Question

You have two capacitors, one with capacitance 15.3× 10 −6 F 15.3×10−6 F and the other of unknown capacitance. You connect the two capacitors in series with a voltage of 355 V 355 V applied across the capacitor pair. You discover that, as a result, the unknown capacitor has a charge of 0.00117 C 0.00117 C . Find its capacitance ? C .

Answer 1

The two capacitors are in series, so the charge on them must be the same.

Using Q = CV for the first capacitance to get the voltage across it.

For C₁ = 15.3 x10^-6 F , and Q₁ = 0.00117 C

so the voltage acorss it => V₁ = Q₁/C₁ = 0.00117 / (15.3 x10^-6 ) = 76.47 volts.

Hence the voltage across the second capacitor is 355- 76.47 = 278.53 volts

Now using Q = CV for this capacitor to get C

C = Q/V = 0.00117 /278.53 = 4.20x10^-6 F

The capacitance C = 4.20x10^-6 F