In the network shown below,the charge accumul | vedclass

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(A) In the steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.The current $I$ flo

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$7.2$

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(A) In the steady state,the capacitor acts as an open circuit,meaning no current flows through the branch containing the capacitor.The current $I$ flows through the outer loop containing the $3 	ext{ V}$ battery and the two $6 \,\Omega$ and $4 \,\Omega$ resistors in series.$I = \frac{V}{R_{eq}} = \frac{3 	ext{ V}}{6 \,\Omega + 4 \,\Omega} = \frac{3}{10} 	ext{ A} = 0.3 	ext{ A}$.The potential difference across the $6 \,\Omega$ resistor (which is in parallel with the capacitor branch) is:$V_{cap} = I 	imes R = 0.3 	ext{ A} 	imes 6 \,\Omega = 1.8 	ext{ V}$.Since the capacitor is connected in parallel to this $6 \,\Omega$ resistor,the potential difference across the capacitor is $1.8 	ext{ V}$.The charge $q$ accumulated in the capacitor is given by:$q = C 	imes V_{cap} = 4 \,\mu	ext{F} 	imes 1.8 	ext{ V} = 7.2 \,\mu	ext{C}$.

In the network shown below,the charge accumulated in the capacitor in steady state will be ........... $\mu C$.

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