Junction rule (KCL): Σ I_in = Σ I_out at any junction (charge conservation). Loop rule (KVL): Σ V = 0 around any closed loop (energy conservation). Apply systematically with chosen sign conventions.
-- NCERT Class 12 Physics, Ch. 3, p. 100Kirchhoff's Laws
8 MCQs9-step worked example
Source: NCERT Current ElectricityOfficial key: NTA-verifiedLast updated: 21 Sep 2026
Kirchhoff's Laws, explained for NEET
The trap: multi-loop network equations are easy to write and easy to get wrong by one sign. A common confusion is writing every IR term as a drop no matter which way you walk the loop, or giving a cell's emf the wrong sign. A second trap: a current comes out negative and it is reported in the direction that was assumed.
The two rules (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98):
- Junction rule — at any junction, the sum of currents entering equals the sum of currents leaving. It rests on conservation of charge (NCERT Points to Ponder, page 105).
- Loop rule — the algebraic sum of changes in potential around any closed loop of resistors and cells is zero.
One sign convention, used everywhere:
- Crossing a resistor R in the direction of the assumed current I: change = −IR. Against the current: +IR.
- Crossing a cell from its negative to its positive terminal: +ε. From positive to negative: −ε.
- Internal resistance r is crossed like any other resistor. NCERT notes that when current flows through a cell from its positive to its negative terminal, the terminal potential difference is V = ε + Ir, not ε − Ir.
Method: assume a direction for each unknown current, use the junction rule to cut the number of unknowns, then write loop equations until you have as many independent equations as unknowns. NCERT Example 3.6 shows that applying the loop rule to the remaining loops gives no additional independent equation. When a network is symmetric, use symmetry first: NCERT Example 3.5 (pages 98–99) solves a cube of 1 Ω edges this way.
Bridge to NEET: the corpus logs multi-loop network questions in the 2023 and 2025 papers; they need simultaneous equations and a consistent sign convention, and a wrong sign produces a distractor.
Watch-out: a negative current means the real current flows opposite to your arrow. Keep the magnitude, flip the direction. If that current then enters a cell at its positive terminal, use V = ε + Ir for that cell.
Can you answer these Kirchhoff's Laws 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
Kirchhoff's junction rule is a consequence of the conservation of which quantity?
Show answer and why every option is right or wrong
Answer: A. Charge does not pile up at a junction, so the current entering equals the current leaving; NCERT states the junction rule rests on conservation of charge (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98; Points to Ponder, page 105).
Why B is wrong: B is wrong because the junction rule is a statement about currents (rate of flow of charge) at a point, not about energy; it is the loop rule that deals with potential changes.
Why C is wrong: C is wrong because no momentum balance is involved; the rule only counts charge flowing into and out of a junction per second.
Why D is wrong: D is wrong because the rule is about charge; carriers having mass does not make mass the conserved quantity the rule is built on.
MCQ 2Easy RecallPractice
According to Kirchhoff's loop rule, the algebraic sum of the changes in potential around any closed loop containing resistors and cells is:
Show answer and why every option is right or wrong
Answer: D. Starting at a point and returning to it, the total change in potential is zero — that is the loop rule (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98).
Why A is wrong: A is wrong because the emfs are only the rises in the loop; once the IR drops are included with their signs, the total change is zero, not the emf sum.
Why B is wrong: B is wrong because the loop current times total resistance is only the sum of the drops; the algebraic sum of rises and drops together is zero.
Why C is wrong: C is wrong because a single resistor's potential difference is only one term; the rule is about the signed sum of every change around the full loop.
MCQ 3Easy RecallPractice
A cell of emf ε and internal resistance r is part of a network. Current I flows through the cell from its positive terminal to its negative terminal. The potential difference between its terminals is:
Show answer and why every option is right or wrong
Answer: C. When current flows through the cell from the positive to the negative terminal, NCERT gives V = ε + Ir (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98).
Why A is wrong: A is wrong because ε − Ir applies when the cell drives current out of its positive terminal through the external circuit; here the current flows the other way through the cell, so the Ir term adds.
Why B is wrong: B is wrong because the terminal potential difference equals ε only in open circuit (no current); a current I through r changes it by Ir.
Why D is wrong: D is wrong because Ir is only the potential difference across the internal resistance; the emf of the cell is left out.
MCQ 4Direct ApplicationPYQ Pattern
Five wires meet at a junction. Currents of 3 A and 2 A flow into the junction through two wires, and a current of 4 A flows out through a third wire. No current flows in the fourth wire. What is the current in the fifth wire?
Show answer and why every option is right or wrong
Answer: D. Current in = 3 A + 2 A = 5 A. Current out so far = 4 A. By the junction rule, the fifth wire must carry 5 − 4 = 1 A out of the junction (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98).
