Condenser Ahas a capacity of 15 mu F wh | vedclass

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Question

As battery is disconnected so the charges on two capacitor remain constant.

$Q_{1}=15 	imes 10^{-6} 	imes 100=1500 \mu C$ and $Q_{2}=10^{-6} 	im

Vedclass · Accepted Answer

$800$

Vedclass · Answer

As battery is disconnected so the charges on two capacitor remain constant.

$Q_{1}=15 	imes 10^{-6} 	imes 100=1500 \mu C$ and $Q_{2}=10^{-6} 	imes 100=100 \mu C$

As dielectric is removed so the capacitance $15 \mu F$ becomes $C_{1}=\frac{15}{15}=1 \mu F$ and

capacitance $1 \mu F$ will be same $C_{2}=1 \mu F$

Common potential $V_{c}=\frac{Q_{1}+Q_{2}}{C_{1}+C_{2}}=\frac{1500+100}{1+1}=800 \mathrm{V}$

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