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Potential Drop across capacitor

Find the potential drop across capacitor \(C_1\)

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Note by Advitiya Brijesh
4 years ago

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\( \frac{C_{2}(E_{1} +E_{2})}{C_{1} + C_{2}} \) Jatin Yadav · 4 years ago

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Swap the positions of E2 and C2. You now have a simple voltage divider. Jimmy Kariznov · 4 years ago

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@Jimmy Kariznov swapping them won't make a difference?

can you please explain a bit more? Hemang Sarkar · 4 years ago

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In steady state, charges stored in the capacitors must be same. Then, we just have to apply Kirchoff's Law. Sambit Senapati · 4 years ago

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@Sambit Senapati Not only at steady state , but at all time instants charges stored in capacitors would be same. Jatin Yadav · 4 years ago

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@Jatin Yadav how? can you explain a bit more? Advitiya Brijesh · 4 years ago

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@Advitiya Brijesh \( q_{cap.} = \int\limits_0^ti_{cap.}(t) dt \). Since , \( i_{cap.}(t) \) is same for both capacitors ( as the circuit is complete ) , \(q_{cap.}\) would also be same for both at any instant .I have assumed that initial charges of both capacitors is zero. (else the question would have been tidious and answer would have been a function of time.) Jatin Yadav · 4 years ago

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@Jatin Yadav yes, of course. Sambit Senapati · 4 years ago

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nice writing :D Rafael Muzzi · 4 years ago

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