logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

A $20 \mu F$ capacitor and $ 40 \; \mu F$ capacitor are connected in series across $400\; V$ supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together. No external voltage is applied. What is potential difference across each capacitor .

$(A)\;\large\frac{1600}{9} \\ (B)\;\large\frac{800}{9} \\ (C)\; 400 \\ (D)\;800 $

Can you answer this question?
 
 

1 Answer

0 votes
In series,
Equivalent $C=\large\frac{20 \times 40}{20 + 40}$
$\qquad= \large\frac{40}{3}$$ \mu F$
Charge on each capacitor $ =\large\frac{40}{3}$$ \times 400 $
$\qquad= \large\frac{16000}{3}$$\mu c$
In series charge on both capacitor is equal.
=> Total charge $=\large\frac{2 \times 16000}{3} $
$\qquad = \large\frac{32000}{3}$$\mu c$
New capacitors are connected in parallel.
Equivalent $C= 20+40=60 \mu F$
Potential $V'= \large\frac{Q'}{C'}$
$\qquad= \large\frac{32000}{3 \times 60}$
$\qquad= \large\frac{1600}{9}$
Hence A is the correct answer.
answered Jan 6, 2014 by meena.p
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...