Capacitors and RC Circuits Explained

CHAPTER FIVE
CAPACITORS
5.1 INTRODUCTION
So far we have limited our study to resistive circuits. In this chapter, we shall introduce
two new and important passive linear circuit elements: the capacitor and the inductor (the
inductor is discussed in detail in Chapter 7). Unlike resistors, which dissipate energy,
capacitors and inductors do not dissipate but store energy, which can be retrieved at a later
time. For this reason, capacitors and inductors are called storage elements. We begin by
introducing capacitors and describing how to combine them in series or in parallel. Later, we
do the same for inductors.
5.2 CAPACITORS
A capacitor is a passive element designed to store energy in its electric field. Besides
resistors, capacitors are the most common electrical components. Capacitors are used
extensively in electronics, communications, computers, and power systems. For example, they
are used in the tuning circuits of radio receivers and as dynamic memory elements in computer
systems.
A capacitor consists of two conducting plates separated by an insulator (or dielectric).
In many practical applications, the plates may be aluminum foil while the dielectric may be air,
ceramic, paper, or mica.
The amount of charge stored, represented by q, is directly proportional to the applied voltage v
so that
q = Cv (5.1)
where C, the constant of proportionality, is known as the capacitance of the capacitor. The unit
of capacitance is the farad (F), in honor of the English physicist Michael Faraday (1791–1867).
From Eq. (5.1), we may derive the following definition.
Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference
between the two plates, measured in farads (F).
Note from Eq. (5.1) that 1 farad = 1 coulomb/volt.
Although the capacitance C of a capacitor is the ratio of the charge q per plate to the
applied voltage v, it does not depend on q or v. It depends on the physical dimensions of the
capacitor. The capacitance is given by
𝑪 =
𝝐𝑨
𝒅
(5.2)

where A is the surface area of each plate, d is the distance between the plates, and ε is the
permittivity of the dielectric material between the plates. Typically, capacitors have values in
the picofarad (pF) to microfarad (μF) range. Figure 5.1 shows the circuit symbols for fixed
and variable capacitors.
Figure 5.1 Circuit symbols for capacitors: (a) fixed capacitor, (b) variable capacitor.
To obtain the current-voltage relationship of the capacitor, we take the derivative of both sides
of Eq. (5.1). Since
i =dq/dt (5.3)
differentiating both sides of Eq. (5.1) gives
i = C dv/dt (5.4)
The voltage-current relation of the capacitor can be obtained by integrating both sides of Eq.
(5.4). We get
𝒗 =
𝟏
𝑪
∫ 𝒊
𝒕
−∞
𝒅𝒕 (5.5)
or
𝒗 =
𝟏
𝑪
∫ 𝒊
𝒕
𝒕𝟎
𝒅𝒕 + 𝒗(𝒕𝟎) (5.6)
where v(t0) = q(t0)/C is the voltage across the capacitor at time t0.
Eq. (5.6) shows that capacitor voltage depends on the past history of the capacitor current.
Hence, the capacitor has memory—a property that is often exploited.
The instantaneous power delivered to the capacitor is
𝒑 = 𝒗𝒊 = 𝑪𝒗
𝒅𝒗
𝒅𝒕
(5.7)
The energy stored in the capacitor is therefore
𝒘 =
𝟏
𝟐
𝑪𝒗𝟐
or 𝒘 =
𝒒𝟐
𝟐𝑪
(5.8)
Eq. (5.8) represents the energy stored in the electric field that exists between the plates of the
capacitor. This energy can be retrieved, since an ideal capacitor cannot dissipate energy
We should note the following important properties of a capacitor:
1. Note from Eq. (5.4) that when the voltage across a capacitor is not changing with time (i.e.,
dc voltage), the current through the capacitor is zero. Thus,
A capacitor is an open circuit to dc.
2. The voltage on the capacitor must be continuous.
The voltage on a capacitor cannot change abruptly.

The capacitor resists an abrupt change in the voltage across it.
3. The ideal capacitor does not dissipate energy. It takes power from the circuit when storing
energy in its field and returns previously stored energy when delivering power to the circuit.
4. A real, nonideal capacitor has a parallel-model leakage resistance. The leakage resistance
may be as high as 100 MΩ and can be neglected for most practical applications.
Example 5.1:
(a) Calculate the charge stored on a 3-pF capacitor with 20 V across it.
(b) Find the energy stored in the capacitor.
Solution:
(a) Since q = Cv, q = 3 × 10−12
× 20 = 60 pC
(b) The energy stored is 𝒘 =
𝟏
𝟐
𝑪𝒗𝟐
=
𝟏
𝟐
× 𝟑 × 𝟏𝟎−𝟏𝟐
× 𝟒𝟎𝟎 = 𝟔𝟎𝟎 𝐩𝐉
Example 5.2:
The voltage across a 5-μF capacitor is v(t) = 10 cos 6000t V Calculate the current through it.
Solution:
By definition, the current is
𝒊(𝒕) = 𝑪
𝒅𝒗
𝒅𝒕
= 𝟓 × 𝟏𝟎−𝟔 𝒅
𝒅𝒕
(𝟏𝟎 𝒄𝒐𝒔 𝟔𝟎𝟎𝟎𝒕)
= −5 × 10−6
× 6000 × 10 sin 6000t = −0.3 sin 6000t A
Practice problems:
1-What is the voltage across a 3-μF capacitor if the charge on one plate is 0.12 mC? How much
energy is stored?
Answer: 40 V, 2.4 mJ.
2-If a 10-μF capacitor is connected to a voltage source with v(t) = 50 sin 2000t V determine the
current through the capacitor.
Answer: cos 2000t A.
5.3 SERIES AND PARALLEL CAPACITORS
We know from resistive circuits that series-parallel combination is a powerful tool for
reducing circuits. This technique can be extended to series-parallel connections of capacitors,
which are sometimes encountered. We desire to replace these capacitors by a single equivalent
capacitor Ceq.
First we obtain the equivalent capacitor Ceq of N capacitors in parallel,

Figure 5.2 (a) Parallel-connected N capacitors, (b) equivalent circuit for the parallel capacitors.
Ceq = C1 + C2 + C3 +· · ·+CN (5.9)
The equivalent capacitance of N parallel-connected capacitors is the sum of the
individual capacitances.
We observe that capacitors in parallel combine in the same manner as resistors in series.
Now we will obtain Ceq of N capacitors connected in series
Figure 5.3 (a) Series-connected N capacitors, (b) equivalent circuit for the series capacitor.
Where
𝟏
𝑪𝒆𝒒
=
𝟏
𝑪𝟏
+
𝟏
𝑪𝟐
+
𝟏
𝑪𝟑
+· · · +
𝟏
𝑪𝑵
(5.10)
The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the
reciprocals of the individual capacitances.
Note that capacitors in series combine in the same manner as resistors in parallel. For N = 2
(i.e., two capacitors in series), Eq. (5.10) becomes
𝟏
𝑪𝒆𝒒
=
𝟏
𝑪𝟏
+
𝟏
𝑪𝟐
Or 𝑪𝒆𝒒 =
𝑪𝟏𝑪𝟐
𝑪𝟏 + 𝑪𝟐
(5.11)
Example 5.6:
Find the equivalent capacitance seen between terminals a and b of the circuit in Fig. 5.4.
Figure 5.4 For Example 6.6.
Solution:
The 20-μF and 5-μF capacitors are in series; their equivalent capacitance is

𝟐𝟎 × 𝟓
𝟐𝟎 + 𝟓
= 𝟒 𝝁𝑭
This 4-μF capacitor is in parallel with the 6-μF and 20-μF capacitors; their combined
capacitance is
4 + 6 + 20 = 30 μF
This 30-μF capacitor is in series with the 60-μF capacitor. Hence, the equivalent capacitance
for the entire circuit is
𝑪𝒆𝒒 =
𝟑𝟎 × 𝟔𝟎
𝟑𝟎 + 𝟔𝟎
= 𝟐𝟎 𝝁𝑭
Practice problems:
1- Find the equivalent capacitance seen at the terminals of the circuit in Figure below.
Answer: 40 μF.
5.4 First Order RC Circuit
Now that we have considered the three passive elements (resistors, capacitors, and
inductors, the inductor is discussed in detail in Chapter 7), we are prepared to consider circuits
that contain various combinations of two or three of the passive elements.
We carry out the analysis of RC and RL circuits by applying Kirchhoff’s laws, as we did
for resistive circuits. The only difference is that applying Kirchhoff’s laws to purely resistive
circuits results in algebraic equations, while applying the laws to RC and RL circuits produces
differential equations, which are more difficult to solve than algebraic equations.
The differential equations resulting from analyzing RC and RL circuits are of the first order.
Hence, the circuits are collectively known as first-order circuits.
A first-order circuit is characterized by a first-order differential equation.
The two types of first-order circuits and the two ways of exciting them add up to the four
possible situations we will study in this chapter.
5.4 THE SOURCE-FREE RC CIRCUIT
A source-free RC circuit occurs when its dc source is suddenly disconnected. The energy
already stored in the capacitor is released to the resistors.

Figure 5.5 A source-free RC circuit.
Consider a series combination of a resistor and an initially charged capacitor, as shown in
Fig. 5.5. Our objective is to determine the circuit response, which, for pedagogic reasons, we
assume to be the voltage v(t) across the capacitor. Since the capacitor is initially charged, we
can assume that at time t = 0, the initial voltage is
v(0) = V0 (5. 12)
with the corresponding value of the energy stored as
𝒘(𝟎) =
𝟏
𝟐
𝑪𝑽𝟎
𝟐
(5.13)
Applying KCL at the top node of the circuit in Fig. 5.5,
iC + iR = 0 (5.14)
By definition, iC = C dv/dt and iR = v/R. Thus,
𝑪
𝒅𝒗
𝒅𝒕
+
𝒗
𝑹
= 𝟎 (5.15)
This is a first-order differential equation, since only the first derivative of v is involved. After
solve it, the capacitor voltage is
𝒗(𝒕) = 𝑽𝟎𝒆−𝒕/𝑹𝑪
(5.16)
This shows that the voltage response of the RC circuit is an exponential decay of the
initial voltage. Since the response is due to the initial energy stored and the physical
characteristics of the circuit and not due to some external voltage or current source, it is called
the natural response of the circuit.
The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the
circuit itself, with no external sources of excitation.
The natural response is illustrated graphically in Fig. 5.6. Note that at t = 0, we have the
correct initial condition as in Eq. (5.12). As t increases, the voltage decreases toward zero. The
rapidity with which the voltage decreases is expressed in terms of the time constant, denoted by
the lower case Greek letter tau, τ.
The time constant of a circuit is the time required for the response to decay by a factor of 1/e or
36.8 percent of its initial value.

