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.

Question

asked Nov 1, 2020 73.2k views

1 Answer

Alasdair McLeay · Answer 1 · 2020-11-07T11:22:04+0000

Answer:

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 = 1.88 μF

C = 1.88* 10^(-6) F

= 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