The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances. This is so, when the current leaves each cell from the positive electrode. If in the combination, the current leaves any cell from the negative electrode, the emf of the cell enters the expression for eeq with a negative sign, as in Eq. (3.47).
-- NCERT Class 12 Physics, Ch. 3, p. 96Key Fact
Parallel cells: the currents leaving the positive electrodes add at the junction, I = I1 + I2
Next, consider a parallel combination of the cells (Fig. 3.14). I1 and I2 are the currents leaving the positive electrodes of the cells. At the point B1, I1 and I2 flow in whereas the current I flows out. Since as much charge flows in as out, we have I = I1 + I2 (3.48)
-- NCERT Class 12 Physics, Ch. 3, p. 96Equations (3.56) and (3.57) can be extended easily. If there are n cells of emf e1, . . . en and of internal resistances r1,... rn respectively, connected in parallel, the combination is equivalent to a single cell of emf eeq and internal resistance req, such that
-- NCERT Class 12 Physics, Ch. 3, p. 97Cells Series Parallel: NEET previous year questions (PYQs) with answers
20 questions in Current Electricity
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More in Current Electricity: 3 exam traps and mistakes · 7 formulas · 6 question patterns from its other lessons.
Sources
NCERT refs: Class 12 Physics Chapter 3, p.95
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.
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