Give the expression of energy stored in an inductance coil carrying current.
To derive the expression for the energy stored in an inductor carrying current, let’s start by recalling the formula for the electromotive force (emf) generated in an inductor. Using this, we’ll determine the rate of work done and then integrate to find the total work.
Suppose an inductor with inductance [latex]L[/latex] is connected to a circuit so that the current flowing through it increases from zero to a maximum value [latex]I[/latex]. Let the current at a given time [latex]t[/latex] be [latex]i[/latex]. The changing current in the inductor generates an opposing emf, expressed as:
[latex]\epsilon = -L \frac{di}{dt}[/latex]
where [latex] L [/latex] is the inductance and [latex]\frac{di}{dt}[/latex] is the rate of change of current.
To sustain current flow through the inductor, the voltage source must do work to overcome this opposing emf. The rate of work done [latex]\frac{dW}{dt}[/latex] is given by:
[latex]\frac{dW}{dt} = -\epsilon \cdot i[/latex]
Thus, the infinitesimal work done [latex] dW [/latex] becomes:
[latex]dW = -(-L \frac{di}{dt}) \cdot i[/latex]
or
[latex]dW = iL \, di[/latex]
To find the total work [latex] W [/latex] required to increase the current from [latex] 0 [/latex] to [latex] I [/latex], we integrate:
[latex]\int_0^W dW = \int_0^I iL \, di[/latex]
which simplifies to:
[latex]W = L \int_0^I i \, di[/latex]
Evaluating this integral, we get:
[latex]W = L \left[ \frac{i^2}{2} \right]_0^I[/latex]
[latex]W = \frac{1}{2} L I^2[/latex]
Therefore, the energy stored in an inductor carrying current [latex]I[/latex]is:
[latex]W = \frac{1}{2} L I^2[/latex]
Note: Inductors store energy in the magnetic field created by the flow of current, while capacitors store energy in an electric field. This is a key difference between the two components that is important to remember.