Two capacitors having capacitances 8 mu F and | vedclass

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(A) Let the two capacitors be $C_1 = 8 \mu F$ and $C_2 = 16 \mu F$ with breaking voltages $V_1 = 20 \ V$ and $V_2 = 80 \ V$.In a series combination,th

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$160$

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(A) Let the two capacitors be $C_1 = 8 \mu F$ and $C_2 = 16 \mu F$ with breaking voltages $V_1 = 20 \ V$ and $V_2 = 80 \ V$.In a series combination,the charge $Q$ on each capacitor is the same.The maximum charge $C_1$ can hold is $Q_1 = C_1 V_1 = 8 \mu F 	imes 20 \ V = 160 \mu C$.The maximum charge $C_2$ can hold is $Q_2 = C_2 V_2 = 16 \mu F 	imes 80 \ V = 1280 \mu C$.Since they are in series,the total charge $Q$ in the combination is limited by the capacitor that reaches its breaking voltage first.Therefore,the maximum charge the combination can store is the minimum of $Q_1$ and $Q_2$.$Q_{\max} = \min(160 \mu C, 1280 \mu C) = 160 \mu C$.

Two capacitors having capacitances $8 \mu F$ and $16 \mu F$ have breaking voltages $20 \ V$ and $80 \ V$ respectively. They are connected in series. What is the maximum charge they can store in this combination in $\mu C$?

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