Charge on the capacitor in the given circuit | vedclass

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(C) In the steady state condition,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.The circu

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(C) In the steady state condition,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 $2 \, \Omega$ and $6 \, \Omega$ resistors.The total resistance of the circuit is $R_{eq} = 2 \, \Omega + 6 \, \Omega = 8 \, \Omega$.The current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{12 \, V}{8 \, \Omega} = 1.5 \, A$.Let the potential at the left junction be $V_L$ and at the right junction be $V_R$. The potential difference across the $6 \, \Omega$ resistor is $V_R - V_L = I 	imes 6 \, \Omega = 1.5 \, A 	imes 6 \, \Omega = 9 \, V$.The capacitor is connected in parallel to the $6 \, \Omega$ resistor. Therefore,the potential difference across the capacitor is equal to the potential difference across the $6 \, \Omega$ resistor,which is $9 \, V$.The charge on the capacitor is $q = C 	imes V = 2 \, \mu F 	imes 9 \, V = 18 \, \mu C$.

Charge on the capacitor in the given circuit in steady state condition is :- ............... $\mu C$

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