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# Potential Drop across capacitor

Find the potential drop across capacitor $$C_1$$

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4 years ago

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

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

swapping them won't make a difference?

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

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

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

how? can you explain a bit more? · 4 years ago

$$q_{cap.} = \int\limits_0^ti_{cap.}(t) dt$$. 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.) · 4 years ago

yes, of course. · 4 years ago