282 Week 8: Faraday’s Law and Induction
8.8: LR Circuits
From here on out, with rare exceptions we will work withinductorsas (self-inductive) circuit elements
just like capacitors and resistors. We will use “The Solenoid” (idealized) as our architypical inductor,
and we will often pretend that they are made with superconductingwire (as a further idealization)
so that they have no resistance to worry about. Real inductors,of course, are made with many turns
of relatively thin wire and can have substantial (non-negligible) resistance as well as self-inductance.
However, their “resistive” properties can always be considered tobe a resistor in series with a pure
zero resistance inductor, so nothing is lost by the idealization as longas we remember to include
their resistance in our circuits.
I(t)+Switch
t = 0V 0LRFigure 105: The architypical direct currentLRcircuit. We generally assume the switch closes at
timet= 0 with the current in the circuitI(0) = 0.
Let us, then, figure out a simple DCLRcircuit, given in figure 105: an inductor in series with
a resistanceR, whichcouldbe the natural resistance of the inductor itself, or an external resistor,
or the combined resistance of an external resistor and the resistance of the inductor. Note well that
we have generated asymbolfor an inductor in an electrical circuit, the squiggly thing that looks
like a coil/solenoid with many turns of wire. We don’t care much about how many turns it has, or
how long it is, or what its cross-sectional area is, or whether or notit contains a magnetic material
(discussed later). All we care about is thecombinedeffect of all of this, the (self) inductanceL(and
possibly its contribution to the total resistanceRof any branch of a circuit it is in).
Obviously no current flows while the switch is open. We imagine closing the switch at timet= 0.
The battery will drive current through the wire. The resistor will oppose this current (Ohm’s Law),
and the inductor willalsooppose this current as long as it isincreasing(Faraday’s Law). At some
finite timetlater, we expect to find some non-zero current in the circuit, one that is changing in
time, and will use this assumption in analyzing the circuit algebraically.
First, however, let’s see what we can figure out using nothing butverbal reasonanddimensional
analysisinstead of algebra and calculus. We begin, as we see, atI(0) = 0. After averylong time,
we rather expect that the current will arrive at some constant value, at which point the back-voltage
generated by the inductor will be zero The voltage gain from the battery will all drop across the
resistor, suggesting that the current will beI∞=V 0 /R. We therefore expect a currentI(t) that
starts at zero and approachesV 0 /Rbefore beginning the problem, and we might guess that it will
approach this current exponentially. All that is left is guessing the exponential time constant.
Well, we have two parameters to play with: RandL. Ohms are Volts/Ampere. Henries are
Volt-Seconds/Ampere. We want a time constant in seconds, so it looks like:
τ=L
R
(610)
will have units of seconds and is the simplest way of getting such a timeout of the three quantities
thatcouldappear in the answer,V 0 ,LandR. If our life depended on just writing down an expression