For the given circuit,calculate the charge on | vedclass

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(C) In steady state,capacitors act as open circuits,meaning no current flows through the branches containing capacitors.The circuit consists of a $10

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$12$

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(C) In steady state,capacitors act as open circuits,meaning no current flows through the branches containing capacitors.The circuit consists of a $10 \ V$ battery and a $4 \ \Omega$ resistor in parallel with a branch containing a $4 \ \Omega$ resistor. However,looking at the circuit,the middle branch has a $4 \ \Omega$ resistor connected directly across the battery terminals (ignoring the internal resistance of the battery as none is specified,or assuming the $1 \ \Omega$ resistor is in series with the battery).Assuming the $1 \ \Omega$ resistor is in series with the $10 \ V$ source,the total resistance of the circuit is $R_{eq} = 1 \ \Omega + 4 \ \Omega = 5 \ \Omega$.The current flowing through the circuit is $I = \frac{V}{R_{eq}} = \frac{10 \ V}{5 \ \Omega} = 2 \ A$.The voltage across the middle $4 \ \Omega$ resistor is $V_{mid} = I 	imes 4 \ \Omega = 2 \ A 	imes 4 \ \Omega = 8 \ V$.Since the capacitor branches are in parallel with this $4 \ \Omega$ resistor,the potential difference across each branch containing capacitors is $8 \ V$.In each branch,there are two $3 \ \mu F$ capacitors in series. The equivalent capacitance of each branch is $C_{eq} = \frac{3 \ \mu F 	imes 3 \ \mu F}{3 \ \mu F + 3 \ \mu F} = 1.5 \ \mu F$.The charge on each branch is $Q = C_{eq} 	imes V = 1.5 \ \mu F 	imes 8 \ V = 12 \ \mu C$.Since the capacitors are in series in each branch,the charge on each individual capacitor is $12 \ \mu C$.

For the given circuit,calculate the charge on each capacitor in steady state in $\mu C$.

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