Two capacitors of capacitances 1 mu F and 2 | vedclass

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(A) Given: $C_1 = 1 \ \mu F, C_2 = 2 \ \mu F, V_1 = 6 \ kV, V_2 = 4 \ kV$.The maximum charge each capacitor can hold is:$Q_1 = C_1 V_1 = 1 \ \mu F 	i

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Given: $C_1 = 1 \ \mu F, C_2 = 2 \ \mu F, V_1 = 6 \ kV, V_2 = 4 \ kV$.The maximum charge each capacitor can hold is:$Q_1 = C_1 V_1 = 1 \ \mu F 	imes 6 \ kV = 6 \ \mu C$.$Q_2 = C_2 V_2 = 2 \ \mu F 	imes 4 \ kV = 8 \ \mu C$.In a series combination,the charge on each capacitor must be the same. Therefore,the maximum charge the combination can withstand is limited by the capacitor with the smaller charge capacity,which is $Q_{max} = 6 \ \mu C$.The equivalent capacitance of the series combination is $C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{1 	imes 2}{1 + 2} = \frac{2}{3} \ \mu F$.The total potential $V_{max}$ the combination can withstand is $V_{max} = \frac{Q_{max}}{C_{eq}} = \frac{6 \ \mu C}{\frac{2}{3} \ \mu F} = 6 	imes \frac{3}{2} \ kV = 9 \ kV$.

Two capacitors of capacitances $1 \ \mu F$ and $2 \ \mu F$ can separately withstand potentials of $6 \ kV$ and $4 \ kV$ respectively. The total potential they together can withstand when they are connected in series is (in $kV$)

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