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(D) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.The circuit effective

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(D) In a steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.The circuit effectively consists of the $12 \, V$ battery connected in series with the $2 \, \Omega$ resistor and the $6 \, \Omega$ resistor.The total resistance of the circuit is $R_{	ext{eq}} = 2 \, \Omega + 6 \, \Omega = 8 \, \Omega$.The current flowing through the circuit is $I = \frac{V}{R_{	ext{eq}}} = \frac{12 \, V}{8 \, \Omega} = 1.5 \, A$.The potential difference across the $6 \, \Omega$ resistor is $V_6 = I 	imes 6 \, \Omega = 1.5 \, A 	imes 6 \, \Omega = 9 \, V$.Since the capacitor is in parallel with the $6 \, \Omega$ resistor,the potential difference across the capacitor is equal to the potential difference across the $6 \, \Omega$ resistor,which is $9 \, V$.The maximum charge on the capacitor is given by $Q_{\max} = C 	imes V$.Given $C = 2 \, \mu F$ and $V = 9 \, V$,we have $Q_{\max} = 2 \, \mu F 	imes 9 \, V = 18 \, \mu C$.

Find the maximum charge on the capacitor $C$ in the following circuit in $\mu C$.

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