First slide

Capacitance

Question

Two capacitors of capacitances 3 $μ$ F and 6 $μ$ F are charged to a potential to 12 v each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

Easy

A

6 V
B

4 V
C

3 V
D

zero

Solution

Charge on I^st condenser,
$Q_{1} = C_{1} V_{1} = 3 \times 12 = 36 μC$
charge on 2^nd condenser,
$Q_{2} = C_{2} V_{2} = 6 \times 12 = 72 μC$
Common potential difference of each condenser
$V = \frac{Q_{2} - Q_{1}}{C_{1} + C_{2}} = \frac{72 - 36}{3 + 6} = \frac{36}{9} = 4 volt$ .

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App