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Q.

N identical uncharged capacitors are connected in series with a battery and a switch. The switch is closed and kept closed until the capacitors are completely charged. Let the total potential energy stored now in all the capacitors be U1. Now, the switch is opened, the battery is removed from the circuit and an identical uncharged capacitor is connected in its place. Now the switch is closed again and kept closed for a long time. The total potential energy stored in all the capacitors is now UF. If UFUI=16, then N is __________.

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answer is 5.

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Detailed Solution

We know that after the first charging, the charge on each capacitor is CVN and U1=N(12C(CVN)2)=CV22N

Now, in the final steady state, let the charge on each of the original N capacitors be (CVNx)

Therefore, the charge on the (N+1)th capacitor is x

So, applying KVL, N(1C(CVNx))=xC             x=CVN+1

So, in the final steady state, the charge on the original N capacitors is QN=CVNCVN+1=CVN(N+1)

And, the charge on the (N+1)th capacitor is CVN+1

Therefore, UF=N(12C(CVN(N+1))2)+12C(CVN+1)2=CV2N(N+1) So, UFUI=1N+1

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