**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 = *2p*f*,
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