Three capacitors of capacitance 1.0 mu F,2.0 | vedclass

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(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{

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$\frac{50}{17} \ V$

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(C) When capacitors are connected in series,the equivalent capacitance $C_{eq}$ is given by:$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$\frac{1}{C_{eq}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{5} = \frac{10 + 5 + 2}{10} = \frac{17}{10} \ \mu F^{-1}$$C_{eq} = \frac{10}{17} \ \mu F$The charge $Q$ on each capacitor in a series combination is the same:$Q = C_{eq} 	imes V = \left( \frac{10}{17} \ \mu F ight) 	imes 10 \ V = \frac{100}{17} \ \mu C$The potential difference $V_2$ across the $2.0 \ \mu F$ capacitor is:$V_2 = \frac{Q}{C_2} = \frac{100/17 \ \mu C}{2 \ \mu F} = \frac{50}{17} \ V$

Three capacitors of capacitance $1.0 \mu F$,$2.0 \mu F$,and $5.0 \mu F$ are connected in series to a $10 \ V$ source. The potential difference across the $2.0 \mu F$ capacitor is

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