Figure 5.6 The voltage response of the RC circuit.
This implies that at t = τ , Eq. (5.16) becomes
𝑽𝟎𝒆−𝝉/𝑹𝑪
= 𝑽𝟎𝒆−𝟏
= 𝟎. 𝟑𝟔𝟖𝑽𝟎
or
τ = RC (5.17)
In terms of the time constant, Eq. (5.16) can be written as
𝒗(𝒕) = 𝑽𝟎𝒆−𝒕/𝝉
(5.18)
Observe from Eq. (5.17) that the smaller the time constant, the more rapidly the voltage
decreases, that is, the faster the response. This is illustrated in Fig. 5.7. A circuit with a small
time constant gives a fast response in that it reaches the steady state (or final state) quickly due
to quick dissipation of energy stored, whereas a circuit with a large time constant gives a slow
response because it takes longer to reach steady state. At any rate, whether the time constant is
small or large, the circuit reaches steady state in five time constants.
Figure 5.7 Plot of v/V0 = e−t/τ for various values of the time constant.
With the voltage v(t) in Eq. (5.18), we can find the current iR(t),
𝒊𝑹(𝒕) =
𝒗(𝒕)
𝑹
=
𝑽𝟎
𝑹
𝒆−𝒕/𝝉
(5.19)
The power dissipated in the resistor is
𝒑(𝒕) = 𝒗. 𝒊𝑹 =
𝑽𝟎
𝟐
𝑹
𝒆−𝟐𝒕/𝝉
(5.20)
The energy absorbed by the resistor up to time t is
𝒘𝑹(𝒕) =
𝟏
𝟐
𝑪𝑽𝟎
𝟐
(𝟏 − 𝒆−𝟐𝒕/𝝉
), 𝝉 = 𝑹𝑪 (5.21)
Notice the energy that was initially stored in the capacitor is eventually dissipated in the
resistor. In summary:
The Key to Working with a Source - free RC Circuit is Finding:
1. The initial voltage v(0) = V0 across the capacitor.

With these two items, we obtain the response as the capacitor voltage vC(t) = v(t) =
v(0)e−t/τ
. Once the capacitor voltage is first obtained, other variables (capacitor current iC,
resistor voltage vR, and resistor current iR) can be determined. In finding the time constant τ =
RC, R is often the Thevenin equivalent resistance at the terminals of the capacitor; that is, we
take out the capacitor C and find R = RTh at its terminals.
Example 5.9: The switch in the circuit in Fig. 5.8 has been
closed for a long time, and it is opened at t = 0. Find v(t) for
t ≥ 0. Calculate the initial energy stored in the capacitor.
Solution:
For t < 0, the switch is closed; the capacitor is an open
circuit to dc, as represented in Fig. 5.9(a). Using voltage division
𝒗𝑪(𝒕) =
𝟗
𝟗 + 𝟑
(𝟐𝟎) = 𝟏𝟓 𝑽, 𝒕 < 𝟎
Since the voltage across a capacitor cannot change
instantaneously, the voltage across the capacitor at t = 0−
is the
same at t = 0, or
vC(0) = V0 = 15 V
For t > 0, the switch is opened, and we have the RC circuit
shown in Fig. 5.9(b). [Notice that the RC circuit in Fig. 5.9(b)
is source free; the independent source in Fig. 5.8 is needed to
provide V0 or the initial energy in the capacitor.] The 1-Ω and
9-Ω resistors in series give
Req = 1 + 9 = 10 Ω
The time constant is
τ = ReqC = 10 × 20 × 10−3
= 0.2 s
Thus, the voltage across the capacitor for t ≥ 0 is
v(t) = vC(0)e−t/τ
= 15e−t/0.2
V
or v(t) = 15e−5t
V
The initial energy stored in the capacitor is
𝒘𝑪(𝟎) =
𝟏
𝟐
𝑪𝒗𝑪
𝟐
(𝟎) =
𝟏
𝟐
× 𝟐𝟎 × 𝟏𝟎−𝟑
× 𝟏𝟓𝟐
= 𝟐. 𝟐𝟓 𝑱
Figure 5.9 For Example 5.9:
(a) t < 0, (b) t > 0.

Practice problems:
(1)Refer to the circuit in Figure below. Let vC(0) = 30 V. Determine vC, vx , and io for t ≥ 0.
Answer: 30e−0.25t
V, 10e−0.25t
V, −2.5e−0.25t
A.
5.5 STEP RESPONSE OF AN RC CIRCUIT
When the dc source of an RC circuit is suddenly applied, the voltage or current source
can be modeled as a step function, and the response is known as a step response.
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.
The step response is the response of the circuit due to a sudden application of a dc voltage or
current source.
Figure 5.10 An RC circuit with voltage step input.
Consider the RC circuit in Fig. 5.10(a) which can be replaced by the circuit in Fig.
5.10(b), where Vs is a constant, dc voltage source. Again, we select the capacitor voltage as the
circuit response to be determined.
We assume an initial voltage V0 on the capacitor, although this is not necessary for the
step response. Since the voltage of a capacitor cannot change instantaneously,
v(0−
) = v(0+
) = V0 (7.22)
where v(0−) is the voltage across the capacitor just before switching and v(0+) is its voltage
immediately after switching. Applying KCL, we have
𝑪
𝒅𝒗
𝒅𝒕
+
𝒗 – 𝑽𝒔.𝒖(𝒕)
𝑹
= 𝟎 (7.23)
where v is the voltage across the capacitor.
Thus,
𝑣(𝑡) = {
𝑉0 , 𝑡 < 0
𝑉𝑠 + (𝑉0 − 𝑉𝑠)𝑒−𝑡/𝜏
, 𝑡 > 0
(5.24)

This is known as the complete response of the RC circuit to a
sudden application of a dc voltage source, assuming the capacitor is
initially charged. The reason for the term “complete” will become
evident a little later. Assuming that Vs > V0, a plot of v(t) is shown
in Fig. 5.11.
If we assume that the capacitor is uncharged initially, we
set V0 = 0 in Eq. (5.24) so that
𝑣(𝑡) = {
0 , 𝑡 < 0
𝑉𝑠 (1 − 𝑒−𝑡/𝜏
), 𝑡 > 0
(5.25)
Rather than going through the derivations above, there is a systematic approach—or
rather, a short-cut method—for finding the step response of an RC or RL circuit. Let us
reexamine Eq. (5.24), which is more general than Eq. (5.25). It is evident that v(t) has two
components. Thus, we may write
v = vf + vn (5.26)
We know that vn is the natural response of the circuit, as discussed in Section 5.2. Now,
vf is known as the forced response because it is produced by the circuit when an external
“force” is applied (a voltage source in this case).
The natural response or transient response is the circuit’s temporary response that will die out
with time.
The forced response or steady-state response is the behavior of the circuit a long time after an
external excitation is applied.
The complete response of the circuit is the sum of the natural response and the forced response.
Therefore, we may write Eq. (5.24) as
v(t) = v(∞) + [v(0) − v(∞)]e−t/τ
(5.27)
where v(0) is the initial voltage at t = 0+
and v(∞) is the final or steady state value. Thus, to find
the step response of an RC circuit requires three things:
Note that if the switch changes position at time t = t0 instead of at t = 0, there is a time
delay in the response so that Eq. (5.27) becomes
v(t) = v(∞) + [v(t0) − v(∞)]e−(t−t0)/τ
(5.28)
where v(t0) is the initial value at t = t+0
. Keep in mind that Eq. (5.27) or (5.28) applies only to
step responses, that is, when the input excitation is constant.
Figure 5.11 Response of an RC circuit with
initially charged capacitor.
1. The initial capacitor voltage v (0).
2. The final capacitor voltage v (∞).
3. The time constant τ.

Example 5.10: The switch in Fig. 5.12 has been in position A for a long time. At t = 0, the
switch moves to B. Determine v(t) for t > 0 and calculate its value at t = 1 s and 4 s.
Solution:
For t < 0, the switch is at position A. Since v is the
same as the voltage across the 5-kΩresistor, the voltage
across the capacitor just before t = 0 is obtained by
voltage division as
𝒗(𝟎−
) =
𝟓
𝟓 + 𝟑
(𝟐𝟒) = 𝟏𝟓 𝑽
Using the fact that the capacitor voltage cannot change instantaneously,
v(0) = v(0−
) = v(0+
) = 15 V
For t > 0, the switch is in position B. The Thevenin resistance connected to the capacitor is
RTh = 4 kΩ, and the time constant is
τ = RThC = 4 × 103
× 0.5 × 10−3
= 2 s
Since the capacitor acts like an open circuit to dc at steady state, v(∞) = 30 V. Thus,
v(t) = v(∞) + [v(0) − v(∞)]e−t/τ
= 30 + (15 − 30)e−t/2
= (30 − 15e−0.5t
) V
At t = 1, v(1) = 30 − 15e−0.5
= 20.902 V
At t = 4, v(4) = 30 − 15e−2
= 27.97 V
Notice that the capacitor voltage is continuous while the resistor current is not.
PRACTICE PROBLEM:
(1) Find v(t) fort > 0 in the circuit in Figure below. Assume the switch has been open for a long
time and is closed at t = 0. Calculate v(t) at t = 0.5.
Answer: −5 + 15e−2t
V, 0.5182 V.

SIX
CHAPTER
MAGNETIC CIRCUITS
6.1 Introduction
All of us are familiar with a magnet. It is a piece of solid body which processes a property
of attracting iron pieces and pieces of some other metals. This is called natural magnet. While
as per the discovery of scientist oersted we can have an electromagnet. Scientist Oersted stated
that any current carrying conductor is always surrounded by a magnetic field. In this chapter we
shall study laws of magnetism, magnetic field due to current carrying conductor,
magnetomotive force, simple series and parallel magnetic circuits.
6.2 Molecular Theory of Magnetization
Not only magnetized but materials like iron, steel are
also complete magnets according to molecular theory. All
materials consist of small magnets internally called
molecular magnets. In unmagnetized materials such magnets
arrange themselves in close d loops as shown in the Fig. 6.1.
So at any joint, effective strength at a point is zero,
due to presence of two unlike poles. Such poles cancel each
other's effect. But if magnetized material is considered or
unmagnetised material subjected to magnetizing force is
considered, then such small molecular magnets arrange
themselves in the direction of magnetizing force, as shown
in the Fig. 6.2.
6.3 laws of Magnetism
There are two fundamental laws of magnetism which are as follows:
Law 1: it state that" like magnetic poles repel and unlike poles attract each other"
Law 2: This law is experimentally proved by scientised Coulomb and hence also known as
Coulomb's law.
The force (F) exerted by one pole on the other pole is,
a) Directly proportional to the product of the pole strengths,
b) Inversely proportional to the square of the distance between them, and
c) Nature of medium surrounding the poles.
Mathematically this law can be expressed as,
Figure 6.1 Molecular magnets in
unmagnetised material
Figure 6.2 Magnetised piece of material

𝑭 =∝
𝑴𝟏𝑴𝟐
𝒅𝟐
M1 and M2 are pole strengths of the poles while d is distance between the poles
𝑭 =
𝑲𝑴𝟏𝑴𝟐
𝒅𝟐 (6.1)
where K depends on the nature of the surroundings and called permeability.
6.4 Magnetic Field
We have seen that magnet has its influence on the surrounding medium. The region
around a magnet within which the influence of the magnet can be experienced is called
magnetic field. Existence of such field can be experienced with the help of compass needle,
iron or pieces of metals or by bringing another magnet in vicinity of a magnet.
6.4.1 Magnetic Lines of Force
In the region surrounding a permanent magnet there
exists a magnetic field, which can be represented by
magnetic flux lines similar to electric flux lines. Magnetic
flux lines, however, do not have origins or terminating
points as do electric flux lines but exist in continuous loops,
as shown in Fig. 6.3. The symbol for magnetic flux is the
Greek letter Φ (phi).
6.4.2 Direction of Magnetic Field
The magnetic flux lines radiate from the north pole to the south pole, returning to the
north pole through the metallic bar. Note the equal spacing between the flux lines within the
core and the symmetric distribution outside the magnetic material. If unlike poles of two
permanent magnets are brought together, the magnets will attract, and the flux distribution will
be as shown in Fig. 6.4. If like poles are brought together, the magnets will repel, and the flux
distribution will be as shown in Fig. 6.5.
FIG. 6.3 Flux distribution for a
permanent magnet.
FIG. 6.4Flux distribution for two
adjacent, opposite poles.
FIG. 6.5 Flux distribution for two
adjacent, like poles.

