A 10μF capacitor is fully charged to a potential difference of 50 V. After removing the source voltage, it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes 20 V. The capacitance of the second capacitor is:

Option 1 - <p>15μF</p>
Option 2 - <p>20μF<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>10μF<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>30μF</p>
5 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
6 months ago
Correct Option - 1
Detailed Solution:

By conservation of charge, (50) (10) = 20 (10+C) ⇒ C=15µF

Thumbs Up IconUpvote Thumbs Down Icon

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Physics Laws of Motion 2025

Physics Laws of Motion 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering