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

Answer:

$0.11$ J

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

$C = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$

$C = \frac{(2.8) (5.7)}{2.8 + 5.7}$

$C = \frac{(2.8) (5.7)}{2.8 + 5.7}$

$C = 1.88$ μF

$C = 1.88\times 10^{-6}$ 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

$U = (0.5) (1.88\times 10^{-6}) (340)^{2}$

$U = 0.11$ J