6.4.3 Properties of Line of Force
Though the lines of force are imaginary, with the help of them
various magnetic effects can be explained very conveniently. Let
us see the various properties of these lines of force.
1) Lines of force are always originating on an N-pole and
terminating on an S-pole, external to the magnet.
2) Each line forms a cloned loop as shown In the Fig. 6.6.
3) Lines of force never intersect each other.
4) The lines of force are like stretched rubberbands and always try to contract in length.
5) The lines of force which are parallel and travelling in the same direction repel each other.
6) Magnetic lines of force always prefer a path offering least opposition.
6.5 Magnetic Flux (Φ)
The total number of lines of force existing in a particular magnetic field is called
magnetic flux. Lines of force can be called lines of magnetic flux. The unit of flux is weber and
flux is denoted by symbol (Φ). The unit weber is denoted as Wb.
1 Weber = 108
lines of force
6.6 Pole Strength
We have seen earlier that force between the poles depends on the pole strengths. As we
are now familiar with flux, we can have idea of pole strength. Every pole has a capacity to
radiate or accept certain number of magnetic lines of force i.e. magnetic flux which is called its
strength.
Unit of pole strength is weber as pole strength is directly related to flux i.e. lines of force.
So when we say unit N-pole, it means a pole is having pole strength of 1 weber.
6.7 Magnetic Flux Density
In the SI system of units, magnetic flux is measured in webers and has the symbol Φ.
The number of flux lines per unit area is called the flux density, is denoted by the capital letter
B, and is measured in teslas. Its magnitude is determined by the following equation:
B = teslas (T)
𝐁 =
𝚽
𝐚
Φ = webers (Wb) (6.2)
a = square meters (m2
)
Fig. 6.6 Line of force complete the
closed path.

where Φ is the number of flux lines passing through the area a (Fig. 6.7). By definition,
1 T = 1 Wb/m2
Example 6.1: For the core of Fig. 6.8, determine the flux density B in teslas.
Solution:
𝑩 =
𝜱
𝒂
=
𝟔×𝟏𝟎−𝟓𝑾𝒃
𝟏.𝟐×𝟏𝟎−𝟑𝒎𝟐
= 𝟓 × 𝟏𝟎−𝟐
𝑻
Practice problems: In Fig. 6.8, if the flux density is 1.2 T and the area is 0.25 in2
, determine
the flux through the core.
Answer: Φ = 1.936 × 10-4
Wb
6.8 Magnetic Field Strength (H)
This gives quantitative measure of strongness or weakness of the magnetic field. Note
that pole strength and magnetic field strength are different. This can be defined as 'the force
experienced by a unit N-pole (i.e. N-pole with 1 Wb of pole strength) when placed at any point
in a magnetic flied is known as magnetic field strength at that point.
It is denoted by H and its unit is newtons per weber i.e. (N/Wb) or amperes per metre (A/m) or
ampere turns per metre (AT/ m). The mathematical expression for calculating magnetic field
strength is,
𝑯 =
𝒂𝒎𝒑𝒆𝒓𝒆 𝒕𝒖𝒓𝒏𝒔
𝒍𝒆𝒏𝒈𝒕𝒉
=
𝑵𝑰
𝒍
𝑨𝑻/𝒎 (6.3)
It be seen that, the magnetic field strength is independent of the type of core material.
Example 6.2: A pole having strength of 0.5×10-3
Wb is placed in a magnetic field at a distance
of 25cm from another pole. It is experiencing a force of 0.5 N. Assume constant of medium
(
𝟏
𝟑𝟔𝝅𝟐×𝟏𝟎−𝟕
). Determine,
a) Magnetic field strength at the point. b) The strength of other pole.
c) Distance at which force experienced will be doubled.
Solution: The given values are,
M1=0.5×10-3
Wb, d =25 cm = 0.25 m, F = 0.5 N, K = (
𝟏
𝟑𝟔𝝅𝟐×𝟏𝟎−𝟕
) = 28144.773
a) Magnetic field strength.
H =
𝐍𝐞𝐰𝐭𝐨𝐧
𝐖𝐛
=
𝐅𝐨𝐫𝐜𝐞 𝐞𝐱𝐩𝐞𝐫𝐢𝐞𝐧𝐜𝐞𝐝
𝐏𝐨𝐥𝐞 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡
=
𝟎.𝟓
𝟎.𝟓×𝟏𝟎−𝟑
=1000 N/Wb
b) According to Coulomb's law.
𝐅 =
𝑲𝑴𝟏𝑴𝟐
𝒅𝟐
⟹ 𝟎. 𝟓 =
𝟎.𝟓×𝟏𝟎−𝟑×𝟐𝟖𝟏𝟒𝟒.𝟕𝟕𝟑×𝐌𝟐
(𝟎.𝟐𝟓)𝟐
Figure 6.7 Concept of magnetic f1ux density.
Fig. 6.8.
a
a

M2 = 2.22x 10-3
Wb ... pole strength of other pole
c) F = 1 N
𝟏 =
𝟎.𝟓×𝟏𝟎−𝟑×𝟐𝟖𝟏𝟒𝟒.𝟕𝟕𝟑×𝟐.𝟐𝟐𝐱 𝟏𝟎−𝟑
𝒅𝟐
d = 0.1767 m = 17.57 cm
6.9 Magnetic Effect of an Electric Current (Electromagnets)
When a coil or a conductor carries a current, it produces the magnetic flux around it.
Then it starts behaving as a magnet. Such a current carrying coil or conductor is called an
electromagnet. This is due to magnetic effect of an electric current. The flux produced and the
flux density can be controlled by controlling the magnitude of current. The direction and shape
of the magnetic field around the coil or conductor depend on the direction of current and shape
of the conduct or through which it is passing.
6.9.1 Magnetic Field due to Straight Conductor
When a straight conductor carries a current, it produces a magnetic field all along its
length. The lines of force are in the form of concentric circles in the planes right angles to the
conductor as shown Fig 6.9. Then current through such conductor will either come out of paper
indicates as cross or will go into the plane of the paper indicates as dot.
.
6.9.1.1 Rules to Determine Direction of Flux around Conductor
1) Right Hand Thumb Rule: It states that, hold the current carrying conductor in the right
hand such that the thumb pointing in the direction of current and parallel to the conductor, then
curled fingers point in the direction of the magnetic field or flux around it. The Fig. 6.10
explained the rule.
Fig. 6.10 Right hind thumb rule.
Fig. 6.9 Magnetic field conductor
due to straight conductor.

2) Corkscrew Rule: Imagine a right handed so- to he along the conductor carrying current
with its axis parallel to the conductor and tip pointing in the direction of the current flow.
Then the direction of the magnetic field is given by the direction in which the screw must
be turned so as to advance in the direction of the current
6.9.2 Magnetic Field due to Circular Conductor i.e. Solenoid
A solenoid is an arrangement in which long conductor is
wound with number of turns close together to form a coil. The
axial length of conductor is much more than the diameter of
turns. The part or element around which the conductor is wound
is called as core of the solenoid. Core may be air or may be
some magnetic material. Solenoid with a steel or iron core is
shown in Fig. 6.11.
The rule to determine the direction of flux and pole of the magnet fanned:
1) The right hand thumb rule: Hold the solenoid in the right hand such that curled fingers
point in the direction of the current through the curled conductor, then the outstretched thumb
along the axis of the solenoid point to the North pole of the solenoid or point the direction of
flux inside the core. This is shown in Fig. 6.12
Fig. 6.12 Direction of flux around a solenoid.
In case of toroid, the core is circular and hence using
right hand thumb rule, the direction of flux in the core,
due to current carrying conductor can be determined.
This is shown in the Fig. 6.13.
2) Corkscrew rule: If axis of the screw is placed along
the axis of the solenoid and if screw is turned in the
direction of the current, then it travels toward the N-pole or
in the direction of the magnetic field inside the solenoid.
6.10 Nature of Magnetic Field of Long Straight Conductor
Fig. 6.13 Direction of flux in toroid.
Fig. 6.11 Solenoid.

Consider a conductor carrying current I ampere of length 'l ' meters. Consider point P in
the vicinity of such conductor. There will be influence of magnetic field on point P which can
be quantified by magnetic field strength H at point P. the magnitude of such magnetic field
strength 'H' can be calculated by using the expression,
𝐇 =
𝐈
𝟒𝛑𝐝
(𝐬𝐢𝐧 𝜶𝟏 + 𝐬𝐢𝐧 𝜶𝟐) (6.4)
For infinitely long conductor i.e. length 'l' is very large, and then α1 and
α2 tend to 90o
.
𝐇 =
𝐈
𝟐𝛑𝐝
𝑨/𝒎
If such 'N' conductors lire grouped together to form a coil or a cable
then field strength due to current I passing through each conductor of
the group can also be calculated by using same expression. The only
change will be the field strength calculated above will get multiplied by
'N'.
𝐇 =
𝐍𝐈
𝟐𝛑𝐝
𝑨𝑻/𝒎
where AT/m mean ampere turns per metre
6.10.1 Magnetic Field Strength due to a Long Solenoid
Similar to the case of long straight conductor, we can decide field strength along the axis
of a long solenoid. Such field strength depends on the number of turns of the conductor around
the core and magnitude of current I passing through the conductor. If 'l' is the length of the
solenoid in metres then H can be determined by the expression,
𝐇 =
𝑵𝑰
𝒍
𝑨𝑻/𝒎 (6.5)
Example 6.3: A current of 2 amps is flowing through each of the conductors in a coil
containing 15 such conductor. If a point pole of unit strength is placed at a perpendicular
distance of 10 cm from the coil, determine the field intensity at that point,
Solution: I =2 A, N =15, d = 10cm = 0.l m.
𝑯 =
𝑵𝑰
𝟐𝝅𝒅
=
𝟏𝟓×𝟐
𝟐𝝅×𝟎.𝟏
= 𝟒𝟕. 𝟕𝟒 𝑨𝑻/𝒎
Example 6.4: A solenoid of 100 cm is wound on a brass tube. If the current through the coil is
0.5 A, calculate the number of turns necessary over the solenoid to produce a field strength of
500 AT/m at the center of the coil.
Solution: l =Length of coil = 100 cm = 1 m, N = Number of turns, I = Current = 0.5 A
Fig. 6.14 Field strength
due to long conductor.
Fig. 6.15 Field strength
due to N conductor.

