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(B) In steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.Therefore,the current i

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(B) In steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.Therefore,the current in the circuit flows only through the $2\Omega$ and $6\Omega$ resistors which are in series with the $2	ext{V}$ battery.The total resistance of the circuit is $R = 2\Omega + 6\Omega = 8\Omega$.The current in the circuit is $I = V / R = 2	ext{V} / 8\Omega = 0.25	ext{A}$.The potential difference across the $6\Omega$ resistor is $V_6 = I 	imes 6\Omega = 0.25	ext{A} 	imes 6\Omega = 1.5	ext{V}$.Since the capacitor branch is in parallel with the $6\Omega$ resistor,the potential difference across the capacitor is equal to the potential difference across the $6\Omega$ resistor.Thus,the potential difference across the capacitor is $1.5	ext{V}$.

Under steady state condition,the potential difference across the capacitor in the circuit is . . . . . . $V$.

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