Three capacitors of 15 µF, 25 μF and 40 µF are connected in parallel across a 250 V supply. Find the amount of energy stored.

Question

Question

This question was previously asked in

PGCIL DT Electrical 12 March 2022 (NR II) Official Paper

Answer 1

The correct answer is option 4):(2.5 Joules)

Concept:

In the Parallel circuit, the equivalent capacitance is the algebraic sum of all the capacitance.
Energy stored in the capacitor = \(1\over 2\) × C × V² Joules

Where

C is the capacitance in farad

V is the Voltage in V

Calculation:

Given

V = 250 V

The capacitors 15 µF, 25 μF and 40 µF

C = 15 + 25 + 40 μF

= 80 μF

Energy stored in the capacitor = \(1\over 2\) × C × V2 Joules

= \(1\over 2\)× 80 × 10^-6 × 250²

= 2.5 Joules