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(D) In the steady state, the capacitors act as open circuits. The current $I$ flows through the outer loop containing the $24 	ext{ V}$ battery and t

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$7$

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(D) In the steady state, the capacitors act as open circuits. The current $I$ flows through the outer loop containing the $24 	ext{ V}$ battery and the $2 	ext{ }\Omega$ resistor in series with the parallel combination of the $(4 	ext{ }\Omega + 6 	ext{ }\Omega)$ resistor branch.First, calculate the total resistance of the circuit:$R_{eq} = 2 + (4 + 6) = 2 + 10 = 12 	ext{ }\Omega$.The current in the circuit is:$I = \frac{V}{R_{eq}} = \frac{24}{12} = 2 	ext{ A}$.The voltage across the parallel branch (between points $x$ and $y$) is:$V_{xy} = I 	imes (4 + 6) = 2 	imes 10 = 20 	ext{ V}$.Now, consider the capacitor branch. The capacitors $C_1 = 4/3 	ext{ }\mu	ext{F}$ and $C_2 = 4 	ext{ }\mu	ext{F}$ are in series across the $20 	ext{ V}$ potential difference.The equivalent capacitance is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{(4/3) 	imes 4}{(4/3) + 4} = \frac{16/3}{16/3} = 1 	ext{ }\mu	ext{F}$.The charge on the capacitors is $Q = C_{eq} V_{xy} = 1 	ext{ }\mu	ext{F} 	imes 20 	ext{ V} = 20 	ext{ }\mu	ext{C}$.The potential difference across the $4 	ext{ }\mu	ext{F}$ capacitor is:$V_{C2} = \frac{Q}{C_2} = \frac{20 	ext{ }\mu	ext{C}}{4 	ext{ }\mu	ext{F}} = 5 	ext{ V}$.The potential difference across the $4 	ext{ }\Omega$ resistor is $V_{yB} = I 	imes 4 = 2 	imes 4 = 8 	ext{ V}$.Using the potential path from $B$ to $A$:$V_B + 8 - V_{C1} - V_A = 0$ (where $V_{C1} = 20 - 5 = 15 	ext{ V}$).Alternatively, $V_A - V_B = V_{xy} - V_{C2} - V_{x B} = 20 - 5 - (2 	imes 6) = 20 - 5 - 12 = 3 	ext{ V}$.Wait, re-evaluating the path: $V_A - V_B = (V_A - V_y) - (V_B - V_y) = 15 - 8 = 7 	ext{ V}$.

Find $V_A - V_B$ for the following given circuit in volts.

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