Question

Two capacitor C1 = 5 µF and C 2 = 3 µF are connected in parallel with a source of 12V. Calculate the total energy stored in the capacitors

Answer 1

Given - Capacitor 1 - 5 µF

Capacitor 2 - 3 µF

Voltage - 12 V

Find - Total energy stored in the capacitor.

Solution - Capacitors connected in parallel given combined capacitance as per the formula -

C = C1 + C2

C = 5 + 3 µF

C = 8 µF = 8*10-⁶ F

Energy stores in capacitor is calculated by the formula -

E = 1/2(CV²)

E = 1/2(8*10-⁶*12²)

E = 1.15*10-³/2

E = 5.8*10-⁴ J

Hence, energy stored in capacitor is 5.8*10-⁴ J