Question

A 2.8 µF capacitor and a 5.7 µF capacitor are connected in parallel across a 340 V potential difference. Calculate the total energy in joules stored in the capacitors.

Answer 1

`0.11` J

Step-by-step explanation:

`C_(1)` = capacitance of first capacitor = 2.8 µF

`C_(2)` = capacitance of second capacitor = 5.7 µF

The capacitors are connected in parallel and their parallel combination is given as

`C = (C_(1) C_(2))/(C_(1) + C_(2))`

`C = ((2.8) (5.7))/(2.8 + 5.7)`

`C = ((2.8) (5.7))/(2.8 + 5.7)`

`C = 1.88` μF

`C = 1.88* 10^(-6)` F

`V` = Potential difference across the parallel combination = 340 Volts

Total energy stored by capacitors is given as

`U = (0.5) C V^(2)`

inserting the values

`U = (0.5) (1.88* 10^(-6)) (340)^(2)`

`U = 0.11` J