Condenser A has a capacity of 15 mu F when it | vedclass

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(B) Since the battery is disconnected,the charges on the two capacitors remain constant.The initial charge on capacitor $A$ is $Q_{1} = C_{A} 	imes V

Vedclass · Accepted Answer

$800$

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(B) Since the battery is disconnected,the charges on the two capacitors remain constant.The initial charge on capacitor $A$ is $Q_{1} = C_{A} 	imes V = 15 \mu F 	imes 100 V = 1500 \mu C$.The initial charge on capacitor $B$ is $Q_{2} = C_{B} 	imes V = 1 \mu F 	imes 100 V = 100 \mu C$.When the dielectric is removed from capacitor $A$,its new capacitance becomes $C_{1} = \frac{C_{A}}{K} = \frac{15 \mu F}{15} = 1 \mu F$.The capacitance of capacitor $B$ remains $C_{2} = 1 \mu F$.When connected in parallel,the common potential $V_{c}$ is given by $V_{c} = \frac{Q_{1} + Q_{2}}{C_{1} + C_{2}}$.Substituting the values,$V_{c} = \frac{1500 \mu C + 100 \mu C}{1 \mu F + 1 \mu F} = \frac{1600}{2} = 800 V$.

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