Lecture 13
Inductance and RL Circuits
Oct. 26, 2011
Joseph Henry
(1797 – 1878)
Writes Henry: "I may however mention one fact which I have
not seen noticed in any work ...when a small battery is
moderately excited... if a wire thirty or forty feet long be used
instead of the short wire, though no spark will be perceptible
when the connection is made, yet when it is broken by
drawing one end of the wire from its cup of mercury, a vivid
spark is produced."
6.3 Inductors (1)
•  An inductor is a passive element designed to store
energy in its magnetic field.
•  An inductor consists of a coil of conducting wire.
Inductance
Photo from Hearst Electronic Products: http://www2.electronicproducts.com/Fundamentals_Inductors_101article-fundamentals_inductors_jul2011-html.aspx
Mathematical Model of Inductors
•  Inductance is the property whereby an inductor exhibits
opposition to the change of current flowing through it, measured
in henrys (H).
di
v=L
dt
and
N2 µ A
L=
l
The unit of inductors is Henry (H), mH (10–3) and µH (10–6).
Remember for the capacitor:
dv
i =C
dt
Inductance & Sparks
The polarity of the voltage is
such as to oppose the change
in current (Lenz’s law).
Inductor Properties
•  The current-voltage relationship of an inductor:
1
i=
L
t
∫ v (t ) d t + i (t )
t0
0
Remember for the capacitor:
•  The power stored by an inductor:
1
w = L i2
2
•  An inductor acts like a short circuit to dc (di/dt = 0) and its
current cannot change abruptly.
Inductor Example
Example 5
The terminal voltage of a 2-H inductor is
v = 10(1-t) V
Find the current flowing through it at t = 4 s
and the energy stored in it within 0 < t < 4 s.
Assume i(0) = 2 A.
Answer:
i(4s) = -18V
w(4s) = 320J
RS = Resistance of
connecting wires
CP = Capacitance
between the coil
windings
RP = Resistance of
the coil windings
Mutual Inductance
Fields are aiding
Fields are opposing
Magnetic flux produced by one coil links the other coil
Inductors Example
Determine vc, iL, and the energy stored in the capacitor and
inductor in the circuit of circuit shown below under dc
conditions.
Answer:
iL = 3A
vC = 3V
wL = 1.125J
wC = 9J
Series Inductors
•  The equivalent inductance of series-connected
inductors is the sum of the individual inductances.
Leq = L1 + L2 + ... + LN
Parallel Inductors
•  The equivalent capacitance of parallel inductors is the
reciprocal of the sum of the reciprocals of the individual
inductances.
1
1
1
1
=
+
+ ... +
Leq L1 L2
LN
Series and Parallel Inductors
Calculate the equivalent inductance for the inductive
ladder network in the circuit shown below:
Answer:
Leq = 25mH
Current and voltage relationship for R, L, C
+
+
+
The Source-Free RL Circuit
•  A first-order RL circuit consists of a inductor L
(or its equivalent) and a resistor (or its equivalent)
By KVL
vL + vR = 0
di
L
+ iR = 0
dt
Inductors law
Ohms law
di
R
= − dt
i
L
i (t ) = I 0 e
−Rt / L
Source-Free RL Circuit
A general form representing a RL
i (t ) = I 0 e
where
•
•
•
−t / τ
L
τ=
R
The time constant τ of a circuit is the time required for the response to decay by a
factor of 1/e or 36.8% of its initial value.
i(t) decays faster for small τ and slower for large τ.
The general form is very similar to a RC source-free circuit.
Comparison between RL and RC circuits
A RL source-free circuit
i (t ) = I 0 e − t / τ
where
τ=
A RC source-free circuit
L
R
v(t ) = V0 e − t /τ
where
τ = RC
Source-Free RL Circuit
The key to working with a source-free RL circuit is
finding:
i (t ) = I 0 e
− t /τ
where
L
τ=
R
1.  The initial voltage i(0) = I0 through the inductor.
2.  The time constant τ = L/R.
Unit-Step Function
•  The unit step function u(t) is 0 for negative values
of t and 1 for positive values of t.
⎧ 0,
u(t ) = ⎨
⎩1,
t<0
t>0
⎧ 0,
u (t − to ) = ⎨
⎩1,
t < to
t > to
⎧ 0,
u (t + to ) = ⎨
⎩1,
t < − to
t > − to
Unit-Step Function
Represent an abrupt change for:
1.  voltage source.
2.  for current source:
Step-response of an RL Circuit
•
The step response of a circuit is its behavior when the excitation is
the step function, which may be a voltage or a current source.
•
Initial current
i(0-) = i(0+) = Io
•
Final inductor current
i(∞) = Vs/R
•
Time constant τ = L/R
t
τ
Vs
Vs
i (t ) = + ( I o − )e u (t )
R
R
−
The Step-Response of a RL Circuit
Three steps to find out the step response of an
RL circuit:
1.  The initial inductor current i(0) at t = 0+.
2.  The final inductor current i(∞).
3.  The time constant τ.
i (t ) = i (∞) + [i (0+) − i (∞)] e
− t /τ
Note: The above method is a short-cut method. You may also
determine the solution by setting up the circuit formula directly
using KCL, KVL , ohms law, capacitor and inductor VI laws.
Step-Response of a RL Circuit
The switch in the circuit shown below has been
closed for a long time. It opens at t = 0.
Find i(t) for t > 0.
i (t ) = 2 + e −10t