Question

A capacitor C1 is connected to a battery and charged to a voltage 180.0 V. It is then disconnected from the battery and connected in parrallel to another capacitor C2 which is initially uncharged. After connection, the potential across C1 drops to 127.5 V. If C1= 30.0 µF find the capacitance of C2.

Answer 1

Given

Capacitor, C₁ = 30 F = 30 * 10^-6 F

Voltage, V = 180 V

Now.

Total charge, Q = C₁*V = 30 * 10^-6 * 180

                          = 5.4 * 10^-3 C

Now, an uncharged capacitor C₂ is added

After connection, Potential across C₁ is 127.5 V

thus, V₁ = 127.5 V

At, steady state, i.e when no charge is flowing, potential across both the capiacitor must be same

Hence,

Potential across C₂ also must be 127.5 V

thus, V₂ = 127.5 V

Now, as we know total charge of the system must remain same.

Hence,

=> C₁*V = C₁*V₁ + C₂*V₂

=> 5.4 * 10^-3 = 30 * 10^-6*127.5 + C₂*127.5

=> ( C₂ + 30 * 10^-6 )* 127.5 = 5.4 * 10^-3

=> C₂ + 30 * 10^-6 = ( 5.4 * 10^-3 ) / 127.5 = 42.35 * 10^-6

=> C₂ = 42.35 * 10^-6 - 30 * 10^-6 = 12.35 * 10^-6 F

or    C₂ = 12.35 F