How much energy is stored in the 0.632 mH inductor of the preceding example when a 30.0 A current flows through it?
Strategy
The energy is given by the equation E_{\text {ind }}=\frac{1}{2} L I^{2}, and all quantities except E_{\text {ind }} are known.
Substituting the value for L found in the previous example and the given current into E_{\text {ind }}=\frac{1}{2} L I^{2} gives
E_{\text {ind }}=\frac{1}{2} L I^{2} (23.44)
=0.5\left(0.632 \times 10^{-3} H \right)(30.0 A )^{2}=0.284 J.
Discussion
This amount of energy is certainly enough to cause a spark if the current is suddenly switched off. It cannot be built up instantaneously unless the power input is infinite.