A capacitor of capacitance 1 mu F withstands | vedclass

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(D) Given: $C_1 = 1.0 \ \mu F$,$V_1 = 6.0 \ kV = 6 	imes 10^3 \ V$.The maximum charge that the first capacitor can hold is $q_1 = C_1 V_1 = 1.0 \ \mu

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$9$

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(D) Given: $C_1 = 1.0 \ \mu F$,$V_1 = 6.0 \ kV = 6 	imes 10^3 \ V$.The maximum charge that the first capacitor can hold is $q_1 = C_1 V_1 = 1.0 \ \mu F 	imes 6 \ kV = 6000 \ \mu C$.Given: $C_2 = 2.0 \ \mu F$,$V_2 = 4.0 \ kV = 4 	imes 10^3 \ V$.The maximum charge that the second capacitor can hold is $q_2 = C_2 V_2 = 2.0 \ \mu F 	imes 4 \ kV = 8000 \ \mu C$.When capacitors are connected in series,the charge on each capacitor must be the same. The system will reach its limit when the capacitor with the smaller maximum charge reaches its capacity.Since $q_1 At this charge,the voltage across the first capacitor is $V_1 = 6 \ kV$ and the voltage across the second capacitor is $V_2' = \frac{q_{max}}{C_2} = \frac{6000 \ \mu C}{2.0 \ \mu F} = 3 \ kV$.The total maximum voltage the system can withstand 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 $2 \ \mu F$ withstands a maximum voltage of $4 \ kV$. What maximum voltage will the system of these two capacitors withstand if they are connected in series? (in $kV$)

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