𝐇 =
𝐍𝑰
𝒍
𝑨𝑻/𝒎 ⟹ 𝟓𝟎𝟎 =
𝐍×𝟎.𝟓
𝟏
⟹ N = 1000 turns
6.11 Force on a Current Carrying Conductor in a Magnetic Field
When a current carrying conductor placed in a magnetic field, that conductor will
experience a mechanical force due to the interaction between the magnetic field and magnetic
field produced by the current carrying conductor. The magnetic field in which it is placed has a
flux pattern as shown in the Fig. 6.16(a). Now current carrying conductor also produces its
own magnetic field around it. Assume the current direction away from observer. This is
clockwise as shown in the Fig. 6.16 (b).
(a) Flux due to magnet (b) Flux due to current carrying conductor
Fig. 6.16 Current carrying conductor in a magnet field.
Now there is presence of two magnetic fields namely due to permanent magnet and due to
current carrying conductor. These two fluxes interact with each other. Such interaction is
shown in the Fig. 6.17 (a). This interaction as seen is in such a way that on one side of the
conductor the two lines help each other, while on other side the two tries to cancel each other.
Due to such interaction on one side of the conductor, there is accumulation of flux lines
(gathering of the flux lines) while on the other side there is weakening of the flux lines.
The resultant flux pattern around the conductor is shown in the Fig. 6.17 (b).
(a) Presence of the two fluxes (b) Resultant flux pattern
Fig. 6.17 Interaction of the two flux lines.
According to properties of the flux lines,
these flux lines will try to shorten themselves.
While doing so, flux lines which are gathered will
exert force on the conductor. So conductor
experiences a mechanical force from high flux

lines area towards low flux Lines area. This is the basic principle on which D.C. electric motors
work and hence also called motoring action.
6.11.1 Fleming's Left Hand Rule
The direction of the force experienced by the current carrying conductor placed in magnetic
field can be determined by a rule called ' Fleming's left Hand Rule'. The rule is explained in
the diagrammatic form in the Fig. 6.18.
6.11.2 Magnitude of Force Experienced by the
Conductor
The magnitude of the force experienced by the conductor depends on the following factors,
1) Flux density (B) of the magnetic field measured in Wb/m2
i.e. Tesla.
2) Magnitude of the current I passing through the conductor in Amperes.
3) Active length 'l' of the conductor in metres.
4) The angle (θ) between the conductor and the flux line of the magnetic field in degree.
The active length of the conductor is that part of the conductor which is actua1ly under the
influence of magnetic field.
The force F is given by,
F = B I l sin θ Newtons (6.6)
Figure 6.19 show different angle between the conductor and the field lines
Fig. 6.19 Force on current carrying conductor.
**How can change the direction of force on current carrying conductor?
Example 6.5: Calculate the force experienced by the conductor if 20 cm long, carrying 50
amperes, placed at right angles to the lines of force of flux density 10×10-3
Wb/m2
.
Solution: The force experienced is given by,
F = B I l sin θ where sin (θ) = 1 as θ = 90o
B= Flux density = 10×10-3
Wb/m2
, l = Active length = 20 cm = 0.2 m, I = current = 50A
F = 10×10-3
×50×0.2 = 0.1 N
6.12 Permeability
If cores of different materials with the same physical dimensions are used in the
electromagnet described in Section 6.9, the strength of the magnet will vary in accordance with
Fig. 6.18 Fleming's left hand rule

the core used. Materials in which flux lines can readily be set up are said to be magnetic and to
have high permeability.
The permeability (µ): is a measure of the ease with which magnetic flux lines can be
established in the material. It is similar in many respects to conductivity in electric circuits.
Diamagnetic: Materials that have permeabilities slightly less than that of free space.
Paramagnetic: Materials that have permeabilities slightly greater than that of free space.
Ferromagnetic: Materials that have very high permeabilities such as iron, nickel, steel, cobalt,
and alloys of these metals
For any magnetic material, there are two permeabilities,
i) Absolute permeability, ii) Relative permeability.
Absolute Permeability (µ): The ratio of magnetic flux density B in a particular medium (other
than vacuum or air) to the magnetic field strength H producing that flux density is called
absolute permeability of that medium. The absolute permeability can be expressed
mathematically as,
𝝁 =
𝑩
𝑯
𝑯/𝒎 (6.7)
Permeability of Free Space or Vacuum (µo): If the magnet is placed in a free space or
vacuum or in air then the ratio of flux density B and magnetic field strength H is called
Permeability of free space or vacuum or air.
The permeability of free space µo (vacuum) is
µo =
𝑩
𝑯
in vacuum = 4π × 10-7
Wb/A.m or H/m
Relative Permeability (µr): The relative permeability is defined as the ratio of flux density
produced in a medium (other than free space) to the flux density produced in tree space, under
the influence of same magnetic field strength and under identical conditions.
𝝁𝒓 =
𝑩
𝑩𝟎
where 𝐇 is same
𝝁𝒓 =
𝝁
𝝁𝟎
How? (6.8)
In general, for ferromagnetic materials, µr ≥100, and for nonmagnetic materials, µr = 1.
6.13 Magnetomotive Force (M.M.F. or F)
The flow of electrons is current which is basically due to electromotive force (e.m.f.). Similarly
the force behind the flow of flux or production of flux in a magnetic circuit is called
magnetomotive force (m.m.f.). The m.m.f. determines the magnetic field strength.
It is the driving force behind the magnetic circuit. It is given by the product of the number of
turns of the magnetizing coil and the current passing through it.

Mathematically it can be expressed as,
m. m. f. = N I ampere turn (6.9)
where N = Number of turns of magnetising coil, and I = Current through coil. Its unit is ampere
turn (AT) or ampere (A).
The magnetic field strength (H) explained in section (6.8) can be defined as magnetomotive
force per unit length. In equation form,
𝑯 =
𝒎.𝒎.𝒇.
𝒍
=
𝑵𝑰
𝒍
(AT/m) (6.10)
H l = N I = m.m.f. ampere turn (6.11)
6.14 Reluctance (S)
In an electric circuit, current flow is opposed by the resistance of the material; similarly there is
opposition by the material to the flow of flux which is called reluctance.
Reluctance: It is defined as the resistance offered by the material to the flow of magnetic flux
through it. It is denoted by 'S'. It is directly proportional to the length of the magnetic circuit
while inversely proportional to the area of cross-section.
𝑺 =
𝒍
𝝁 𝒂
𝑨/𝑾𝒃 (6.12)
Where 'l ' in 'm' while 'a' in 'm2
'
The reluctance can be also expressed as the ratio of magnetomotive force to the flux produced.
𝑺 =
𝒎.𝒎.𝒇.
𝒇𝒍𝒖𝒙
=
𝑵𝑰
𝜱
𝑨𝑻/𝑾𝒃 𝒐𝒓 𝑨/𝑾𝒃 (6.13)
flux =
𝒎.𝒎.𝒇.
𝒓𝒆𝒍𝒖𝒄𝒕𝒂𝒏𝒄𝒆
𝐨𝐫 𝚽 =
𝐅
𝐒
(6.14)
Sometimes, the above equation is called the “Ohm’s Law of Magnetic Circuit” because it
resembles a similar expression in electric circuits i.e.
current = e.m.f. /resistance or I = V /R
6.15 Permeance
The permeance of the magnetic circuit is defined as the reciprocal of the reluctance. It is
defined as the property of the magnetic circuit due to which it allows flow of the magnetic flux
through it
Permeance = 1/Reluctance
It is measured in weber per ampere (Wb/A)
6.16 Magnetic Hysteresis
It may be defined as the lagging of magnetisation or induction
flux density (B) behind the magnetising force (H). Alternatively,
Fig. 6.20

it may be defined as that quality of a magnetic substance, due to which energy is dissipated in
it, on the reversal of its magnetism. Let us take an unmagnetised bar of iron AB and magnetise
it by placing it within the field of a solenoid (Fig. 6.20). The field H (= NI/l) produced by the
solenoid is called the magnetising force. The value of H can be increased or decreased by
increasing or decreasing current through the coil. Let H be increased in steps from zero up to a
certain maximum value and the corresponding values of flux density (B) be noted. If we plot
the relation between H and B, a curve like OA, as shown in Fig. 6.21, is obtained. The material
becomes magnetically saturated for H = OM and has at that time a maximum flux density of
Bmax established through it.
If H is now decreased gradually (by decreasing solenoid
current), flux density B will not decrease along AO, as might be
expected, but will decrease less rapidly along AC. When H is zero, B is not but has a definite
value Br = OC. It means that on removing the magnetising force H, the iron bar is not
completely demagnetised. This value of B (= OC) measures the retentivity or remanence of
the material and is called the remanent or residual flux density Br. To demagnetise the iron bar,
we have to apply the magnetising force in the reverse direction. When H is reversed (by
reversing current through the solenoid), then B is reduced to zero at point D where H = OD.
This value of H required to wipe off residual magnetism is known as coercive force (Hc) and is
a measure of the coercivity of the material i.e. its ‘tenacity’ with which it holds on to its
magnetism.
6.17 Ampere’s Circuital Law
As mentioned in the introduction to this chapter, there is a broad similarity between the
analyses of electric and magnetic circuits.
If we apply the “cause” analogy to Kirchhoff’s voltage law (∑↻ 𝑉=0), we obtain the following:
∑↻ 𝐦. 𝐦. 𝐟. = 𝟎 (for magnetic circuits) (6.15)
which, in words, states that the algebraic sum of the rises and drops of the m.m.f. around a
closed loop of a magnetic circuit is equal to zero; that is, the sum of the rises in m.m.f. equals
the sum of the drops in m.m.f. around a closed loop.
Equation (6.15) is referred to as Ampère’s circuital law.
As an example of Eq. (6.15), consider the magnetic circuit appearing in Fig. 6.22 constructed
of three different ferromagnetic materials. Applying Ampère’s circuital law, we have
∑↻ 𝐦. 𝐦. 𝐟. = 𝟎
+𝑵𝑰
⏟
𝑹𝒊𝒔𝒆
‒ 𝑯𝒂𝒃𝒍𝒂𝒃
⏟
𝑫𝒓𝒐𝒑
‒ 𝑯𝒃𝒄𝒍𝒃𝒄
⏟
𝑫𝒓𝒐𝒑
‒ 𝑯𝒄𝒂𝒍𝒄𝒂
⏟
𝑫𝒓𝒐𝒑
= 𝟎
Fig. 6.21

𝑵𝑰
⏟
Impressed
mmf
= 𝑯𝒂𝒃𝒍𝒂𝒃 + 𝑯𝒃𝒄𝒍𝒃𝒄 + 𝑯𝒄𝒂𝒍𝒄𝒂
⏟
mmf drops
All the terms of the equation are known except the
magnetizing force for each portion of the magnetic
circuit, which can be found by using the B-H curve if the flux density B is known.
6.18 THE FLUX Φ
If we continue to apply the relationships described in
the previous section to Kirchhoff’s current law, we
will find that the sum of the fluxes entering a
junction is equal to the sum of the fluxes leaving a
junction; that is, for the circuit of Fig. 6.23,
Φa = Φb + Φc (at junction a)
6.19 Magnetic Circuits
The magnetic circuit can be defined as, the closed path traced by the magnetic lines of
force i.e. flux. Such a magnetic circuit is associated with different magnetic quantities as
m.m.f., flux reluctance, permeability etc.
Consider simple magnetic circuit shown in the Fig 6.24 (a). This circuit consists of an iron core
with cross sectional area of 'a' m2
with a mean length of 'l' m, (this is mean length of the
magnetic path which flux is going to trace.) A coil of N turns is wound on one of the sides of
the square core which is excited by a supply. This supply drives a current I through the coil.
This current carrying coil produces the flux (Φ) which completes its path through the core as
shown in the Fig 6.24 (a). This is analogous to simple electric circuit in which a supply i.e.
e.m.f. of E volts drives a current I which completes its path through a dosed conductor having
resistance R. This analogous electrical circuit is shown in the Fig 6.24 (b).
The main quantities are.
I = current flowing through the coil, N = Number of turns, Φ = Flux in weber,
B = Flux density in the core, μ = Absolute permeability of the magnetic material
μr = Relative permeability of the magnetic material.
The main relations are, given by,
Magnetic field strength inside the solenoid is: 𝑯 =
𝑵𝑰
𝒍
(AT/m)
Now flux density is, B = μ H 𝑩 =
𝝁𝒐 𝝁𝒓 𝑵 𝑰
𝒍
Now as area of cross-section is 'a' m2
, total flux in core is,
FIG. 6.22 Series magnetic circuit of
three different materials.
FIG. 6.23 Flux distribution of a
series-parallel magnetic network.

