Question

A capacitor of
capacitance 4uF is
charged from 10V
to 20V. The
increased in stored
energy will be​

Answer 1

Answer:

we know that,

c=4×10^-6F

v2=20v, V1=10v

Now, The increased in stored energy will be

U=1/2×cv²

U=1/2×4×10^-6×(20-10)²

U=2×10^-6×10²

U=2×10^-4

U=0.2m energy...

Explanation:

Hope it helps you frnd......

Answer 2

Given :

Capacitance = 4 μF

Initial voltage = 10 V

Final voltage = 20 V

To find :

The increase in the stored energy .

Solution :

Energy , U = ( 1 / 2 ) * C * V * V

Now , the increased in stored energy will be

ΔU = ( 1 / 2 ) * C * ( Change in voltage )^2

=> ΔU = ( 1 / 2 ) * 4 * 10^(-6) * ( 20 - 10 )²

=> ΔU = 2 * 10^(-6) * 10²

=> ΔU = 2 * 10^(-4)  J

The increase in the stored energy is 2 * 10^(-4)  J .