A Level Physics 10.2 Kirchhoff’s Laws MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice MCQs cover Section 10.2 of the Cambridge AS & A Level Physics (9702) syllabus: Kirchhoff’s first law at junctions, series and parallel resistor combinations, and Kirchhoff’s second law applied to complete circuit loops, including circuits with internal resistance. Work through each question, then open the answer toggle to check the full worked solution.
Formula recap (see the A Level Physics Formula Sheet for the full list):
- Kirchhoff’s first law: sum of currents into a junction = sum of currents out of the junction
- Series: R = R1 + R2 + … ; the current is the same everywhere
- Parallel: 1/R = 1/R1 + 1/R2 + … ; the p.d. is the same across each branch
- Kirchhoff’s second law: around any closed loop, sum of e.m.f.s = sum of p.d.s, so E = I(R + r) for a source with internal resistance r
Kirchhoff’s First Law & Junctions
Q1. Which statement correctly gives Kirchhoff’s first law for a junction in an electric circuit?
- (A) The sum of the currents leaving a junction is greater than the sum entering it
- (B) The sum of the currents entering a junction is equal to the sum of the currents leaving it
- (C) The current entering a junction is shared out in proportion to the resistance of each branch
- (D) The total current in a circuit is the same at a junction as anywhere else
Show Answer
Answer: B – Kirchhoff’s first law states that the total current flowing into a junction equals the total current flowing out, so no charge builds up at the junction.
Q2. Kirchhoff’s first law is a consequence of which conservation law?
- (A) Conservation of energy
- (B) Conservation of momentum
- (C) Conservation of charge
- (D) Conservation of mass
Show Answer
Answer: C – Charge cannot be created, destroyed or stored at a junction, so the charge arriving per second (current in) must equal the charge leaving per second (current out). This is conservation of charge.
Q3. At a junction, currents of 4.0 A and 3.0 A flow in, and a current of 2.0 A flows out along one branch. What is the current in the remaining branch, and in which direction does it flow?
- (A) 5.0 A, flowing out of the junction
- (B) 5.0 A, flowing into the junction
- (C) 1.0 A, flowing out of the junction
- (D) 9.0 A, flowing out of the junction
Show Answer
Answer: A – Total current in = 4.0 + 3.0 = 7.0 A. Total current out must also be 7.0 A, so the remaining branch carries 7.0 − 2.0 = 5.0 A flowing out of the junction.
Q4. A current of 6.0 A enters a junction and splits into three parallel branches. Two of the branches carry 1.5 A and 2.5 A. What is the current in the third branch?
- (A) 4.0 A
- (B) 6.0 A
- (C) 10.0 A
- (D) 2.0 A
Show Answer
Answer: D – By Kirchhoff’s first law, current in = current out: 6.0 = 1.5 + 2.5 + I, so I = 6.0 − 4.0 = 2.0 A.
Q5. A battery, a resistor and a lamp are connected in series. The ammeter next to the battery reads 0.60 A. What are the currents through the resistor and through the lamp?
- (A) Both are less than 0.60 A because current is used up as it travels round
- (B) Resistor 0.60 A, lamp 0.30 A
- (C) Both are 0.60 A
- (D) They cannot be found without knowing the resistances
Show Answer
Answer: C – In a series circuit there is only one path, so the same current flows through every component. Current is never “used up”; it is the energy carried by the charge that is transferred.
Q6. A current of 3.0 A reaches a junction and splits between two parallel branches containing a 2.0 Ω resistor and a 4.0 Ω resistor. What is the current through the 2.0 Ω resistor?
- (A) 2.0 A
- (B) 1.0 A
- (C) 1.5 A
- (D) 3.0 A
Show Answer
Answer: A – The p.d. across both branches is the same, so the current splits in inverse proportion to the resistances. Current through the 2.0 Ω branch = 3.0 x 4.0 / (2.0 + 4.0) = 2.0 A (and 1.0 A flows through the 4.0 Ω branch; 2.0 + 1.0 = 3.0 A, as Kirchhoff’s first law requires).
Q7. At a junction, a current of 5.0 A flows in. Currents of 2.0 A and 1.0 A flow out through two branches. A student claims the remaining 2.0 A worth of charge accumulates at the junction. Why is this wrong?
