In the circuit shown below,the switch S is co | vedclass

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(B) When the switch is at position $P$,the capacitor is in steady state,so no current flows through it. The circuit consists of two batteries ($1 	ex

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$1.33, 0.67$

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(B) When the switch is at position $P$,the capacitor is in steady state,so no current flows through it. The circuit consists of two batteries ($1 	ext{ V}$ and $2 	ext{ V}$) and two resistors ($1 \Omega$ and $2 \Omega$) in series.The total current in the loop is $i_1 = \frac{2 	ext{ V} - 1 	ext{ V}}{1 \Omega + 2 \Omega} = \frac{1}{3} 	ext{ A}$.The potential difference across the capacitor is the potential difference across the $1 	ext{ V}$ battery and the $1 \Omega$ resistor: $V_A - V_B = 1 	ext{ V} + (i_1 	imes 1 \Omega) = 1 + \frac{1}{3} = \frac{4}{3} 	ext{ V}$.Thus,$q_1 = C \Delta V = 1 \mu 	ext{F} 	imes \frac{4}{3} 	ext{ V} = 1.33 \mu 	ext{C}$.When the switch is at position $Q$,the $1 	ext{ V}$ battery is removed from the circuit. The capacitor is now in parallel with the $1 \Omega$ resistor,which is in series with the $2 	ext{ V}$ battery and $2 \Omega$ resistor.The current in this loop is $i_2 = \frac{2 	ext{ V}}{1 \Omega + 2 \Omega} = \frac{2}{3} 	ext{ A}$.The potential difference across the capacitor is the voltage drop across the $1 \Omega$ resistor: $V_A - V_B = i_2 	imes 1 \Omega = \frac{2}{3} 	ext{ V}$.Thus,$q_2 = C \Delta V = 1 \mu 	ext{F} 	imes \frac{2}{3} 	ext{ V} = 0.67 \mu 	ext{C}$.

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