The stored charge in the capacitor in steady | vedclass

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(C) In steady state, the capacitor acts as an open circuit. No current flows through the branch containing the capacitor.
Let the potential at the no

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) In steady state, the capacitor acts as an open circuit. No current flows through the branch containing the capacitor.
Let the potential at the node between the $5 \Omega$ and $4 \Omega$ resistors be $V_1$, and the potential at the node between the $4 \Omega$ and $10 \Omega$ resistors be $V_2$.
Since the capacitor is in series with the $10 \Omega$ resistor, the potential across the capacitor is equal to the potential at node $V_2$ (assuming the bottom wire is at $0 V$).
Using nodal analysis, the circuit simplifies to a voltage divider network. The equivalent resistance of the network is calculated, and the voltage at the node connected to the capacitor is found to be $0.4 V$.
Thus, the charge $Q = C 	imes V = 100 \mu F 	imes 0.4 V = 40 \mu C$.

The stored charge in the capacitor in steady state of the following circuit is . . . . . . $\mu C$.

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