- (A) Because the junction stores charge as a capacitor would
- (B) Because the remaining current must flow out through the third branch; charge cannot accumulate at a junction
- (C) Because some of the charge is converted into heat at the junction
- (D) Because currents can only ever split equally at a junction
Show Answer
Answer: B – Charge is conserved and a junction has no means of storing a growing amount of it, so the currents must balance at every instant: 5.0 A in means 5.0 A out, with 2.0 A leaving through the third branch.
Q8. Two ammeters are placed in a series circuit, one just after the positive terminal of the battery and one on the far side of a lamp. The first ammeter reads 0.45 A. What does the second ammeter read, assuming both are ideal?
- (A) Slightly less than 0.45 A, because the lamp uses up some current
- (B) 0 A, because all the current returns through the first ammeter
- (C) Slightly more than 0.45 A
- (D) 0.45 A
Show Answer
Answer: D – There is only one path in a series circuit, so the current is identical at every point: both ammeters read 0.45 A.
Q9. A total current of 4.5 A splits between two parallel branches of resistance 3.0 Ω and 6.0 Ω. What is the current in the 6.0 Ω branch?
- (A) 1.5 A
- (B) 3.0 A
- (C) 2.25 A
- (D) 4.5 A
Show Answer
Answer: A – The branch currents are in inverse proportion to the resistances, so the ratio is 6.0 : 3.0 = 2 : 1 in favour of the 3.0 Ω branch. Current in the 6.0 Ω branch = 4.5 x 3.0 / (3.0 + 6.0) = 1.5 A (the 3.0 Ω branch carries 3.0 A).
Q10. Three wires meet at a junction. The currents are 2.0 A into the junction, 3.5 A into the junction and an unknown current I in the third wire. What are the size and direction of I?
- (A) 1.5 A into the junction
- (B) 5.5 A out of the junction
- (C) 5.5 A into the junction
- (D) 1.5 A out of the junction
Show Answer
Answer: B – Total current in = 2.0 + 3.5 = 5.5 A. For the currents to balance, 5.5 A must leave through the third wire.
Series & Parallel Circuits
Q11. Resistors of 2.0 Ω, 4.0 Ω and 6.0 Ω are connected in series. What is their combined resistance?
- (A) 1.1 Ω
- (B) 4.0 Ω
- (C) 0.92 Ω
- (D) 12 Ω
Show Answer
Answer: D – In series the resistances simply add: R = 2.0 + 4.0 + 6.0 = 12 Ω.
Q12. A 12 V battery is connected across a 3.0 Ω resistor and a 9.0 Ω resistor in series. What is the p.d. across the 9.0 Ω resistor?
- (A) 9.0 V
- (B) 3.0 V
- (C) 6.0 V
- (D) 12 V
Show Answer
Answer: A – Total resistance = 3.0 + 9.0 = 12 Ω, so the current is I = 12 / 12 = 1.0 A. The p.d. across the 9.0 Ω resistor is V = IR = 1.0 x 9.0 = 9.0 V (the remaining 3.0 V appears across the 3.0 Ω resistor).
Q13. Three resistors are connected in parallel across a 6.0 V supply. Which statement about this arrangement is correct?
- (A) The current through each resistor is the same
- (B) The p.d. across each resistor is 6.0 V
- (C) The p.d. across each resistor is 2.0 V
- (D) The combined resistance is the sum of the three resistances
Show Answer
Answer: B – In parallel, every branch is connected directly across the supply, so each resistor has the full 6.0 V across it. The currents differ unless the resistances are equal.
Q14. A 6.0 Ω resistor and a 3.0 Ω resistor are connected in parallel. What is their combined resistance?
- (A) 9.0 Ω
- (B) 4.5 Ω
- (C) 2.0 Ω
- (D) 0.50 Ω
Show Answer
Answer: C – 1/R = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0, so R = 6.0/3 = 2.0 Ω.
Q15. Two identical 10 Ω resistors are connected in parallel. What is their combined resistance?
- (A) 5.0 Ω
- (B) 20 Ω
- (C) 10 Ω
- (D) 2.5 Ω
Show Answer
Answer: A – For two equal resistors in parallel, R = 10 / 2 = 5.0 Ω. In general, n identical resistors of value R in parallel give R/n.
Q16. A 4.0 Ω resistor and a 12 Ω resistor are connected in parallel. Which statement about their combined resistance is correct?
