What happens when we
switch an inductor on?
Consider this circuit that consists of a DC power supply
of p.d. V, a switch, S, an inductor, L, and a
resistor, R. The circuit looks like this:
When the switch is first closed, the inductor has a
current I passing through it. The magnetic field is being built
up according to the relationship:
This is the combination of Faraday’s and Lenz’s Laws.
Lenz’s Law gives us the minus sign that tells us that the induced EMF
opposes the voltage from the battery. Immediately the switch is turned
on, the induced voltage has the same value as the battery voltage, but
the opposite sign.
We also know that the EMF is related to the current by:
It is a series circuit, so the current is the same
all the way round the circuit. By Kirchhoff II we can write:
The battery voltage V remains the same. The
voltage across the inductor has the same value as the induced voltage,
E. We can
write:
And we can rewrite this as:
The voltage drop across the resistor is IR, so we
can write the Kirchhoff II statement as:
Immediately the switch is turned on, the voltage across
the resistor is 0 as all the voltage is across the inductor.
A short time later the balance is altered, as the induced
emf is reduced, the voltage across the resistor is increased. The sum
of the two voltages is V.
After a couple of seconds or so, the reverse voltage made
by the inductor is zero. Therefore the voltage across the resistor is
the battery voltage, V.
Measuring the transient is not at all easy unless you
have a datalogger. The graph for the induced voltage looks like this:
For the voltage across the resistor, it looks like this:
The graph of current against time follows the same
pattern as the voltage across the resistor to give:
Since the equation has a differential term (dI/dt),
the mathematical trick of calculus tells us that the form of the graphs
will be exponential.
Consider the reverse voltage decay. As with all
exponential functions, there is a constant fraction of the voltage
remaining for each time period. We can say that there is a halflife,
the time taken for the voltage to drop from say 12 V to 6 V. In the
next halflife, the voltage falls from 6 V to 3 V and so on.
As with capacitors, there is a time constant. With
capacitors we saw that the product of ohms and farads was seconds. With
inductance, we see that the henries divided by ohms makes
seconds. Let’s give the time constant the physics code
t (“tau”, a Greek lower case
letter ‘t’).
So we can write an equation for the decay of the reverse
voltage. For a capacitor, we saw that the equation was of the form:
So we can write a similar equation for the inductor.
This tidies up to give us:
There is no reason at all that we can’t write the
equation as:
The term
t
is the time constant, which we work out by dividing the inductance by
the resistance. In this form, the equation is identical to the
equation for the capacitor.
As with capacitance, the time constant can be defined as
the time taken for the voltage to fall to about 37 % of its original
value.
It takes about 5 time constants for the reverse voltage
to be close to zero, i.e., the full supply voltage to be applied. This
is often called the transient time. A real inductor (a coil of
wire) will not be perfect; it will have a definite resistance, so there
will be a time constant. If we have an external resistor, we would
need to take the resistance of the inductor into account by adding it to
the resistance of the inductor.
Worked Example
A universal motor has a resistance of 60 ohms
and an inductance of 100 mH. It is connected to a 230 V DC
supply. Work out:
(a)
The time constant of the drop
in reverse voltage;
(b)
The time needed for the forward
voltage to reach 100 V;
(c)
The time needed for the forward
voltage to be about 230 V. 
Answer
1
t = 100 × 10^{3}
÷ 60 = 1.67 × 10^{3} s
2
Reverse voltage = 130 V
130 = 230 × e^{t/1.67× 10^3}
Þ ln 130 = ln 230 +
t/1.67 × 10^{3}
t = (ln 130 – ln 230) × 1.67 × 10^{3}
t = (0.571) × 1.67 × 10^{3}
= 9.53 × 10^{4} s = 0.95 ms
3
Time = 5
t = 8.35 × 10^{3}
s.

You can see from this that the inductive fall in the
reverse voltage is a very shortterm phenomenon.
When doing a calculation like this, make sure that you
know whether you are dealing with a reverse voltage (back emf) or the
forward voltage. It was easy to put the 100 V into the equation in the
example above.
In reality, it is impossible to measure the reverse
voltage directly. This is because we are assuming that the inductor is
perfect. In reality it is not. It will always have a resistance
associated with it. If we measure the voltage, we would see an
exponential rise in the forward voltage. So we have to infer the
exponential fall in the reverse voltage.
Another problem about inductive rise is that it takes
place over a very short time period. 