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


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



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


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


chapter 10 inductance

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