In the circuit shown in the figure,C_1 = C_2 | vedclass

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(D) In a steady state,capacitors act as open circuits. The circuit simplifies to a series-parallel combination of resistors.Let the nodes be defined s

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$B$ and $C$ both

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(D) In a steady state,capacitors act as open circuits. The circuit simplifies to a series-parallel combination of resistors.Let the nodes be defined such that the potential difference across $C_1$ is $V_1$ and across $C_2$ is $V_2$.The upper branch has resistances $1 \Omega, 2 \Omega, 3 \Omega$ in series,total $6 \Omega$. The lower branch has $2 \Omega, 1 \Omega, 3 \Omega$ in series,total $6 \Omega$.The total resistance $R_{eq} = (6 \Omega || 6 \Omega) = 3 \Omega$.The total current from the $120 V$ source is $I = 120 / 3 = 40 A$.This current splits equally into the two branches: $I_{upper} = 20 A$ and $I_{lower} = 20 A$.Potential at the node between $1 \Omega$ and $2 \Omega$ (upper) is $120 - (20 	imes 1) = 100 V$. Potential at the node between $2 \Omega$ and $1 \Omega$ (lower) is $120 - (20 	imes 2) = 80 V$. Thus,$V_{C1} = 100 - 80 = 20 V$.Charge $Q_1 = C_1 V_1 = 2 \mu F 	imes 20 V = 40 \mu C$.Potential at the node between $2 \Omega$ and $3 \Omega$ (upper) is $100 - (20 	imes 2) = 60 V$. Potential at the node between $1 \Omega$ and $3 \Omega$ (lower) is $80 - (20 	imes 1) = 60 V$. Thus,$V_{C2} = 60 - 60 = 0 V$.Charge $Q_2 = C_2 V_2 = 2 \mu F 	imes 0 V = 0$.Therefore,$C_2$ has zero charge and $C_1$ has $40 \mu C$ charge. Option $D$ is correct.

In the circuit shown in the figure,$C_1 = C_2 = 2 \mu F$. Find the charge stored in the capacitors.

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