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As shown in the figure, in steady state, the charge stored in the capacitor is ______ ×10⁻⁶ C. [Figure: three branches between a common left wire and a common right wire. Top: cell E=10 V in series with r=10 Ω. Middle: R=100 Ω. Bottom: capacitor C=1.1 μF in series with R'=200 Ω.]

Asked in JEE Main 27th July 2nd Shift 2022 · Capacitors in DC circuits

Figure: Capacitors in DC circuits
Answer: 10

Step-by-step solution

Steady state: the capacitor branch carries no current, so R' has no drop.

Loop current: I=(10)/(10+100)=1/(11) A.

Capacitor voltage = voltage across R =(100)/(11) V.

Q=1.1×(100)/(11)=10 μC=10×10⁻⁶ C.

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