@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.)

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

swapping them won't make a difference?

can you please explain a bit more?

Log in to reply

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

Log in to reply

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

Log in to reply

how? can you explain a bit more?

Log in to reply

Log in to reply

yes, of course.

Log in to reply

nice writing :D

Log in to reply