What are the charges stored in the 1 ,mu F an | vedclass

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Question

(B) In steady state (after the capacitors are fully charged),no current flows through the branches containing capacitors. The capacitors act as open c

Vedclass · Accepted Answer

$4 \,\mu C$ and $8 \,\mu C$,respectively

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(B) In steady state (after the capacitors are fully charged),no current flows through the branches containing capacitors. The capacitors act as open circuits.The circuit simplifies to a series combination of the $1 \,k\Omega$ resistor and the $2 \,k\Omega$ resistor connected to the $6 \,V$ battery.The total resistance of the circuit is $R_{	ext{total}} = 1 \,k\Omega + 2 \,k\Omega = 3 \,k\Omega = 3000 \,\Omega$.The steady-state current in the circuit is $i = \frac{E}{R_{	ext{total}}} = \frac{6 \,V}{3000 \,\Omega} = 2 	imes 10^{-3} \,A = 2 \,mA$.The potential difference across the $2 \,k\Omega$ resistor is $V_{2k} = i 	imes R = (2 	imes 10^{-3} \,A) 	imes (2000 \,\Omega) = 4 \,V$.Since the branches containing the capacitors are in parallel with the $2 \,k\Omega$ resistor,the potential difference across both the $1 \,\mu F$ and $2 \,\mu F$ capacitors is equal to the potential difference across the $2 \,k\Omega$ resistor,which is $4 \,V$.The charge stored in the $1 \,\mu F$ capacitor is $Q_1 = C_1 	imes V = (1 \,\mu F) 	imes (4 \,V) = 4 \,\mu C$.The charge stored in the $2 \,\mu F$ capacitor is $Q_2 = C_2 	imes V = (2 \,\mu F) 	imes (4 \,V) = 8 \,\mu C$.Therefore,the charges are $4 \,\mu C$ and $8 \,\mu C$,respectively.

What are the charges stored in the $1 \,\mu F$ and $2 \,\mu F$ capacitors in the circuit below,once the currents become steady?

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