Why A is wrong: A is wrong because it counts only the incoming currents and ignores the 4 A already leaving through the third wire.
Why B is wrong: B is wrong because it gets the size right but the direction wrong: 1 A more must leave to balance 5 A in against 4 A out; if it entered, 6 A would flow in against 4 A out.
Why C is wrong: C is wrong because it adds all currents (3 + 2 + 4) without signs; the junction rule balances currents in against currents out.
MCQ 5Direct ApplicationPYQ Pattern
A cell of emf 12 V and internal resistance 1 Ω and a cell of emf 6 V and internal resistance 1 Ω are joined in a single loop with a 4 Ω resistor. The cells oppose each other (positive terminal joined to positive terminal). What is the current in the loop?
Show answer and why every option is right or wrong
Answer: A. Walking the loop in the direction the 12 V cell drives current: +12 − 6 − I(1) − I(1) − I(4) = 0, so 6 = 6I and I = 1 A (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98).
Why B is wrong: B is wrong because it leaves out both internal resistances (6/4 = 1.5 A); internal resistances are crossed like any other resistor in the loop.
Why C is wrong: C is wrong because it adds the emfs (12 + 6 = 18 V, 18/6 = 3 A); in the loop walk the opposing cell is crossed from + to −, so its emf enters as −6 V.
Why D is wrong: D is wrong because it drops the 6 V cell from the loop equation (12/6 = 2 A); every cell in the loop contributes a signed emf term.
MCQ 6Direct ApplicationPYQ Pattern
Going from point A to point B along one branch of a network, you first cross a 2 Ω resistor carrying a current of 3 A in the direction A to B, and then an ideal cell of emf 5 V, entering it at its negative terminal and leaving at its positive terminal. What is V_A − V_B?
Show answer and why every option is right or wrong
Answer: C. V_B = V_A − (3)(2) + 5, so V_B − V_A = −6 + 5 = −1 V and V_A − V_B = 1 V. Crossing a resistor along the current gives −IR; crossing a cell from − to + gives +ε (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98).
Why A is wrong: A is wrong because it gives the cell the wrong sign (−6 − 5 = −11, so V_A − V_B = 11 V); crossing from the negative to the positive terminal is a rise of +5 V.
Why B is wrong: B is wrong because it flips both signs (+6 − 5 = +1 for V_B − V_A, so V_A − V_B = −1 V); on this path the resistor is a drop and the cell is a rise.
Why D is wrong: D is wrong because it treats the resistor as a rise (+6 + 5 = 11 for V_B − V_A, so V_A − V_B = −11 V); moving along the current direction through a resistor is a drop of IR.
MCQ 7CalculationPYQ Pattern
Points X and Y are joined by three parallel branches. Branch 1: an ideal cell of emf 9 V (positive terminal towards X) in series with a 3 Ω resistor. Branch 2: an ideal cell of emf 3 V (positive terminal towards X) in series with a 3 Ω resistor. Branch 3: a 6 Ω resistor. What is the current in the 6 Ω resistor?
Show answer and why every option is right or wrong
Answer: B. Assume currents I₁ and I₂ flow up branches 1 and 2 into X, and I₃ flows down the 6 Ω resistor: I₃ = I₁ + I₂. Loop 1–3: 9 − 3I₁ − 6I₃ = 0. Loop 2–3: 3 − 3I₂ − 6I₃ = 0. Adding the loop equations: 12 − 3(I₁ + I₂) − 12I₃ = 0 → 12 = 15I₃ → I₃ = 0.8 A. Then I₁ = 1.4 A and I₂ = −0.6 A (the 3 V branch actually carries 0.6 A downward). Check: 1.4 − 0.6 = 0.8 A (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98).
Why A is wrong: A is wrong because it adds the branch current sizes 1.4 A + 0.6 A without signs; I₂ comes out negative, so the junction rule gives 1.4 − 0.6 = 0.8 A.
Why C is wrong: C is wrong because 1.4 A is the current in the 9 V branch, not in the 6 Ω resistor; part of it returns through the 3 V branch.
Why D is wrong: D is wrong because it treats the whole network as one series loop (9 + 3 = 12 V across 3 + 3 + 6 = 12 Ω); the two cells sit in parallel branches, so separate loop equations are needed.
MCQ 8CalculationPractice
Twelve resistors of 1 Ω each form the edges of a cube. A battery of emf 10 V with negligible internal resistance (treat 10 V as exact) is connected across two diagonally opposite corners of the cube. What is the equivalent resistance of the network between those corners?
Show answer and why every option is right or wrong
Answer: B. By symmetry, each of the 3 edges leaving the first corner carries I, each of the 6 middle edges carries I/2, and each of the 3 edges entering the far corner carries I. Loop rule along one path: 10 − I(1) − (I/2)(1) − I(1) = 0 → I = 4 A. Total current = 3I = 12 A, so R_eq = 10/12 = (5/6) Ω (NCERT Class 12 Physics Part I, Chapter 3, Example 3.5, pages 98–99).
