Question

You have two capacitors, one with capacitance 15.3 × 10-6 F and the other of unknown capacitance. You connect the two capacitors in series with a voltage difference of 399 V applied across the capacitors. You discover that, as a result, the unknown capacitor has a charge of 0.00103 C. Find its capacitance.

Answer 1

Suppose the known capacitance is C₁ = 15.3*10^-6 F,     the unknown capacitance is C₂. As they are connected in series the equivalent capacitance is,

C = (C₁C₂)/(C₁+C₂)

As the applied voltage is V = 399 Volts, the charge stored in the equivalent capacitor is, Q = CV

We know when two capacitors are connected in series, equal amount of charge is stored in them and the amount of charge is also equal to the charge stored in the equivalent capacitor.

Hence Q = 0.00103 C, V = 399 V

1/C = 1/C₁ + 1/C₂ = V/Q

=> 1/C₂ = V/Q - 1/C₁ = (399/0.00103) - (10⁶/15.3) = 0.322*10⁶

=> C₂ = 3.1*10^-6 F