Waste less time on Facebook — follow Brilliant.
×

Potential Drop across capacitor

Find the potential drop across capacitor \(C_1\)

[hide=Try it!] Interesting Problem![/hide]

Note by Advitiya Brijesh
4 years, 3 months ago

No vote yet
8 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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

Jatin Yadav - 4 years, 3 months ago

Log in to reply

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

Jimmy Kariznov - 4 years, 3 months ago

Log in to reply

swapping them won't make a difference?

can you please explain a bit more?

Hemang Sarkar - 4 years, 3 months ago

Log in to reply

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

Sambit Senapati - 4 years, 3 months ago

Log in to reply

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

Jatin Yadav - 4 years, 3 months ago

Log in to reply

how? can you explain a bit more?

Advitiya Brijesh - 4 years, 3 months ago

Log in to reply

@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, 3 months ago

Log in to reply

yes, of course.

Sambit Senapati - 4 years, 3 months ago

Log in to reply

nice writing :D

Rafael Muzzi - 4 years, 3 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...