# Potential Drop across capacitor

Find the potential drop across capacitor $$C_1$$

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4 years, 11 months ago

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$$\frac{C_{2}(E_{1} +E_{2})}{C_{1} + C_{2}}$$

- 4 years, 11 months ago

Swap the positions of E2 and C2. You now have a simple voltage divider.

- 4 years, 11 months ago

swapping them won't make a difference?

can you please explain a bit more?

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 11 months ago

how? can you explain a bit more?

- 4 years, 11 months 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, 11 months ago

yes, of course.

- 4 years, 11 months ago

nice writing :D

- 4 years, 11 months ago