Question

• A 1 capacitance is charged using a constant of 10 for 20 seconds . What is the energy finally stored by the capacitor
  

Answer 1

Solution: Given:

Capacitance, C = 1 farad

Current, I = 10 A

Time, t = 20 s

Formula used:

Charge, q = current x time

Energy stored in the capacitor

$U =$ $\frac{q^2}{2C}$

So, the charge, q = 10 x 20 = 200 C

Energy stored is

$U=\frac{200\times 200}{2\times 1}\\ \\ U=20000 J$

Answer:

The energy stored in the capacitor is 20000 J.