𝚽 =
𝒎.𝒎.𝒇.
𝒓𝒆𝒍𝒖𝒄𝒕𝒂𝒏𝒄𝒆
=
𝑭
𝑺
Where NI = Magnetomotive force m.m.f. in AT = Hl
𝑺 =
𝒍
𝝁𝒐𝝁𝒓 𝒂
𝑨/𝑾𝒃 = Reluctance offered by the magnetic path.
(a) (b)
Fig. 6.24 (a) Magnetic circuit (b) Electrical equivalent
Example 6.6: An iron ring of circular cross sectional area of 3.0 cm2
and mean diameter of 20
cm is wound with 500 turns of wire and carry a current of 2.09 A to produce magnetic flux of
0.5 mWb in the ring. Determine the permeability of the material.
Solution: The given values are:
a = 3 cm2
= 3 × 10-4
m2
, d = 20 cm. N = 500, I = 2A, Φ = 0.5 m Wb
Now, l = π×d =π ×20 = 62.8318 cm = 0.628318 m
𝑺 =
𝒍
𝝁𝒐𝝁𝒓 𝒂
=
𝟎.𝟔𝟐𝟖𝟑𝟏𝟑
𝟒𝝅×𝟏𝟎−𝟕×𝝁𝒓×𝟑×𝟏𝟎−𝟒
=
𝟏.𝟔𝟔𝟔𝟕×𝟏𝟎𝟗
𝝁𝒓
(6.6.1)
𝑺 =
𝒎.𝒎.𝒇.
𝒇𝒍𝒖𝒙
=
𝑵𝑰
𝜱
=
𝟓𝟎𝟎×𝟐
𝟎.𝟓
= 𝟐 × 𝟏𝟎𝟔
𝑨𝑻/𝑾𝒃 (6.6.2)
Equating (6.6.1) and (6.6.2),
𝟐 × 𝟏𝟎𝟔
=
𝟏.𝟔𝟔𝟔𝟕×𝟏𝟎𝟗
𝝁𝒓
⟹ μr = 833.334.
6.19.1 Series Magnetic Circuits
In practice magnetic circuit may be composed of various materials of different
permeabilitics, of different lengths and of different cross-sectional areas. Such a circuit is called
composite magnetic circuit. When such parts are connected one after the other the circuit is
called series magnetic circuit. Consider a circular ring made up of different materials of lengths
l1, l2 and l3 and with cross-sectional areas a1, a2, and a3 with absolute permeabilities μ1, μ2, μ3
as shown in the Fig. 6.25.
Let coil wound on a ring has N turns carrying a
current of I amperes.

The total m.m.f. available is NI AT. This will set the flux 'Φ' which is same through all the
three element of the circuit.
This is similar to three resistances connected in series in electrical circuit and connected to
e.m.f. carrying same current I through all of them.
Therefore; the total reluctance of the magnetic circuit is,
Total 𝑺𝑻 = 𝑺𝟏 + 𝑺𝟐 + 𝑺𝟑 =
𝒍𝟏
𝝁𝟏𝒂𝟏
+
𝒍𝟐
𝝁𝟐𝒂𝟐
+
𝒍𝟑
𝝁𝟑𝒂𝟑
Total 𝚽 =
𝑻𝒐𝒕𝒂𝒍 𝒎.𝒎.𝒇.
𝑻𝒐𝒕𝒂𝒍 𝒓𝒆𝒍𝒖𝒄𝒕𝒂𝒏𝒄𝒆
=
𝑵𝑰
𝑺𝑻
=
𝑵𝑰
(𝑺𝟏+𝑺𝟐+𝑺𝟑)
NI = ST Φ = (S1 + S2 + S3) Φ ⟹ NI = S1 Φ + S2 Φ + S3 Φ
(m.m.f.)T = (m.m.f.)1 + (m.m.f.)2 + (m.m.f.)3
The total m.m.f. also can be expressed as,
(m.m.f.)T = H1l1 + H2l2 + H3l3
Where 𝑯𝟏 =
𝑩𝟏
𝝁𝟏
, 𝑯𝟐 =
𝑩𝟐
𝝁𝟐
, 𝑯𝟑 =
𝑩𝟑
𝝁𝟑
So for a series magnetic circuit we can remember,
1) The magnetic flux through all the parts is same.
2) The equivalent reluctance is sum of the reluctance of different parts.
3) The resultant m.m.f. necessary is sum of the mades in same individual part.
Example 6.7: The electromagnet of Fig. 6.26 has picked up a
section of cast iron. Determine the current I required to establish
the indicated flux in the core.
If lab = lcd = lef = lfa = 101.6×10-3
m, lbc = lde = 12.7×10-3
m,
Area (throughout) = 6.452×10-4
m2
, Φ = 3.5 × 10–4
Wb
μr (sheet steel) ≅6161.57, μr (cast iron) ≅ 269.57 At/m
Solution: we can determine the length for each material rather
than work with the individual sections:
lefab =101.6×10-3
+101.6×10-3
+101.6×10-3
= 304.8×10-3
m
lbcde = 12.7×10-3
+ 101.6×10-3
+ 12.7×10-3
= 127×10-3
m
The flux density for each section is B
𝑩 =
𝜱
𝒂
=
𝟑. 𝟓 × 𝟏𝟎−𝟒
𝟔. 𝟓𝟒𝟐 × 𝟏𝟎−𝟒
= 𝟎. 𝟓𝟒𝟐 𝑻
The flux density for each section is B
and the magnetizing force is 𝐻 =
𝐵
𝜇
Fig. 6.25 A series Magnetic circuit
FIG. 6.26 Electromagnet
for Example 6.7.

H (sheet steel) =
𝑩
𝝁𝒓𝝁𝟎
=
𝟎.𝟓𝟒𝟐
𝟔𝟏𝟔𝟏.𝟓𝟕×𝟒𝝅×𝟏𝟎−𝟕
≅ 𝟕𝟎 𝑨𝑻/𝒎
H(cast iron)
𝑩
𝝁𝒓𝝁𝟎
=
𝟎.𝟓𝟒𝟐
𝟐𝟔𝟗.𝟓𝟕×𝟒𝝅×𝟏𝟎−𝟕
≅ 𝟏𝟔𝟎𝟎 𝑨𝑻/𝒎
Determining Hl for each section yields
Hefab lefab = (70 At/m) (304.8 × 10-3
m) = 21.34 AT
Hbcde lbcde = (1600 At/m)(127 ×10-3
m) = 203.2 AT
The magnetic circuit equivalent and the electric circuit analogy for the system of Fig. 6.26
appear in Fig. 6.27.
Applying Ampere's circuital law,
NI = Hefab lefab + Hbcde lbcde = 21.34 At + 203.2 At = 224.54 At
And (50 t) I = 224.54
At so that 𝐼 =
224.54
50
= 4.49 𝐴
Example 6.8: Determine the secondary current I2 for the transformer of Fig. 6.28 if the
resultant clockwise flux in the core is 1.5 ×10-5
Wb. μr (sheet steel) ≅3978.87
Solution: This is the first example with two magnetizing forces to consider..
The flux density throughout is
𝑩 =
𝜱
𝑨
=
𝟏.𝟓×𝟏𝟎−𝟓 𝑾𝒃
𝟎.𝟏𝟓×𝟏𝟎−𝟑 𝒎𝟐
= 𝟏𝟎 × 𝟏𝟎−𝟐
𝑻 = 𝟎. 𝟏𝟎 𝑻
and
H (sheet steel) =
𝑩
𝝁𝒓𝝁𝟎
=
𝟎.𝟏
𝟑𝟗𝟕𝟖.𝟖𝟕×𝟒𝝅×𝟏𝟎−𝟕
≅ 𝟐𝟎 𝑨𝑻/𝒎
Applying Ampere's circuital law,
N1I1 = N2I2 = Habcdalabcda
(60 t)(2 A) - (30 t)(I2) = (20 At/m)(0.16 m)
120 At - (30 t)I2 = 3.2 At
And (30 t)I2 = 120 At - 3.2 At
or 𝑰𝟐 =
𝟏𝟏𝟔.𝟖𝑨𝒕
𝟑𝟎𝒕
= 𝟑. 𝟖𝟗 𝑨
6.19.2 Series Circuit with Air Gap
Before continuing with the illustrative examples, let us
consider the effects that an air gap has on a magnetic
circuit. Consider a ring having mean length of iron
part as 'li' as shown in the Fig. 6.29.
FIG. 6.27 (a) Magnetic circuit
equivalent and (b) electric circuit analogy
for the electromagnet of Fig. 6.25.
Figure 6.28 Transformer for Example 6.8.

Total m.m.f. = N I AT
Total reluctance ST = Si + Sg
Where Si = Reluctance of iron path, Sg = Reluctance of air gap
𝑺𝒊 =
𝒍𝒊
𝝁𝒂𝒊
, 𝑺𝒈 =
𝒍𝒈
𝝁𝟎𝒂𝒊
The cross-sectional area of air gap is assumed to be equal to area of the iron ring.
𝑺𝑻 =
𝒍𝒊
𝝁𝒂𝒊
+
𝒍𝒈
𝝁𝟎𝒂𝒊
𝜱 =
𝒎.𝒎.𝒇.
𝑹𝒆𝒍𝒖𝒄𝒕𝒂𝒏𝒄𝒆
=
𝑵𝑰
𝑺𝑻
Or Total m.m.f. = m.m.f. for iron + m.m.f. for air gap
NI = Si Φ + Sg Φ AT for ring
Example 6.9: An iron ring 8 cm mean diameter is made up of round iron of diameter 1cm and
permeability of 900, has air gap of 2mm wide. It consists of winding with 400 turns carrying a
current of 3.5A. Determine
i) m.m.f. ii) total reluctance iii) the flux iv) flux density of ring
Solution: The ring and the winding is shown in the Fig 6.30.
Diameter of ring d = 8 cm,
length of iron = π d ‒ length of air gap
li = π ×(8×10-2
) ‒ 2×10-3
= 0.2493 m
Key Point: While calculating iron length do not forget to
subtract length of air gap from total mean length
lg = length of air gap = 2 × 10-3
m, diameter of iron 1 cm
area of cross section a =
𝝅
𝟒
d2
=
𝝅
𝟒
(1 × 10-2
)2
= 7.853 × 10-5
m2
Area of cross section of air gap and ring is to be assumed same.
i) Total m.m.f. produced = N I = 400× 3.5 = 1400 AT
ii) Total reluctance ST = Si + Sg
𝑺𝒊 =
𝒍𝒊
𝝁𝟎𝝁𝒓𝒂
given μr = 900
=
𝟎.𝟐𝟒𝟗𝟑
𝟒𝝅×𝟏𝟎−𝟕×𝟗𝟎𝟎×𝟕.𝟖𝟓𝟑×𝟏𝟎−𝟓 = 𝟐𝟖𝟎𝟔𝟗𝟒𝟕. 𝟔𝟏𝟓 𝑨𝑻/𝑾𝒃
𝑺𝒈 =
𝒍𝒈
𝝁𝟎𝒂
as μr = 1for air
𝑺𝒈 =
𝟐×𝟏𝟎−𝟑
𝟒𝝅×𝟏𝟎−𝟕×𝟕.𝟖𝟓𝟑×𝟏𝟎−𝟓 = 𝟐𝟎. 𝟐𝟔𝟔𝟕 × 𝟏𝟎𝟔
𝑨𝑻/𝑾𝒃
Figure 6.29 a ring with an air gap
Figure 6.30

