Inductors and calculus
Inductors do not have a stable "resistance"
as conductors do. However, there is a definite mathematical
relationship between voltage and current for an inductor, as
follows:
You should recognize the form of this
equation from the capacitor chapter. It relates one variable
(in this case, inductor voltage drop) to a rate of change
of another variable (in this case, inductor current). Both
voltage (v) and rate of current change (di/dt) are
instantaneous: that is, in relation to a specific point
in time, thus the lower-case letters "v" and "i". As with
the capacitor formula, it is convention to express
instantaneous voltage as v rather than e, but
using the latter designation would not be wrong. Current
rate-of-change (di/dt) is expressed in units of amps per
second, a positive number representing an increase and a
negative number representing a decrease.
Like a capacitor, an inductor's behavior is
rooted in the variable of time. Aside from any resistance
intrinsic to an inductor's wire coil (which we will assume
is zero for the sake of this section), the voltage dropped
across the terminals of an inductor is purely related to how
quickly its current changes over time.
Suppose we were to connect a perfect
inductor (one having zero ohms of wire resistance) to a
circuit where we could vary the amount of current through it
with a potentiometer connected as a variable resistor:
If the potentiometer mechanism remains in a
single position (wiper is stationary), the series-connected
ammeter will register a constant (unchanging) current, and
the voltmeter connected across the inductor will register 0
volts. In this scenario, the instantaneous rate of current
change (di/dt) is equal to zero, because the current is
stable. The equation tells us that with 0 amps per second
change for a di/dt, there must be zero instantaneous voltage
(v) across the inductor. From a physical perspective, with
no current change, there will be a steady magnetic field
generated by the inductor. With no change in magnetic flux (dΦ/dt
= 0 Webers per second), there will be no voltage dropped
across the length of the coil due to induction.
If we move the potentiometer wiper slowly in
the "up" direction, its resistance from end to end will
slowly decrease. This has the effect of increasing current
in the circuit, so the ammeter indication should be
increasing at a slow rate:
Assuming that the potentiometer wiper is
being moved such that the rate of current increase
through the inductor is steady, the di/dt term of the
formula will be a fixed value. This fixed value, multiplied
by the inductor's inductance in Henrys (also fixed), results
in a fixed voltage of some magnitude. From a physical
perspective, the gradual increase in current results in a
magnetic field that is likewise increasing. This gradual
increase in magnetic flux causes a voltage to be induced in
the coil as expressed by Michael Faraday's induction
equation e = N(dΦ/dt). This self-induced voltage across the
coil, as a result of a gradual change in current magnitude
through the coil, happens to be of a polarity that attempts
to oppose the change in current. In other words, the induced
voltage polarity resulting from an increase in
current will be oriented in such a way as to push against
the direction of current, to try to keep the current at its
former magnitude. This phenomenon exhibits a more general
principle of physics known as Lenz's Law, which
states that an induced effect will always be opposed to the
cause producing it.
In this scenario, the inductor will be
acting as a load, with the negative side of the
induced voltage on the end where electrons are entering, and
the positive side of the induced voltage on the end where
electrons are exiting.
Changing the rate of current increase
through the inductor by moving the potentiometer wiper "up"
at different speeds results in different amounts of voltage
being dropped across the inductor, all with the same
polarity (opposing the increase in current):
Here again we see the derivative
function of calculus exhibited in the behavior of an
inductor. In calculus terms, we would say that the induced
voltage across the inductor is the derivative of the current
through the inductor: that is, proportional to the current's
rate-of-change with respect to time.
Reversing the direction of wiper motion on
the potentiometer (going "down" rather than "up") will
result in its end-to-end resistance increasing. This will
result in circuit current decreasing (a negative
figure for di/dt). The inductor, always opposing any change
in current, will produce a voltage drop opposed to the
direction of change:
How much voltage the inductor will produce
depends, of course, on how rapidly the current through it is
decreased. As described by Lenz's Law, the induced voltage
will be opposed to the change in current. With a
decreasing current, the voltage polarity will be
oriented so as to try to keep the current at its former
magnitude. In this scenario, the inductor will be acting as
a source, with the negative side of the induced
voltage on the end where electrons are exiting, and the
positive side of the induced voltage on the end where
electrons are entering. The more rapidly current is
decreased, the more voltage will be produced by the
inductor, in its release of stored energy to try to keep
current constant.
Again, the amount of voltage across a
perfect inductor is directly proportional to the rate of
current change through it. The only difference between the
effects of a decreasing current and an increasing
current is the polarity of the induced voltage. For
the same rate of current change over time, either increasing
or decreasing, the voltage magnitude (volts) will be the
same. For example, a di/dt of -2 amps per second will
produce the same amount of induced voltage drop across an
inductor as a di/dt of +2 amps per second, just in the
opposite polarity.
If current through an inductor is forced to
change very rapidly, very high voltages will be produced.
Consider the following circuit:
In this circuit, a lamp is connected across
the terminals of an inductor. A switch is used to control
current in the circuit, and power is supplied by a 6 volt
battery. When the switch is closed, the inductor will
briefly oppose the change in current from zero to some
magnitude, but will drop only a small amount of voltage. It
takes about 70 volts to ionize the neon gas inside a neon
bulb like this, so the bulb cannot be lit on the 6 volts
produced by the battery, or the low voltage momentarily
dropped by the inductor when the switch is closed:
When the switch is opened, however, it
suddenly introduces an extremely high resistance into the
circuit (the resistance of the air gap between the
contacts). This sudden introduction of high resistance into
the circuit causes the circuit current to decrease almost
instantly. Mathematically, the di/dt term will be a very
large negative number. Such a rapid change of current (from
some magnitude to zero in very little time) will induce a
very high voltage across the inductor, oriented with
negative on the left and positive on the right, in an effort
to oppose this decrease in current. The voltage produced is
usually more than enough to light the neon lamp, if only for
a brief moment until the current decays to zero:
For maximum effect, the inductor should be
sized as large as possible (at least 1 Henry of inductance). |