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(B) $1$. First, calculate the total resistance of the circuit. The resistors $1 	ext{ k}\Omega$ and $2 	ext{ k}\Omega$ are in series with the $3 	e

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$0.5$

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(B) $1$. First, calculate the total resistance of the circuit. The resistors $1 	ext{ k}\Omega$ and $2 	ext{ k}\Omega$ are in series with the $3 	ext{ k}\Omega$ resistor. Total resistance $R_{eq} = 1 	ext{ k}\Omega + 2 	ext{ k}\Omega + 3 	ext{ k}\Omega = 6 	ext{ k}\Omega = 6000 \ \Omega$.$2$. The current $I$ flowing through the circuit is $I = \frac{E}{R_{eq}} = \frac{3 	ext{ V}}{6000 \ \Omega} = 0.5 	imes 10^{-3} 	ext{ A} = 0.5 	ext{ mA}$.$3$. The potential at point $B$ relative to $A$ is $V_{AB} = I 	imes R_{1k} = 0.5 	ext{ mA} 	imes 1 	ext{ k}\Omega = 0.5 	ext{ V}$. Thus, $V_A - V_B = 0.5 	ext{ V}$.$4$. The potential at point $C$ relative to $A$ is determined by the voltage divider rule for the capacitors. Since the capacitors are in series and no $DC$ current flows through them, the potential $V_C$ is determined by the ratio of their capacitances. $V_A - V_C = V_{AD} 	imes \frac{C_2}{C_1 + C_2}$, where $V_{AD} = I 	imes (1 	ext{ k}\Omega + 2 	ext{ k}\Omega) = 0.5 	ext{ mA} 	imes 3 	ext{ k}\Omega = 1.5 	ext{ V}$.$5$. $V_A - V_C = 1.5 	ext{ V} 	imes \frac{2 \mu	ext{F}}{1 \mu	ext{F} + 2 \mu	ext{F}} = 1.5 	imes \frac{2}{3} = 1.0 	ext{ V}$.$6$. Now, $V_{BC} = V_B - V_C = (V_A - V_C) - (V_A - V_B) = 1.0 	ext{ V} - 0.5 	ext{ V} = 0.5 	ext{ V}$.

Consider the circuit given here. The potential difference $V_{BC}$ between the points $B$ and $C$ is (in $V$)

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