ST = 2806947.615 + 20.2667×106
= 23.0737×106
AT/Wb
ii) 𝜱 =
𝒎.𝒎.𝒇.
=
𝑵𝑰
𝑺𝑻
=
𝟏𝟒𝟎𝟎
𝟐𝟑.𝟎𝟕𝟑𝟕×𝟏𝟎𝟔
= 𝟔. 𝟎𝟔𝟕 × 𝟏𝟎−𝟓
𝑾𝒃
iii) 𝒇𝒍𝒖𝒙 𝒅𝒆𝒏𝒔𝒊𝒕𝒚 𝑩 =
𝜱
𝒂
=
𝟔.𝟎𝟔𝟕×𝟏𝟎−𝟓
𝟕.𝟖𝟓𝟑×𝟏𝟎−𝟓
= 𝟎. 𝟕𝟕𝟐𝟓 𝑾𝒃/𝒎𝟐
6.19.3 Parallel Magnetic Circuits
In case of electric circuits, resistances can be connected in parallel. Current through each of
such resistances is different while voltage across all of them is same. Similarly different
reluctances may be in parallel in case of magnetic circuits. A magnetic circuit which has more
than one path for the flux is known as a parallel magnetic circuit.
Consider a magnetic circuit shown in the Fig. 6.31 (a). At point A the total flux Φ, divides into
two parts Φ1 and Φ2.
Φ = Φ1 + Φ2
The fluxes Φ1 and Φ2 have, their paths completed through ABCD and AFED respectively.
This is similar to division of current in case of parallel connection of two resistances in an
electric circuit. The analogous electric circuit is shown in the Fig. 6.30 (b).
(a) Magnetic circuit (b) Equivalent electrical circuit
Fig. 6.31 A parallel magnetic circuit
The mean length of path ABCD = l1, The mean length of the path AFED = l2
The mean length of the path AD = lc
The reluctance of the path ABCD = S1, The reluctance of path AFED = S2
The reluctance of path AD = Sc
The total m.m.f. produced = N I AT
𝒇𝒍𝒖𝒙 =
𝒎.𝒎.𝒇.
m.m.f. = Φ × S
For path ABCDA, NI = Φ1 S1 + Φ Sc
For path AFEDA, NI = Φ2 S2 + Φ Sc

Where 𝑺𝟏 =
𝒍𝟏
𝝁𝒂𝟏
, 𝑺𝟐 =
𝒍𝟐
𝝁𝒂𝟐
, and 𝑺𝒄 =
𝒍𝒄
𝝁𝒂𝒄
Generally a1 = a2 = ac area of cross section
For parallel circuit,
Total m.m.f. = m.m.f. required by central limb + m.m.f. required by any of outer limb
NI = (N I)AD + (N I)ABCD or (NI)AFED
NI = Φ Sc + (Φ1S1 or Φ2S2)
As in the electric circuit e.m.f. across parallel branches is same, in the magnetic circuit the
m.m.f. across parallel branches is same.
Thus same m.m.f. produces different fluxes in the two parallel branches. For such parallel
branches, Φ1S1 = Φ2S2
Hence while calculating total m.m.f., the m.m.f. of only one of the two parallel branches must
be considered.
6.19.4 Parallel Magnetic Circuit with Air Gap
Consider a parallel magnetic circuit with air gap
in the central limb as shown in the Fig. 6.32.
The analysis of this circuit is exactly similar
to the parallel circuit discussed above. The only
change is the analysis of central limb. The central
limb is series combination of iron path and air
gap. The central limb is made up of,
path GD = iron path = lc
path GA = air gap = lg
The total flux produced is Φ. It gets divided at A into Φ1 and Φ2.
Φ = Φ1 + Φ2
The reluctance of central limb is now,
Sc = Si + Sg =
𝒍𝒄
𝝁𝒂𝒄
+
𝒍𝒈
𝝁𝟎𝒂𝒄
Hence m.m.f. of central limb is now,
(m.m.f.)AD = (m.m.f.)GD + (m.m.f.)GA
Thus the electrical equivalents circuit for such case
become as shown in the Fig.6.33.
Similarly there may be air gaps in the side limbs but the method of analysis remains the same.
Example 6.10: The magnetic circuit shown in Fig. 6.34 is constructed of wrought iron. The
cross-section of central limb is 8 cm2
and of each other limbs, 5 cm2
. If the coil on the center
Fig. 6.33 Electrical equivalent circuit
Fig . 6.32 Parallel circuit with air gap

limb is wound with 1000 turns, calculate the exciting current required to set up a flux of 1.2
mWb in the center limb. Width of each air gap is 1mm. points on the B/H curve of wrought
iron are as follows:
Fig. 6.34
Solution: Given; lc =length of central limb = 10 cm= 0.1 m
ac = 8cm2
= 8×l0-4
m2
, Φc = 12 mWb =1.2x10-3
Wb
li = Length of iron path of side limb = 25 cm = 0.25 m (on each side)
lg = Length of air gap= 1 mm = 1×10-3
m
ai = 5 cm2
= 5×10-4
m2
This is the example of parallel magnetic circuit
The flux in central limb 1.2 mWb gets divided into two equal paths as shown in Fig. 6.34.
Flux in side limbs =
𝟏.𝟐
𝟐
, i.e. Φi = 0.6 mWb
Flux density in central limb is, 𝑩𝒄 =
𝜱𝒄
𝒂𝒄
=
𝟏.𝟐×𝟏𝟎−𝟑
𝟖×𝟏𝟎−𝟒
= 𝟏. 𝟓 𝑻𝒆𝒔𝒍𝒂 𝒐𝒓 𝑾𝒃/𝒎𝟐
Flux density in air gap is, 𝑩𝒈 =
𝜱𝒊
𝒂𝒊
=
𝟎.𝟔×𝟏𝟎−𝟑
𝟓×𝟏𝟎−𝟒
= 𝟏. 𝟐 𝑻𝒆𝒔𝒍𝒂
Flux density in side limb is, 𝑩𝒊 =
𝜱𝒊
𝒂𝒊
=
𝟎.𝟔×𝟏𝟎−𝟑
𝟓×𝟏𝟎−𝟒
= 𝟏. 𝟐 𝑻𝒆𝒔𝒍𝒂 𝒐𝒓
applying Kirchhoff m.m.f. low to ABCD e c the loop,
m.m.f. = Hc lc+Hg lg +Hi li,
Hc lc = m.m.f. required by central limb, Hg lg = m.m.f. required by air gap
Hi li = m.m.f. required by iron path on any one side
i) Central limb
Bc = 1.5 Tesla
From B-H table given corresponding, Hc = 2000 AT/m

Hc lc = 2000× 0.1 = 200 AT
ii) Side limb Hi = 1.2 Tesla
From B-H table given corresponding, Hi = 625 AT/m
Hi li = 625× 0.25 = 156.25 AT
iii) The air gap Bg = μ0 Hg ⇛ 𝐻𝑔 =
𝐵𝑔
𝜇0
Hg = 954929.65 AT/m
Hg lg = 954.9296 × (1 × 10- 3
)
Total m.m.f. required NI = 200 +156.25 + 954.9296 = 1311.17 AT
Current I =
1311.17
𝑁𝑜.𝑜𝑓 𝑡𝑢𝑟𝑛𝑠
=
1311.1 7
1000
I = 1.31 A
Example 6.11: A cast steel structure is made of a rod of square section 2.5 cm× 2.5 cm as
shown in the Fig. 6.35. What is the current that should be passed in a 500 turn coil on the left
limb so that flux of 2.5 mWb is made to pass in the right limb. Assume permeability as 750 and
neglect leakage.
Solution: This is parallel magnetic circuit. The total flux produced gets distributed into two
parts having reluctance S1 and S2
S1 = Reluctance of centre limb, S2 = Reluctance of right side
𝑺𝟏 =
𝒍𝟏
𝝁𝒓𝝁𝟎𝒂𝟏
=
𝟐𝟓×𝟏𝟎−𝟐
𝟒𝝅×𝟏𝟎−𝟒
×𝟕𝟓𝟎×𝟐.𝟓×𝟐.𝟓×𝟏𝟎−𝟒 = 𝟒𝟐𝟒. 𝟒𝟏𝟑 × 𝟏𝟎𝟑
𝑨𝑻/𝑾𝒃
𝑺𝟐 =
𝒍𝟐
𝝁𝒓𝝁𝟎𝒂𝟏
=
𝟒𝟎×𝟏𝟎−𝟐
𝟒𝝅×𝟏𝟎−𝟒
×𝟕𝟓𝟎×𝟐.𝟓×𝟐.𝟓×𝟏𝟎−𝟒 = 𝟔𝟕𝟗. 𝟎𝟔𝟏 × 𝟏𝟎𝟑
𝑨𝑻/𝑾𝒃
For branch AB and CD, m.m.f. is same
m.m.f. = S1 Φ1= S2 Φ2
And Φ2 = 2.5 mWb
𝚽𝟏 =
𝚽𝟐𝐒𝟐
𝐒𝟏
=
𝟐.𝟓×𝟏𝟎−𝟑×𝟔𝟕𝟗.𝟎𝟔𝟏×𝟏𝟎𝟑
𝟒𝟐𝟒.𝟒𝟏𝟑×𝟏𝟎𝟑
= 𝟒 𝒎𝑾𝒃
Φ = Φ1 + Φ2 = 2.5 + 4 = 6.5 mWb
Total m.m.f. required is Sum of the m.m.f. required for AEFB and that for either central or side
limb.
SAEFB = S2 = 679.061×103
AT/Wb
m.m.f. for AEFB = SAEFB × Φ = 679.061× 10 3
× 6.5× 10-3
= 4413.8965 AT
Total m.m.f. = 4413.8965+ Φ1 S1 = 4413.8965+ 4 × 10-3
× 424.413x 103
= 6111.548 AT
But Total m.m.f. = N I
I =
𝟔𝟏𝟏𝟏.𝟓𝟒𝟖
𝟓𝟎𝟎
= 12.223 A
6.20 Kirchhoff's Laws for Magnetic Circuit
Fig. 6.35