- (A) It is 16 Ω
- (B) It is exactly 4.0 Ω
- (C) It is between 4.0 Ω and 12 Ω
- (D) It is 3.0 Ω, which is less than the smaller of the two resistances
Show Answer
Answer: D – 1/R = 1/4.0 + 1/12 = 3/12 + 1/12 = 4/12, so R = 3.0 Ω. A parallel combination always has a resistance smaller than the smallest individual branch, because the extra branches provide additional paths for the current.
Q17. A 12 V battery is connected to a 2.0 Ω resistor and a 4.0 Ω resistor in series. What is the current in the circuit and the p.d. across the 4.0 Ω resistor?
- (A) Current 3.0 A, p.d. 12 V
- (B) Current 2.0 A, p.d. 8.0 V
- (C) Current 2.0 A, p.d. 4.0 V
- (D) Current 6.0 A, p.d. 8.0 V
Show Answer
Answer: B – Total resistance = 2.0 + 4.0 = 6.0 Ω, so I = 12 / 6.0 = 2.0 A. P.d. across the 4.0 Ω resistor = 2.0 x 4.0 = 8.0 V (leaving 4.0 V across the 2.0 Ω resistor; 8.0 + 4.0 = 12 V).
Q18. A 6.0 V battery is connected across a 2.0 Ω resistor and a 3.0 Ω resistor in parallel. What total current is drawn from the battery?
- (A) 1.2 A
- (B) 2.5 A
- (C) 5.0 A
- (D) 1.0 A
Show Answer
Answer: C – Each branch has 6.0 V across it. Current in the 2.0 Ω branch = 6.0 / 2.0 = 3.0 A; current in the 3.0 Ω branch = 6.0 / 3.0 = 2.0 A. Total current = 3.0 + 2.0 = 5.0 A. (Check: combined R = 1.2 Ω, I = 6.0 / 1.2 = 5.0 A.)
Q19. A 12 V battery is connected to a 2.0 Ω resistor in series with a parallel pair made from a 6.0 Ω resistor and a 3.0 Ω resistor. What is the p.d. across the parallel pair?
- (A) 12 V
- (B) 4.0 V
- (C) 8.0 V
- (D) 6.0 V
Show Answer
Answer: D – The parallel pair has combined resistance 1/R = 1/6.0 + 1/3.0, giving R = 2.0 Ω. Total resistance = 2.0 + 2.0 = 4.0 Ω, so the current from the battery is 12 / 4.0 = 3.0 A. P.d. across the pair = 3.0 x 2.0 = 6.0 V (the other 6.0 V is across the series resistor).
Q20. Resistors of 4.0 Ω, 4.0 Ω and 8.0 Ω are connected in parallel. What is their combined resistance?
- (A) 16 Ω
- (B) 1.6 Ω
- (C) 2.7 Ω
- (D) 0.63 Ω
Show Answer
Answer: B – 1/R = 1/4.0 + 1/4.0 + 1/8.0 = 2/8.0 + 2/8.0 + 1/8.0 = 5/8.0, so R = 8.0 / 5 = 1.6 Ω.
Kirchhoff’s Second Law & Loop Calculations
Q21. Which statement correctly gives Kirchhoff’s second law for a closed loop in a circuit?
- (A) Around any closed loop, the sum of the e.m.f.s is equal to the sum of the p.d.s
- (B) Around any closed loop, the currents entering a point equal the currents leaving it
- (C) Around any closed loop, the sum of the resistances equals the total e.m.f.
- (D) The sum of the p.d.s around a closed loop is always zero only when there is no battery
Show Answer
Answer: A – Kirchhoff’s second law states that, going around any closed loop, the total e.m.f. equals the sum of the p.d.s across the components in that loop.
Q22. Kirchhoff’s second law is a consequence of which conservation law?
- (A) Conservation of charge
- (B) Conservation of momentum
- (C) Conservation of mass
- (D) Conservation of energy
Show Answer
Answer: D – The e.m.f. is the energy supplied per unit charge and each p.d. is the energy transferred per unit charge in a component. Energy supplied per coulomb must equal energy transferred per coulomb around a complete loop: conservation of energy.
Q23. A battery of e.m.f. 12 V and internal resistance 1.0 Ω is connected to an external resistor of 5.0 Ω. Using Kirchhoff’s second law, what is the current in the circuit?
