Tutorial 5 Extension - Derivation of Equation for Energy in an Inductor

Consider a current increasing from 0 to I amps.  In a short time interval dt the current increases from i to i + di, where di is the tiny increase in the instantaneous current.

We know that the induced EMF is given by:

We have missed the minus sign out to keep things simple.

We know that energy = current × voltage × time.  So we can write:

So this is the energy in the very small sample of time, and di is the very small increase in current.  To get the total increase in current, we need to integrate this expression.

So our final formula is:

Back to Tutorial 5A

Derivation of the equation for reactance

Consider a perfect inductor of inductance L connected up in an AC circuit as shown:

The voltage is V, and the current is I.  The current alternates with a frequency f.

We know that a sinusoidal alternating voltage waveform is described by the equation:

We haven’t got a second wave to worry about, so phase can be kept out for the moment.

We also know that:

We know that inductance is defined by:

By differentiation we can write:

So we can say:

The maximum value of the EMF is when t = 0.  This results in:

This can be rearranged to make the current the subject of the formula:

We also know that the voltage = maximum voltage – reverse EMF.  When t = 0, the voltage is 0, so we can write:

This is helpful because it gets rid of that minus sign.  So we can now write:

We know that reactance is the ratio of the voltage to the current, so we can rearrange further to say:

Since w = 2pf, we can finally write:

So we have derived an expression for the reactance of an inductor.  Now go and have a cup of coffee.  Click HERE to go back to Tutorial 5 B