Similar to the electrical circuit Kirchhoff's Laws can be used to analyse complex magnetic
circuit. The laws can be stated as below:
6.20.1 Kirchhoff's Flux Law
The total magnetic flux arriving at any junction in a magnetic circuit is equal to the total
magnetic flux leaving that junction.
At a junction, ∑ 𝚽 = 𝟎
The law in fact is used earlier to analyse parallel magnetic circuit at a junction A shown in the
Fig.6.31(a), where Φ = Φ1 + Φ2
6.20.2 Kirchhoff's M.M.F. Law
The resultant m.m.f. around a closed magnetic circuit is equal to the algebraic sum of the
products of the flux and the reluctance of each part of the closed circuit i.e. for a closed
magnetic circuit
∑ 𝒎. 𝒎. 𝒇. = ∑ 𝚽𝐒
m.m.f. also can be calculated as H×l where H is field strength and "l' is mean length
m.m.f. = Hl
Alternatively the same law can be stated as:
The resultant m.m.f. around any closed loop of magnetic circuit is equal to the algebraic sum of
the products of the magnetic field strength and the length of each part of the circuit i.e. for a
closed magnetic circuit.
∑ 𝒎. 𝒎. 𝒇. = ∑ 𝐇𝒍
6.21 Magnetic Leakage and Fringing
Most of the applications which are using magnetic
effects of an electric current are using flux in air gap
for their operation. Such devices are generators,
motors, measuring instruments like ammeter,
voltmeter etc. Such devices consist of magnetic
circuit with an air gap and flux in air gap is used to
produce the required effect.
Such flux which is available in air gap and is utilized to produce the desired effect is called
useful flux denoted by Φu. It is expected that whatever is the flux produced by the magnetizing
coil, it should complete its path through the iron and air goop. So part of the flux completes its
path through the air or medium in which coil an d magnetic circuit is placed.
Fig. 6.36 leakage and useful flux

Key Point: Such flux which leaks and complete its path through surrounding air or
medium instead of the desired path is called the leakage flux.
The fig. 6.35 shows the useful and leakage flux.
6.21.1 Leakage Coefficient or Hopkinson's Coefficient
The ratio of the total flux (ΦT) to the useful flux (Φu) is defined as the leakage
coefficient of Hopkinson's coefficient or leakage factor of that magnetic circuit.
It is denoted by λ,
𝛌 =
𝐭𝐨𝐭𝐚𝐥 𝐟𝐥𝐮𝐱
𝐮𝐬𝐞𝐟𝐮𝐥 𝐟𝐥𝐮𝐱
=
𝚽𝐓
𝚽𝐮
The value of 'λ' is always greater than 1 as ΦT is always more than Φu. It generally varies
between 1.1 and 1.25. Ideally its value should be 1.
6.21.2 Magnetic Fringing
When flux enters into the air gap, it passes through the air gap in terms of parallel flux
lines. There exists a force of repulsion between the magnetic lines of force which are parallel
and having same direction. Due to this repulsive force there is tendency of the magnetic flux to
bulge out (spread out) at the edge of the air gap. This tendency of flux to bulge out at the edges
of the air gap is called magnetic fringing.
It has following two effects:
1) It increases the effective cross-sectional area of the air gap.
2) It reduces the flux density in the air gap.
So leakage, fringing and reluctance, in practice should be as
small as possible.
Key Point: This is possible by choosing good magnetic
material and making the air gap as narrow as possible
Example 6.12: A cast iron ring of 40 cm mean length and circular cross-section of 5 cm
diameter is wound with a coil. The coil carrier a current of 3 A and produces a flux of 3 mWb
in the air gap. The length of the air gap is 2 mm. The relative permeability of the cast iron is
800. The leakage coefficient is 1.2. Calculate number of turns of the coil.
Solution: Given, lt = 40 cm = 0.4 m, lg = 2×10-3
m, li = lt ‒ lgap = 0.4 ‒ 2 × 10-3
=0.398m
a =
𝝅
𝟒
d2
=
𝝅
𝟒
(5 × 10-2
)2
= 19.634 × 10-4
m2
cross-section area
I = 3A, Φg =2×10-3
Wb, μr = 800, λ =1.2
𝛌 =
𝚽𝐓
𝚽𝐠
leakage coefficient, ΦT = 2.4 × 10-3 Wb
Fig. 6.37 Magnetic fringing

Reluctance of iron path 𝑺𝒊 =
𝒍𝒊
𝝁𝒓𝝁𝟎𝒂
=
𝟎.𝟑𝟗𝟖
𝟒𝝅×𝟏𝟎−𝟕×𝟖𝟎𝟎×𝟏𝟗.𝟔𝟑𝟒×𝟏𝟎−𝟒
= 𝟐𝟎𝟏𝟔𝟐𝟗. 𝟏𝟔 𝐀𝐓/𝐖𝐛
𝜱𝑻 =
𝒎. 𝒎. 𝒇.
=
(𝒎. 𝒎. 𝒇. )𝒊𝒓𝒐𝒏 𝒑𝒂𝒕𝒉
𝑺𝒊
m.m.f. for Iron path = 483.909 AT
Reluctance of air gap 𝐒𝐠 =
𝒍𝒈
𝝁𝟎𝒂
=
𝟐×𝟏𝟎−𝟑
𝟒𝝅×𝟏𝟎−𝟕×𝟏𝟗.𝟔𝟑𝟒×𝟏𝟎−𝟒
= 810608.86 AT/Wb
m.m.f. for air gap = 1621.2177 AT
Total m.m.f. required NI = (m.m.f.) iron+ (m.m.f.) air gap = 483.909+1621.2177
= 2105.1267 AT
N = 2105.1267/I i.e. N = 702 turns

CHAPTER SEVEN
INDUCTOR
ELECTROMAGNETIC INDUCTION AND
7.1 Introduction
Up till now we have discussed the basic properties, concept of magnetism and magnetic
circuits. Similarly we have studied the magnetic effects of an electric current. But we have not
seen the generation of e.m.f. with the help of magnetism. The e.m.f. can be generated by
different ways, by chemical action, by heating thermocouples etc. But the most popular and
extensively used method of generating an e.m.f. is based on electromagnetism.
After the magnetic effects of an electric current, attempts were made to produce electric
current with the help of magnetism rather than getting magnetism due to current carrying
conductor. In 1831, an English Physicist, Michael Faraday succeeded in getting e.m.f. from
magnetic flux. The phenomenon by which e.m.f. is obtained from flux ill called
electromagnetic induction.
7.2 Faraday's Experiment
Let us study first the experiment conducted by Faraday to get understanding of electromagnetic
induction. Consider a coil having 'N' turn is connected
to a galvanometer as shown in the Fig. 7.1.
Galvanometer indicates flow of current in the circuit, if
any. A permanent magnet is moved relative to coil,
such that magnetic lines of force associated with coil
get changed. Whenever, there is motion of permanent
magnet, galvanometer deflects indicating flow of current
through the circuit. The deflection continues as long as motion of magnet exists.
With this experiment Faraday stated laws called Faraday's Laws of Electromagnetic Induction.
Key Point: This phenomenon of cutting of flux lines by the conductor to get the induced
e.m.f. in the conductor or coil is called Electromagnetic Induction.
Thus, to have induced e.m.f. there must exist.
1) A coil or conductor. 2) A magnetic field (permanent magnet or electromagnet).
3) Relative motion between conductor and magnetic flux (achieved by moving conductor with
respect to flux or moving with respect to conductor.)
Key Point: the e.m.f. exists as along as relative motion persists.
7.3 Faraday's Laws of Electromagnetic Induction
Fig. 7.1 Faraday's experiment

From the experiment discussed above, Michael Faraday a British scientist stated two laws of
electromagnetic induction.
7.3.1 First Law
Whenever the number of magnetic lines of force (flux) linking with a coil or circuit changes an
e.m.f. gets induced in that coil or circuit.
7.3.2 Second Law
The magnitude of the induced e.m.f. is directly proportional to the rate of change of flux
linkages (flux × number of turns of coil= N × Φ).
As per the first law, e.m.f. will get induced in the coil and as per second law the magnitude of
e.m.f. is proportional to the rate of change of flux linkages.
𝒆 = 𝑵
𝒅𝜱
𝒅𝒕
(7.1)
Now as per Lenz's law (discussed later), the induced e.m.f. sets up a current in such a direction
so as to oppose the very cause producing it.. Thus such an induced e.m.f. is mathematically
expressed along with its sign as,
𝒆 = −𝑵
𝒅𝜱
𝒅𝒕
𝐯𝐨𝐥𝐭𝐬 (7.2)
7.4 Nature of the Induced E.M.F.
E.M.F. gets induced in a conductor, whenever there exists change in flux with that conductor
according to Faraday's law.
Depending upon the nature of methods the induced e.m.f. is classified as.
1) Dynamically induced e.m.f. 2) Statically induced e.m.f.
7.5 Dynamically Induced E.M.F.
The dynamically induced e.m.f. or motional induced e.m.f. is an induced e.m.f. which is due to
physical movement of coil conductor with respect to flux, or movement of magnet with respect
to stationary coil conductor.
7.5.1 Magnitude of Dynamically Induced E.M.F.
Consider e conductor of length l metres moving in the air gap between the poles of the magnet.
If plane of the motion of the conductor is parallel to the plane of the magnetic field then
there is no cutting of flux lines and there cannot be any induced e.m.f. in the conductor
such condition is shown in the Fig. 7.2(a).

Fig. 7.2 (a) No cutting of flux (b) Maximum cutting of flux
In second case as shown in the Fig. 7.2(b), the velocity direction i.e. motion of conductor
is perpendicular to the flux. Hence whole length of conductor is cutting the flux line hence
there is maximum possible induced e.m.f. in the conductor. Under such condition plane of flux
and plane of motion are perpendicular to each other.
Consider a conductor moving with velocity v m/s such that is plane of motion or direction
of velocity is perpendicular to the direction of flux lines as shown in Fig. 7.3 (a).
Fig. 7.3
But if conductor is moving with a velocity v but at a certain angle θ measured with
respect to direction of the field (plane of the flux) as shown in the Fig. 7.3 (b) then component
of velocity which is v sinθ is perpendicular to the direction of flux and hence responsible for
the induced (e.m.f.). The other component v cosθ is parallel to the plane of the flux and hence
will not contribute to the dynamically induced e.m.f.
Under this condition magnitude of induced e.m.f. is given by,
e = B l v sinθ (7.3)
where θ is measured with respect to plane of the flux.
Example 7.1: A conductor of 2 m length with a uniform velocity of 1.27 m/s under magnetic
field having a flux density of 1.2 Wb/m2
(tesla). Calculate the magnitude of induced m.m.f. if
conductor moves:
i) At right angle to axis of field. ii) At an angle of 60° to the direction of field.
Solution:
i) The magnitude of induced e.m.f.
e = B l v for θ = 90°
e = 1.2×2 × 1.27 = 3.048 volts
ii) e = B l v sinθ where θ = 60°
= 1.2×2×1.27 × sin 60 = 2.6397 volts
Example 7.2: A coil carries 200 turn gives rise a flux of 500 uWb when carrying a certain
current. If this current is reversed in 1/10 th of a second, Find the average e.m.f. induced in the
coil.

