Three capacitors C1 = 10.7 µF, C2 = 23.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

E = 1336.71875 J

Explanation:

We are given;. Capacitance of Capacitor 1; C1 = 10.7 µF

Capacitor 2; C2 = 23.0 µF

Capacitor 3; C3 = 29.3 µF

Supply voltage;V = 125 V

Formula for capacitance in series is;

Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3)

Thus, equivalent capacitance is;

C(eq) = (1/10.7) + (1/23) + (1/29.3) = 0.1711 µF = 0.1711 × 10^(6) F

Now, the formula for maximum energy stored is;

E = ½ × C(eq) × V²

E = ½ × 0.1711 × 10^(-6) × 125²

E = 1336.71875 J

12000j is work done

im sorry i dont know