Question

A 100muF capacitor is charged to 100 V. After the charging, battery is disconnected. The capacitor is then connected in parallel to another capacitor. The final voltage is 20 V. Calculate cap the capacity of second capacitor.

Answer 1

Answer:

The answer of this question is 500muF.

Answer 2

The capacity of second capacitor is $= 400 \mu F$

Explanation:

Given :

Capacitance of first capacitor $C_{1} = 100 \times 10^{-6}$ F

Voltage between first capacitor plate $V_{1} = 100$ V

Charge on first capacitor is given by,

  $C = \frac{Q}{V}$

  $Q _{1} = CV$

  $Q_{1} = 100 \times 100 \times 10^{-6}$

  $Q_{1} = 0.01$ C

For second capacitor charge is given by,

  $Q_{2} = 20\times 100 \times 10^{-6}$

  $Q_{2} = 0.002$ C

Difference of charge is 0.008 C

So find the new capacity of second capacitor,

  $C _{2} = \frac{Q_{diff} }{V}$

  $C _{2} = \frac{0.008}{20} = 0.0004$

  $C _{2} = 400 \times 10^{-6} = 400 \mu F$