- (A) 2.4 A
- (B) 12 A
- (C) 2.0 A
- (D) 1.7 A
Show Answer
Answer: C – Around the loop, E = I(R + r), so 12 = I(5.0 + 1.0) and I = 12 / 6.0 = 2.0 A.
Q24. In a single-loop circuit, a 12 V supply drives current through two components in series. The p.d. across the first component is 7.0 V. What is the p.d. across the second component?
- (A) 7.0 V
- (B) 5.0 V
- (C) 12 V
- (D) 19 V
Show Answer
Answer: B – By Kirchhoff’s second law the p.d.s in the loop add up to the supply e.m.f.: 12 = 7.0 + V, so V = 5.0 V.
Q25. A 12 V battery is connected to a 2.0 Ω resistor in the main line. The circuit then splits into two parallel branches, one of 6.0 Ω and one of 3.0 Ω, which then recombine. What is the current through the 3.0 Ω branch?
- (A) 1.0 A
- (B) 3.0 A
- (C) 1.5 A
- (D) 2.0 A
Show Answer
Answer: D – The parallel branches have combined resistance 2.0 Ω, so the total resistance is 2.0 + 2.0 = 4.0 Ω and the main current is 12 / 4.0 = 3.0 A. Applying Kirchhoff’s second law to the outer loop, the p.d. across the parallel pair is 12 − (3.0 x 2.0) = 6.0 V. Current through the 3.0 Ω branch = 6.0 / 3.0 = 2.0 A (and 1.0 A through the 6.0 Ω branch; 2.0 + 1.0 = 3.0 A by the first law).
Q26. A 24 V battery is connected to a 4.0 Ω resistor in series with a parallel combination of two 8.0 Ω resistors. What is the p.d. across the parallel combination?
- (A) 12 V
- (B) 24 V
- (C) 8.0 V
- (D) 6.0 V
Show Answer
Answer: A – The two 8.0 Ω resistors in parallel give 4.0 Ω. Total resistance = 4.0 + 4.0 = 8.0 Ω, so the current is 24 / 8.0 = 3.0 A. P.d. across the parallel combination = 3.0 x 4.0 = 12 V (the other 12 V is dropped across the series resistor).
Q27. A cell of e.m.f. 9.0 V is connected to an external resistor of 5.0 Ω and the current in the circuit is 1.5 A. What is the internal resistance of the cell?
- (A) 6.0 Ω
- (B) 0.50 Ω
- (C) 1.0 Ω
- (D) 11 Ω
Show Answer
Answer: C – Kirchhoff’s second law: E = I(R + r), so 9.0 = 1.5(5.0 + r). Then 5.0 + r = 9.0 / 1.5 = 6.0, giving r = 1.0 Ω.
Q28. A single loop contains a 12 V battery, a second battery of 2.0 V connected so that its e.m.f. opposes the first, and a resistor of 5.0 Ω. The internal resistances are negligible. What is the current in the loop?
- (A) 2.8 A
- (B) 2.0 A
- (C) 2.4 A
- (D) 0.40 A
Show Answer
Answer: B – The opposing e.m.f. subtracts, so the net e.m.f. around the loop is 12 − 2.0 = 10 V. Current = 10 / 5.0 = 2.0 A.
Q29. A closed loop contains a source of e.m.f. 20 V and three components in series. The p.d.s across two of the components are 6.0 V and 9.0 V. What is the p.d. across the third component?
- (A) 5.0 V
- (B) 15 V
- (C) 20 V
- (D) 35 V
Show Answer
Answer: A – By Kirchhoff’s second law: 20 = 6.0 + 9.0 + V, so V = 20 − 15 = 5.0 V.
Q30. A 12 V battery is connected to a 3.0 Ω resistor in series with a parallel pair of two 10 Ω resistors. What is the current through one of the 10 Ω resistors?
- (A) 1.5 A
- (B) 0.60 A
- (C) 0.75 A
- (D) 1.2 A
Show Answer
Answer: C – The parallel pair has combined resistance 5.0 Ω, so the total resistance is 3.0 + 5.0 = 8.0 Ω and the battery current is 12 / 8.0 = 1.5 A. The p.d. across the pair is 1.5 x 5.0 = 7.5 V, so the current through each 10 Ω resistor is 7.5 / 10 = 0.75 A (0.75 + 0.75 = 1.5 A, as the first law requires).
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