A2.8 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.

J

Explanation:

= capacitance of first capacitor = 2.8 µF

= capacitance of second capacitor = 5.7 µF

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

μF

F

= Potential difference across the parallel combination = 340 Volts

Total energy stored by capacitors is given as

inserting the values

J

the answer would be 32*f