Why A is wrong: A is wrong because it puts all 12 edges in parallel (1/12 Ω); the edges form three groups in series, not one parallel set.
Why C is wrong: C is wrong because it adds only the first and last groups of three edges (1/3 + 1/3) and forgets the six middle edges, which each carry I/2.
Why D is wrong: D is wrong because it treats the middle group as three parallel edges (1/3 + 1/3 + 1/3); there are six middle edges, giving 1/6 Ω for that group.
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How do you solve a Kirchhoff's Laws question? A worked example
- 1
Given
Points X and Y are joined by three parallel branches:• Branch 1: ideal cell, emf ε₁ = 12 V (positive terminal towards X), in series with R₁ = 2 Ω.• Branch 2: ideal cell, emf ε₂ = 6 V (positive terminal towards X), in series with R₂ = 2 Ω.• Branch 3: resistor R₃ = 4 Ω.
- 2
Required
The current in each branch, with its actual direction.
- 3
Concept
Kirchhoff's junction rule (charge conservation) and loop rule (the algebraic sum of potential changes round a closed loop is zero), with one fixed sign convention: along the current through a resistor, −IR; through a cell from − to +, +ε (NCERT Class 12 Physics Part I, Chapter 3, Section 3.12, pages 97–98).
- 4
Formula
• Junction X: I₃ = I₁ + I₂ (I₁, I₂ assumed to flow up branches 1 and 2 into X; I₃ assumed to flow from X to Y through R₃).• Loop (branch 1 + branch 3): ε₁ − I₁R₁ − I₃R₃ = 0• Loop (branch 2 + branch 3): ε₂ − I₂R₂ − I₃R₃ = 0
- 5
Substitution
• 12 − 2I₁ − 4I₃ = 0 → I₁ = 6 − 2I₃• 6 − 2I₂ − 4I₃ = 0 → I₂ = 3 − 2I₃• I₃ = I₁ + I₂
- 6
Calculation
I₃ = (6 − 2I₃) + (3 − 2I₃) = 9 − 4I₃ → 5I₃ = 9 → I₃ = 1.8 A.
I₁ = 6 − 2(1.8) = 2.4 A. I₂ = 3 − 2(1.8) = −0.6 A.
Check with the junction rule: 2.4 + (−0.6) = 1.8 A. Check the outer loop (branch 1 up, branch 2 down): 12 − 2(2.4) + 2(−0.6) − 6 = 12 − 4.8 − 1.2 − 6 = 0.
The emfs and resistances are problem-defined exact values, so they do not limit the significant figures; the answers are exact to the digits shown. - 7
Final answer
• Branch 1 (12 V): 2.4 A, flowing from Y up to X.• Branch 2 (6 V): 0.6 A, flowing from X down to Y — opposite to the assumed arrow, so the current enters the 6 V cell at its positive terminal.• Branch 3 (4 Ω): 1.8 A, flowing from X to Y.
- 8
Common trap
Reporting the 6 V branch as 0.6 A in the assumed (upward) direction, or adding 2.4 A + 0.6 A = 3.0 A for the 4 Ω resistor. The minus sign is information: it reverses the direction and must be kept in the junction equation. Consistency check on branch 2: V_X − V_Y = 1.8 × 4 = 7.2 V, which equals the 6 V emf plus the 0.6 × 2 = 1.2 V drop across R₂ — the branch potential difference exceeds the emf because current is pushed into the cell's positive terminal.
- 9
Similar NEET-style question
"Points X and Y are joined by three branches: a 10 V ideal cell with 2 Ω, a 2 V ideal cell with 2 Ω (both positive terminals towards X), and a 2 Ω resistor. Find the current in each branch."
Strategy: I₃ = I₁ + I₂; 10 − 2I₁ − 2I₃ = 0; 2 − 2I₂ − 2I₃ = 0. Adding: 12 − 2I₃ − 4I₃ = 0 → I₃ = 2.0 A. Then I₁ = 3.0 A and I₂ = −1.0 A, so the 2 V branch actually carries 1.0 A from X to Y. Check: 3.0 + (−1.0) = 2.0 A.
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What to remember before solving Kirchhoff's Laws questions
More in Current Electricity: 3 exam traps and mistakes · 7 formulas · 5 question patterns from its other lessons.
Kirchhoff's Laws questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Kirchhoff's Laws?
1 recurring pattern from past papers — click to collapse
Multi-loop network requiring Kirchhoff's voltage and current laws to solve simultaneous equations.
Multi StepMedium
Common distractors
forgets sign convention
Mixes signs around loop
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.
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