Question

A $\mu$ capacitors is charged to $200$volt and then the battery is disconnected. When it is connected in parallel to another uncharged capacitor, the potential difference between the plates of both is $40$volt. The capacitance of the other capacitors is:-

• A. $200\mu F$
• B. $400\mu F$
• C. $300\mu F$
• D. $1600\mu F$

Answer 1

Answer: (C)

A capacitors $\mu$ is changed to $200V$ then, the battery is disconnected.

when it is connected in parallel to another unchanged, capacitor. The potenial difference between the plates of both $40V$

Solution

Current flowing will be constant. So, charged accumulated before and after will be same.

$Q=CV=C_1{V}_1+C_2{V}_{2}200\times 100=40(200+C_2)$

As capacitors are connected in parallel voltage across then will be same So,

$V_1=V_2$

on simplification

$\frac{200\times 100}{40}=200+C_2{C}_2=500-200=300\mu F$