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. One way of achieving this goal is to use large inductors. What inductance L would be needed to store energy 3.0 (kilowatt-hours) in a coil carrying current I= 300 A?
U = 1/2 LI^2
L = 2E / I^2
3.0 kWh = 1.08*10^7
The Attempt at a Solution
2(1.08 * 10^7) / 900 A^2 = 24,000 H