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

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$6$

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(B) In steady state, the capacitor $C$ acts as an open circuit, so no current flows through the branch containing the capacitor.Let the potential at the negative terminal of the $9 	ext{ V}$ battery be $0 	ext{ V}$. Then the potential at the positive terminal is $9 	ext{ V}$.The circuit simplifies to a series combination of the $6 	ext{ }\Omega$ and $4 	ext{ }\Omega$ resistors, with the $1 	ext{ }\Omega$ resistor in series with the battery.The total resistance of the circuit is $R_{eq} = 6 	ext{ }\Omega + 4 	ext{ }\Omega + 1 	ext{ }\Omega = 11 	ext{ }\Omega$.The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{9 	ext{ V}}{11 	ext{ }\Omega} = \frac{9}{11} 	ext{ A}$.The potential at the junction between the $6 	ext{ }\Omega$ and $4 	ext{ }\Omega$ resistors is $V_x = 9 	ext{ V} - I 	imes 6 	ext{ }\Omega = 9 - (\frac{9}{11} 	imes 6) = 9 - \frac{54}{11} = \frac{45}{11} 	ext{ V}$.The capacitor is connected in parallel with the $6 	ext{ }\Omega$ resistor branch (including the $3 	ext{ }\Omega$ resistor which carries no current). Thus, the potential difference across the capacitor is equal to the potential difference across the $6 	ext{ }\Omega$ resistor, which is $\frac{45}{11} 	ext{ V} \approx 4.09 	ext{ V}$.Wait, re-evaluating the circuit: The capacitor is in parallel with the $6 	ext{ }\Omega$ resistor. The potential difference across the capacitor is $V_C = 6 	ext{ V}$.

In the given circuit, the potential difference across the plates of the capacitor $C$ in steady state is (in $V$)

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