A circuit consists of a 12 V ideal battery with internal resistance $r = 0.2 Ω$ connected to two resistors $R_{1} = 4 Ω$ and $R_{2} = 6 Ω$ which are in parallel. The combination is then connected in series with a resistor $R_{3} = 8 Ω$ . Using Kirchhoff's laws, find (i) the current supplied by the battery, (ii) the current through each resistor $R_{1}, R_{2}, R_{3}$ , and (iii) the total power delivered by the battery.

hard 5 marks Class 12 · Physics · Current Electricity

Answer

(i)

I_{bat} = 0.93 A

; (ii)

I_{1} = 0.56 A, I_{2} = 0.37 A, I_{3} = 0.93 A

; (iii)

P_{total} = 10.6 W

The correct answer is (i) $I_{bat} = 0.93 A$ ; (ii) $I_{1} = 0.56 A, I_{2} = 0.37 A, I_{3} = 0.93 A$ ; (iii) $P_{total} = 10.6 W$ — If the question expects power dissipated in the external circuit only, subtract internal loss: $P_{int} = I^{2} r = 1.13 2^{2} \times 0.2 = 0.256 W$ , so external power $= 13.58 - 0.256 = 13.32 W$ , approximately $13.3 W$ .

Step-by-step solution

First find the equivalent resistance of the parallel part: $1/ R_{p} = 1/ R_{1} + 1/ R_{2} = 1/4 + 1/6 = (3 + 2) /12 = 5/12$ , so $R_{p} = 12/5 = 2.4 Ω$ .
Total external resistance seen by the battery is $R_{ext} = R_{p} + R_{3} = 2.4 + 8 = 10.4 Ω$ .
Include internal resistance: total series resistance $R_{tot} = R_{ext} + r = 10.4 + 0.2 = 10.6 Ω$ .
Battery current from Ohm's law: $I_{bat} = E / R_{tot} = 12/10.6 = 1.132 A$ ? Wait, recalc: $12/10.6 = 1.132$ A. But this is the current through the whole series, which includes $R_{3}$ and the parallel combination.
Current through $R_{3}$ is the same as battery current: $I_{3} = 1.132 A$ .
Voltage across the parallel network is $V_{p} = I_{3} \times R_{p} = 1.132 \times 2.4 = 2.717 V$ .
Now split this voltage across $R_{1}$ and $R_{2}$ using $I = V / R$ : $I_{1} = V_{p} / R_{1} = 2.717/4 = 0.679 A$ , $I_{2} = V_{p} / R_{2} = 2.717/6 = 0.453 A$ .
Check KCL at the node: $I_{1} + I_{2} = 0.679 + 0.453 = 1.132 A$ , which matches $I_{3}$ , confirming consistency.
Total power supplied by the battery is $P = E \times I_{bat} = 12 \times 1.132 = 13.58 W$ .
If the question expects power dissipated in the external circuit only, subtract internal loss: $P_{int} = I^{2} r = 1.13 2^{2} \times 0.2 = 0.256 W$ , so external power $= 13.58 - 0.256 = 13.32 W$ , approximately $13.3 W$ .

Common mistake: Treating the parallel branch current as the battery current without accounting for series resistor

Concept tested: Kirchhoff's laws in combined circuits

Published 13 September 2026 · How this solution was prepared and checked

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