Three capacitors, each of capacitance 120 pF, are each charged to 500 V and then connected in series. Determine (a) the potential difference between the end plates, (b) the charge on each capacitor, and (c) the energy stored in the system. Ans. a.) 1.5kV b.)60nC c.) 45*10^-6 J

Question

Three capacitors, each of capacitance 120 pF, are each charged to 500 V and then connected in series. Determine (a) the potential difference between the end plates, (b) the charge on each capacitor, and (c) the energy stored in the system.       Ans. a.) 1.5kV  b.)60nC  c.) 45*10^-6 J

I NEED THE SOLUTION ON HOW TO GET THIS PLLLLLSSS.

Damon · Accepted Answer

C = Q/V
If each is charged to 500 volts and they are connected in series, then nothing happens, they each remain at 500 volts because no current flows.
The total voltage is 3 * 500 = 1500 Volts

Q = C V = 120 * 10^-12 * 500 = 60,000*10^-12 = 60^10^-9 farad

energy in EACH = (1/2) Q V = 30*10^-9 * (500) = 15 * 10^-6
multiply by 3 = 45 *10^-6 Joules

henry2, · Answer

a. V = 3 * 0.5kv = b. Q = C*V = 120*10^-12 * 500 = 60*10^3 * 10^-12 = 60^10^-9 C = 60 nC. c. Ceq = 120/3 = 40 pF. Eneroy = Ceq*V/2 = 40*10^-12 * 1500^2/2 = 45*10^-6 J.

Anonymous · Answer

Haw did you get the (3)?