You have two capacitors, one with capacitance 15.9 ?F and the other of unknown c

Question

You have two capacitors, one with capacitance 15.9 ?F and the other of unknown capacitance. You connect the two capacitors in series with a voltage difference of 397-V applied across the capacitors. You discover that, as a result, the unknown capacitor has a charge of 1.19 mC. Find its capacitance.

You have two capacitors, one with capacitance 15.9 F and the other of unknown capacitance. You connect the two capacitors in series with a voltage difference of 397-V applied across the capacitors. You discover that, as a result, the unknown capacitor has a charge of 1.19 mC. Find its capacitance. Number

Explanation / Answer

Charge is same when capacitors are connected in series

So , Q = Ceq*V = C1*C2/(C1+C2) *V

1.19*10-3 = 15.9*C2/(15.9+C2)*397

3*10-6 (15.9 + C2) *10-6 = 15.9 C2 *10-12

47.7 + 3C2 = 15.9C2

C2 = 3.70*10-6 F =3.7 muF (ans)

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