Two capacitors of capacity 4 mu F and 6 mu F | vedclass

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(B) Given: $C_1 = 4 \mu F$,$C_2 = 6 \mu F$,and $V = 500 \ V$.In a series combination,the equivalent capacitance $C_{	ext{eq}}$ is given by:$C_{	ext{

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$300$

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(B) Given: $C_1 = 4 \mu F$,$C_2 = 6 \mu F$,and $V = 500 \ V$.In a series combination,the equivalent capacitance $C_{	ext{eq}}$ is given by:$C_{	ext{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{4 	imes 6}{4 + 6} = \frac{24}{10} = 2.4 \mu F$.The total charge $Q$ stored in the series combination is:$Q = C_{	ext{eq}} V = 2.4 \mu F 	imes 500 \ V = 1200 \mu C$.Since capacitors are in series,the charge on each capacitor is the same,so $Q_1 = Q = 1200 \mu C$.The potential difference across the $4 \mu F$ capacitor $(V_1)$ is:$V_1 = \frac{Q_1}{C_1} = \frac{1200 \mu C}{4 \mu F} = 300 \ V$.

Two capacitors of capacity $4 \mu F$ and $6 \mu F$ are connected in series to a $500 \ V$ battery. The potential difference across the $4 \mu F$ capacitor is: (in $V$)

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