Download - Chapter 10 INDUCTANCE
Chapter 10 INDUCTANCE
Recommended Problems:
9,15,16,17,18,19,21,22,23,26,27,29,31,35,47,48,51,52,53,69,
71,72.
In this chapter we are going to discuss the following topics:
Self Inductance
RL Circuits
Energy in Magnetic Field
LC Circuits
Self Inductance R
L
S Ib Iin
Consider the circuit shown in the
Figure.
When the switch is closed, the
current, and so the magnetic field,
through the circuit increases from
zero to a specific value.
The increasing magnetic flux induces an emf. By Lenz's law, this
induced emf opposes the change in flux.
The effect of this induced emf is to retard the change of the
original current, that is, retard its increasing.
The same phenomena occurred when the switch is opened where
the current in this case decreases from a specific value to zero.
The emf induced due to the decreasing of the magnetic flux now
tends to oppose the decreasing of the original current.
This phenomena is called the self induction since the changing
flux through the circuit arises from the circuit itself. The emf
induced due to this phenomena is called the self-induced emf.
If the emf induced in a circuit is due to the changing of the
magnetic flux set up by another circuit we have the mutual
induction phenomena.
To obtain a quantitative description of the self induction, we know
from Faraday's law that the induced emf is proportional to the time
rate of the magnetic flux, i.e.,
dt
dNL
m IBandBmBut
dt
dILL
The constant L is called the self-inductance, or simply the
inductance of the coil. The SI unit of inductance is Henry (H),
which, from the last equation, is equivalent to AV.s1H1
Now comparing the last two equations
dt
dNL
m
dt
dIL
I
NL m
As it is clear, L depends on the geometric features of the coil.
It should be noted that all elements in a circuit
have some inductance but it is too small to be
significant except that of a coil. A coil that has
significant inductance is called inductor, and is
represented in the circuits by the symbol
Example 32.1 Find the inductance of an ideal solenoid of N
turns and length l. Solution: Knowing that, inside the solenoid B is uniform and
given by
Il
NnIB oo A
l
NIBA om 0cos
l
ANA
l
NI
I
NL o
o
2
• Test Your Understanding (1)
A coil with zero resistance has its
ends labeled a and b. The
potential at a is higher than at b.
Which of the following could be
consistent with this situation?
a) The current is increasing and is directed from a to b;
b) The current is decreasing and is directed from a to b;
c) The current is increasing and is directed from b to a;
d) The current is constant and is directed from b to a.
a b
RL Circuits Consider the RL circuit shown.
Suppose that the switch is thrown
to point 1 at t=0 . Applying
Kirchhoff's loop rule to the circuit at
time t we get
It is not difficult to verify that the solution of the differential
equation given in above is
L
1 R
2
S
0dt
dILIR
teII
1max RI
maxwith
R
Land Is the time constant of the RL circuit
It is clear that at t=0, I=0, while as t→, t=Imax. This means that:
that is, the inductor acts as if it were an open circuit at t=0,
and acts as if it were a wire with negligible resistance at →.
If the battery is suddenly removed, by throwing the switch to point
2 in the circuit and applying Kirchhoff's rule again we get
0dt
dILIR
t
eII
max
The variations of I with time are plotted in in the figure shown.
t
q
Im
(a)
t
I
Io
(b)
As it is clear from the graph (a), the
current takes some time to reach its
maximum value.
The graph of Figure (b) tells that the
current takes some time to reach it zero
value.
In another word, the inductor has the
effect to hinder the current from
reaching its final value for some time.
t
I
Closing the switch
Opening the switch
Red line representing I vs t without self induction
Blue line representing I vs t with self induction
If one plot the variation of I vs time when closing or opening a
switch in a circuit with and without an inductor we obtain
Example 32.3 Consider the
circuit shown. Find the time
constant of the circuit, the current in
the circuit at t=2ms , and compare
the P.D across the resistor with that
across the inductor.
Solution
a) The time constant is given by the Equation
b) The current is
6
12 V 30 mH
S
ms0.50.6
100.3 3
R
L
A66.016
121 5
2
max
eeIIt
c) The P.D. across the resistor is given by
teRIIRV oR
1
While the P.D. across the inductor is given by
tt
eRIeI
Ldt
dILV o
oL VRIVV oLR 12
• Test Your Understanding (2)
For the circuit shown in the figure, just
after closing the switch S, across which
of the following is the voltage equal to
the emf of the battery?
(a) R. (b) L.
(c) Both R and L. (d) Neither R nor L.
Referring to the previous question, after closing the S for a long
time, across which of the following is equal to the emf of the
battery?
(a) R. (b) L.
