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=3.0 kWh (kilowatt-hours) in a coil carrying current I = 300 A?

Answer 1

Concept and reason

The concept required to solve the given problem is the inductance of the coil. First write the formula of energy stored in an inductor. Next, rearrange the equation for the inductance of a coil. Finally, calculate the inductance of the coil.

Fundamentals

The energy stored in an inductor is given by,

$E = \frac{1}{2}L{I^2}$

Here, $L$ is the inductance of the inductor and $I$ is the current.

(a)

The energy stored in an inductor is given by,

$E = \frac{1}{2}L{I^2}$

Here, $L$ is the inductance of the inductor and $I$ is the current.

Rearrange the above equation for the inductance.

$L = \frac{{2E}}{{{I^2}}}$ …… (1)

Substitute $3.0{\rm{ kWh}}$ for $E$ and $300{\rm{ A}}$ for $I$ in equation (1).

$\begin{array}{c}\\L = \frac{{2E}}{{{I^2}}}\\\\ = \frac{{2\left( {3.0{\rm{ kWh}}} \right)\left( {\frac{{1000{\rm{ W}}}}{{1{\rm{ kW}}}}} \right)\left( {\frac{{60 \times 60{\rm{ s}}}}{{1{\rm{ h}}}}} \right)}}{{{{\left( {300{\rm{ A}}} \right)}^2}}}\\\\ = 240{\rm{ H}}\\\end{array}$

Ans: Part a

The inductance of the inductor to store the required energy should be $240{\rm{ H}}$.