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JEE Main 2026 · 8 Apr · Shift 2, Physics question 49

The stored charge in the capacitor in steady state of the following circuit is ________ μC.

A ladder network: a 12 V cell on the left; along the top 5 Ω, 4 Ω and 10 Ω in series; along the bottom 2 Ω and 2 Ω; vertical rungs of 12 Ω, 10 Ω and 4 Ω; a 100 μF capacitor closes the right end.
Answer and worked solution

Official answer: 200

Steady state. No current flows through the capacitor branch, so the 10 Ω resistor carries none. The capacitor voltage is the voltage across the 4 Ω rung.

Node potentials. Battery: top-left 12 V, bottom-left 0. Solving KCL at the four inner nodes gives:

  • top: 7 V and 5 V;
  • bottom: 2 V and 3 V.

So the 4 Ω rung (between 5 V and 3 V) has 2 V across it.

Charge. Q=100 μF×2 V=200 μQ=100\ \mu\text{F}\times2\text{ V}=200\ \muC.

Answer: 200

Topics:RC circuitsKirchhoff's laws
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