Question

Hello,

I have three questions on Induction Effects. Thanks in advance for your assistance.

Q1.) Consider the inductor in the context of an energy storage device. The electric-power industry would like to find efficient ways to store surplus energy generated during low-demand hours to help meet customer requirements during high-demand hours. Perhaps superconducting coils can be used. How much energy, in kWh, could be stored in a 310 H inductor carrying a current of 370 A ?

Q2.) The electric-power industry is interested in finding a way to store electric energy during times of low demand for use during peak-demand times. In addition to using superconducting coils, a different way of achieving this goal is to use large inductors. What inductance L would be needed to store energy E=3.0kWh (kilowatt-hours) in a coil carrying current I=300A?

Q3.) An air-filled toroidal solenoid has a mean radius of 11.0 cm and a cross-sectional area of 5.00 cm2 . When the current is 14.0 A , the energy stored is 0.370 J . How many turns does the winding have?

Thanks for your help!

Answer 1

1)
Energy stored in an Inductor is Given by, U = 1/2*LI^2
U = 1/2*310 * 370^2
U = 21219500 J
Now Converting J to KWh
U = 5.89 KWh

2)
E = 3.0 KWh
E = 3.0 * 3.6 * 10^6 J
E = 1.08 * 10^7 J

Energy stored in an Inductor is Given by, E = 1/2*LI^2
1.08 * 10^7 = 1/2 * L * 300^2
L = 240 H

3)
r = 11.0 cm = 0.11 m
A = 5.0 * 10^-4 m^2
I = 14.0 A
E = 0.370 J

E = 1/2*L*I^2
0.370 = 1/2*L * 14.0^2
L = 3.78 * 10^-3 H

We know,
L = (u*N^2*A)/(2*π*r)
3.78 * 10^-3 = (2*10^-7 * N^2 * 5.0 * 10^-4)/(0.11)
N = 2039