(c) Both R and L. (d) Neither R nor L.
R
L
S
• Test Your Understanding (3)
• Test Your Understanding (4)
The circuit shown includes sinusoidal
voltage source such that the magnetic
field in the inductor is constantly
changing. The inductor is a simple air-
core solenoid. The switch in the circuit
is closed and the lightbulb glows
steadily. An iron rod is inserted into the
interior of the solenoid, which
increases the magnitude of the
magnetic field in the solenoid. As this
happens, the brightness of the lightbulb
(a) increases, (b) decreases,
(c) Goes off, (d) unaffected.
Example (problem 23) The
switch in the figure is open
for t <0 and then closed at time t =0. Find the currents if the circuit at t=0 and a long time after closing the switch.
1 H
8
10 V
I2
S
4
4
I1
I3
Solution
At t=0, the inductor treated as an open circuit 02 I
A25.18
1031 II
After a long time the inductor treated as a wire
7.6
12
324eqR
A5.17.6
101 eqII AII 5.0andA0.1 23
Energy in Magnetic Field
Let us start from the equation of the RL circuit, i.e.,
0dt
dILIR
Multiplying the above equation by I
02 dt
dILIRII
The 1st term represents the power of the battery, while the 2nd
term represents the power delivered to the resistor the 3rd term
represents the power delivered to the inductor, i.e.,
dt
dILI
dt
dUPL IIdLdU
I
0
221 LLUm
LRtoL eRIRI
dt
dUP 222
0
22 dteRIdU LRto
22
2
1
2oo LI
R
LRIU
Example 32.4 Consider once
again the RL circuit shown with the switch is thrown to point 2 after being on position 1 for a long time. Show that all the energy initially stored in the
magnetic field of the inductor appears
as internal energy in the resistor as the
current decays to zero.
1 R
2
S
Solution: It is known that the current decay as
L
Rt
eII o
Consider the circuit shown with the
capacitor is charged with Qmax.
Oscillations in an LC Circuit
C L
S
After closing S the charge will flow
through the inductor. At some time let
the charge in the capacitor to be q and
the current in the inductor to be I. The
total energy in the circuit at this time is
LCtotal UUU 221
2
2LI
C
q
Deriving the above Eq. with respect to time
dt
dIL
dt
dq
Cdt
dUtotal2
21
2
2
1
0butdt
dUtotal 0dt
dIIL
dt
dq
C
q
2
2
andButdt
qd
dt
dI
dt
dqI 0
2
2
dt
qdL
C
q
01
or2
2
qLCdt
qd
This differential Eq.is the SHM equation with its solution
tQq cosmax
To find the constant we know that 00max tatQq
LC
1with tQq cosmax
To find the current we have
tItQdt
dqI sinsin maxmax maxmaxwith QI
Im
I, Q
t
Qm
T
2T
Knowing that
T
2
t
TQq
2cosmax
tT
II2
sinmax
Im
I, Q
t
Qm
T
2T
221
2
2LI
C
QUUU LCtotal
Let’s go back to the energy expression
Substituting for Q and I we get
tLItC
QUtotal 22
max212
2max sincos2
maxmaxmax
1b Q
LCQIut
ttC
Qt
C
Qt
C
QUtotal 22
2max2
2max2
2max sincos
2sin
2cos
2
2max2
12max
2LI
C
QUtotal
UL
UC UL
t
UC Um
Example 32.3 Consider the Circuit
show. First S1 is open and S2 is closed such
that the capacitor is charged. Now if S2 is
opened to remove the battery and then S1
is closed to connect the capacitor with the
inductor. a) Find of the circuit.
b) Find Qmax and Imax.
c) Find I(t) and Q(t). Solution
9 pF
2.81 mH
S1
12 V
S2
a) The frequency is given by
Hz
LC
6
123103.6
1091081.2
11
b) The maximum charge on the capacitor is the initial charge
before opening S2, i.e.,
CCQ 1012 1008.112109max
And for the current we have
AQI 46maxmax 1079.610103.6 1008.1
c) Using the obtained results we get
ttQtq610cos)( 103.6cos1008.1max
ttItq6479.6sin)( 103.6sin10max
• Test Your Understanding (5)
At an instant of time during the oscillations of an LC circuit, the
current is at its maximum value. At this instant, the voltage across
the capacitor:
a) is different from that across
the inductor b) is zero
c) has its maximum value d) is impossible to determine
Referring to the previous question. at an instant of time during the
oscillations of an LC circuit, the current is at its momentarily zero.
At this instant, the voltage across the capacitor:
a) is different from that across
the inductor b) is zero
c) has its maximum value d) is impossible to determine
• Test Your Understanding (6)