The stored charge in the capacitor in steady state of the following circuit is μC. [Figure: a ladder circuit. A 12 V cell forms the left side. Four vertical rungs stand to its right, in order: 12 Ω, 10 Ω, 4 Ω and a 100 μF capacitor. The top rail has no resistor between the cell and the 12 Ω rung, t…

Question

The stored charge in the capacitor in steady state of the following circuit is μC. [Figure: a ladder circuit. A 12 V cell forms the left side. Four vertical rungs stand to its right, in order: 12 Ω, 10 Ω, 4 Ω and a 100 μF capacitor. The top rail has no resistor between the cell and the 12 Ω rung, then 5 Ω between the 12 Ω and 10 Ω rungs, 4 Ω between the 10 Ω and 4 Ω rungs, and 10 Ω between the 4 Ω rung and the capacitor. The bottom rail has no resistor between the cell and the 12 Ω rung, 2 Ω between the 12 Ω and 10 Ω rungs, 2 Ω between the 10 Ω and 4 Ω rungs, and a plain wire from the 4 Ω rung to the capacitor.]

Accepted Answer

200. In steady state no current flows in the capacitor branch, so the top 10 Ω carries no current and the capacitor voltage equals the voltage across the 4 Ω rung. Right section: top 4 Ω + rung 4 Ω + bottom 2 Ω = 10 Ω, in parallel with the 10 Ω rung gives 5 Ω. The 12 Ω rung sits directly across the cell and does not change the current elsewhere. The rest is the top 5 Ω, the 5 Ω combination and the bottom 2 Ω in series: 12 Ω across 12 V, so 1 A flows through the top 5 Ω. Voltage across the 10 Ω rung =1×5=5 V, so the 10 Ω right section carries 0.5 A. V_4Ω=0.5×4=2 V, so Q=100 μF×2 V=200 μC.

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