Question

± energy stored in an inductor
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
part a
what inductance ( l ) would be needed to store energy ( e = 3.0mathrm{kwh} ) (kilowatt - hours) in a coil carrying current ( i = 200mathrm{a} )?

Explanation:

Step1: Convert energy unit

Convert \(E = 3.0\mathrm{kWh}\) to joules.
Since \(1\mathrm{kW}=10^{3}\mathrm{W}\), \(1\mathrm{h} = 3600\mathrm{s}\), then \(E=3.0\times10^{3}\mathrm{W}\times3600\mathrm{s}=1.08\times 10^{7}\mathrm{J}\)

Step2: Use energy formula for inductor

The energy stored in an inductor is \(E=\frac{1}{2}LI^{2}\).
We need to solve for \(L\), so \(L=\frac{2E}{I^{2}}\)
Given \(I = 200\mathrm{A}\), substitute \(E = 1.08\times 10^{7}\mathrm{J}\) and \(I = 200\mathrm{A}\) into the formula:
\(L=\frac{2\times1.08\times 10^{7}\mathrm{J}}{(200\mathrm{A})^{2}}\)
\(L=\frac{2.16\times 10^{7}\mathrm{J}}{40000\mathrm{A}^{2}} = 540\mathrm{H}\)

Answer:

\(540\)