A capacitor of capacitance 1 mu F withstands | vedclass

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(C) Given: $C_1 = 1.0 \mu F$,$V_1 = 6 kV = 6000 V$.$C_2 = 2.0 \mu F$,$V_2 = 4 kV = 4000 V$.The maximum charge that $C_1$ can hold is $q_1 = C_1 V_1 =

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$9$

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(C) Given: $C_1 = 1.0 \mu F$,$V_1 = 6 kV = 6000 V$.$C_2 = 2.0 \mu F$,$V_2 = 4 kV = 4000 V$.The maximum charge that $C_1$ can hold is $q_1 = C_1 V_1 = 1.0 \mu F 	imes 6 kV = 6000 \mu C$.The maximum charge that $C_2$ can hold is $q_2 = C_2 V_2 = 2.0 \mu F 	imes 4 kV = 8000 \mu C$.In a series combination,the charge on each capacitor must be the same. Therefore,the maximum charge the combination can hold is limited by the capacitor with the smaller maximum charge,which is $q_{max} = 6000 \mu C$.When the charge on the combination is $6000 \mu C$,the voltage across $C_1$ is $V_1' = 6 kV$ and the voltage across $C_2$ is $V_2' = q_{max} / C_2 = 6000 \mu C / 2.0 \mu F = 3000 V = 3 kV$.The total maximum voltage across the series combination is $V_{total} = V_1' + V_2' = 6 kV + 3 kV = 9 kV$.

$A$ capacitor of capacitance $1 \mu F$ withstands a maximum voltage of $6 kV$,while a capacitor of capacitance $2.0 \mu F$ withstands a maximum voltage of $4 kV$. If the two capacitors are connected in series,then the two capacitors combined can take up a maximum voltage of ... $kV$.

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