Solution: The magnitude of induced e.m.f. is,
𝒆 = 𝑵
𝒅𝜱
𝒅𝒕
where dΦ is change in flux linkages i.e. change in NΦ. Now in this problem flux is 500×10-6
for given current. After reversing this current, flux will reverse its direction. So flux becomes
(-500 × 10-6
).
dΦ = Φ2 – Φ1 = -500 × 10-6
– (500 × 10-6
) = –1× 10-3
This happens in lime dt = 0.1 sec
Average e.m.f. = 𝒆 = −𝑵
𝒅𝜱
𝒅𝒕
= −𝟐𝟎𝟎
(−𝟏×𝟏𝟎−𝟑)
𝟎.𝟏
= 𝟐 𝒗𝒐𝒍𝒕𝒔
7.5.2 Direction of Dynamically Induced E.M.F.
The direction of Induced e.m.f. can be decided by using two rules.
1) Fleming's Right Hand Rule
As discussed earlier, the Fleming's Left Hand Rule is used to get
direction of force experienced by conductor carrying current
placed in a magnetic field while Fleming's Right Hand Rule is to
be used to get direction of induced e.m.f. when conductor is
moving in a magnetic field. According to Fleming's right hand
rule, outstretch the three fingers of right hand namely the thumb,
fore finger and the middle finger, perpendicular to each other.
Arrange the right hand so that first finger point in the direction of
flux lines (from N to S) and thumb in the direction of motion of conductor with respect to the
flux then the middle finger will point in the direction of the induced e.m.f. (or current).
Consider the conductor moving in a magnetic field as shown in the Fig.7.4.
2) Lenz's Law
Thus rule is based on the principles derived by German Physicist Heinrich Lenz. The
Lenz's law states that,
'The direction of an induced e.m.f. produced by the electromagnetic induction is such that it sets
up a current which always opposes the cause that is responsible for inducing the e.m.f.'
In short the induced e.m.f. always opposes the cause producing it, which is represented by a
negative sign, mathematically in its expression.
𝒆 = −𝑵
𝒅𝜱
𝒅𝒕
𝐯𝐨𝐥𝐭𝐬
Consider a solenoid as shown in the Fig. 7.5 (a) and (b). When a bar magnet is moved and the
N-pole of magnet is facing a coil. According to Lenz's Law, the direction of current due to
Fig. 7.4

induced e.m.f. is so as to oppose the cause. So e.m.f. will set up a current through coil in such a
way that the end of solenoid facing bar magnet will become N-pole when magnet move toward
coil. Hence two like poles will face each other experiencing force of repulsion. While if the
same bar magnet is moved away from the coil, then induced e.m.f. will set up a current in the
direction which will cause, the end of solenoid facing bar magnet to behave as S-pole. Because
of this two unlike poles face each other and there will be force of attraction.
(a) (b)
Fig. 7.5 Lenz's Jaw
7.6 Statically Induced E.M.F.
Key Point: Statically induced e.m.f., it's an e.m.f. induced in a coil without physical
moving the coil or the magnet.
Explanation: To have an induced e.m.f. there must be change in flux associated with a coil.
Such a change in flux can be achieved without any physical movement by increasing and
decreasing the current (alternating current) producing the flux rapidly, with time. Such
alternating current means it changes its magnitude periodically with time. This produces the
flux which is also alternating i.e. changing with time. Thus there exists dΦ/dt associated with
coil placed in the vicinity of an electromagnet. This is responsible for producing an e.m.f. in the
coil. This is called statically induced e.m.f. The concept of statically induced e.m.f. is shown in
the Fig. 7.6.
Fig. 7.6 Concept of statically induced e.m.f.
Such an induced e.m.f. can be observed in case of a device known as transformer.
The statically induced e.m.f. is further classified as,
1) Self induced e.m.f. and 2) Mutually induced e.m.f.

7.7 Self Induced E.M.F.
According to Faraday's law, due to rate of change of flux linkages there will be induced
e.m.f. in the coil. So without physically moving coil or flux there, is induced e.m.f. in the coil .
The phenomenon is called self induction. The e.m.f. induced in a coil due to the change of its
own flux linked with it is called self induced e.m.f.
7.7.1 Self Inductance
According to Lenz's law the direction of this induced e.m.f. will be so as to oppose the cause
producing it. The cause is the current I hence the self induced e.m.f. will try to set up a current
which is in opposite direction to that of current I. When current is increased, self induced e.m.f.
reduces the current tries to keep it to its original value and vice versa. So any change in current
through coil is opposed by the coil.
7.7.2 Magnitude of Self Induced E.M.F.
The coefficient of self Inductance and denoted by 'L' is constant
𝑳 =
𝑵𝜱
𝑰
It can be defined as flux linkages per ampere current in it. Its unit is henry (H),
Self induced e.m.f., 𝒆 = −𝑳
𝒅𝑰
𝒅𝒕
The coefficient of self inductance is also defined as the e.m.f. induced in volts when the current
in the circuit changes uniformly at the rate of one ampere per second.
7.7.3 Expressions for Coefficient of Self Inductance (L)
𝑳 =
𝑵𝟐
𝝁𝒂
𝒍
=
𝑵𝟐
𝒍
𝒉𝒆𝒏𝒓𝒊𝒆𝒔
Where l = length of magnetic circuit
a = area of cross-section of magnetic circuit through which flux is passing.
Example 7.3: If a coil has 500 turns is linked with a flux of 50 mWb, when carrying a current
of 125 A. Calculate the inductance of the coil. If this current is reduced to zero uniformly in 0.1
sec, calculate the self induced e.m.f. in the coil.
Solution: The inductance is given by, 𝑳 =
𝑵𝜱
𝑰
Where N = 500, Φ = 50 mWb = 50 × l0-3
Wb, I = 125 A
𝑳 =
𝟓𝟎𝟎×𝟓𝟎×𝟏𝟎−𝟑
𝟏𝟐𝟓
= 0.2 H
𝒆 = −𝑳
𝒅𝑰
𝒅𝒕
= −𝑳 [
𝐅𝐢𝐧𝐚𝐥𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐈−𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝐯𝐚𝐥𝐮𝐞 𝐨𝐟 𝐈
𝐓𝐢𝐦𝐞
] = −𝟎. 𝟐 × (
𝟎−𝟏𝟐𝟓
𝟎.𝟏
) = 𝟐𝟓𝟎 𝐕

This is positive because current is decreased. So this 'e' will try to opposite this decrease,
means will try to increase current and will help the growth of the current.
Example 7.4: A coil is wound uniformly on an iron core. The relative permeability of the iron
is 1400. The length of the magnetic circuit is 70 cm. The cross-sectional area of the core is 5
cm2
. The coil has 1000 turns. Calculate,
i) Reluctance of magnetic circuit ii) Inductance of coil in henries.
iii) E.M.F. induced in coil if a current of 10 A is uniformly reversed in 0.2 seconds.
Solution: μr = 1400, L = 70 cm = 0.7 m, N = 1000, a = 5 cm2
= 5× 10-4
m2
, μ0 = 4π ×l0-7
i) 𝑺 =
𝒍
𝝁𝒓𝝁𝟎𝒂
=
𝟎.𝟕
𝟏𝟒𝟎𝟎×𝟒𝝅×𝟏𝟎−𝟕×𝟓×𝟏𝟎−𝟒
= 𝟕. 𝟗𝟓𝟕 × 𝟏𝟎𝟓
𝑨𝑻/𝑾𝒃
ii) 𝑳 =
𝑵𝟐
𝑺
=
(𝟏𝟎𝟎𝟎)𝟐
𝟕.𝟗𝟓𝟕×𝟏𝟎𝟓 = 𝟏. 𝟐𝟓𝟔𝟔 𝑯
iii) A current of + 10 A is made ‒ 10 A in 0.2 sec.
𝒅𝑰
𝒅𝒕
=
−𝟏𝟎−𝟏𝟎
𝟎.𝟐
= −𝟏𝟎𝟎
𝒆 = −𝑳
𝒅𝑰
𝒅𝒕
= −𝟏. 𝟐𝟓𝟔𝟔 × (−𝟏𝟎𝟎) = 𝟏𝟐𝟓. 𝟔𝟔 𝑽
7.8 Mutually Induced E.M.F.
If the flux produced by one coil is getting linked with another coil and due to change in this
flux produced by first coil, there is induced e.m.f. in the second coil, then such an e.m.f. is
called mutually induced e.m.f.
7.8.1 Magnitude of Mutually Induced E.M.F.
Let I1 Current flowing through coil A
N1 = Number of turns of coil A N2 = Number of turns of coil B
Φ1 = Flux produced due to current I1 in webers, Φ2 = Flux linking with coil B
According to Faraday's law, the induced e.m.f. in coil B is,
𝒆𝟐 = −𝑵𝟐
𝒅𝜱𝟐
𝒅𝒕
Negative sign indicates that this e.m.f. will set up a current which will oppose the change
of flux linking with it.
𝒆𝟐 = −𝑴
𝒅𝑰𝟏
𝒅𝒕
𝒗𝒐𝒍𝒕𝒔
where M is coefficient of mutual inductance and equal to (
𝑵𝟐𝜱𝟐
𝑰𝟏
). Coefficient of mutual
inductance is defined as the property by which e.m.f. gets induced in the second coil
because of change in current through first coil.

7.8.3 Expressions of the Mutual Inductance
𝑴 =
𝑵𝟏
𝟐
𝑵𝟐
𝟐
𝝁𝒂
𝒍
=
𝑵𝟏
𝟐
𝑵𝟐
𝟐
𝒍
𝒉𝒆𝒏𝒓𝒊𝒆𝒔
7.8.4 Coefficient of Coupling or Magnetic Coupling Coefficient
𝑴 = 𝑲 √𝑳𝟏𝑳𝟐 where, 𝑲 = √𝑲𝟏𝑲𝟐 or 𝑲 =
𝑴
√𝑳𝟏𝑳𝟐
The K is called coefficient of coupling. If entire flux produced by one coil links with other then
K = K1 = K2 and maximum mutual inductance existing between the coil is 𝑴 = √𝑳𝟏𝑳𝟐
Example 7.5: Two coils A and B are kept in parallel planes, such that 70 % of the flux
produced by coil A links with coil B . Coil A has 10,000 turns. Coil B has 12,000 turns. A
current of 4 A in coil A produces a flux of 0.04 mWb while a current of 4A in coil B produces
a flux of 0.08 mWb. Calculate,
i) Self inductances LA and LB, ii) Mutual inductance M, iii) Coupling coefficient.
Solution: The given values are,
NA= 10,000, NB = 12,000, ΦB = 0.7 ΦA, KA =
𝚽𝐁
𝚽𝐀
= 0.7,
ΦA = 0.04 ×10-3 Wb for IA = 4 A, ΦB = 0.08 ×10-3 Wb for IB = 4 A.
i) Self Inductance, 𝑳𝑨 =
𝑵𝑨𝜱𝑨
𝑰𝑨
=
𝟏𝟎𝟎𝟎𝟎×𝟎.𝟎𝟒×𝟏𝟎−𝟑
𝟒
= 𝟎. 𝟏 𝑯 , 𝑳𝑩 =
𝑵𝑩𝜱𝑩
𝑰𝑩
=
𝟏𝟐𝟎𝟎𝟎×𝟎.𝟎𝟖×𝟏𝟎−𝟑
𝟒
= 𝟎. 𝟐𝟒 𝑯
ii) Mutual Inductance M =
𝑵𝑩𝜱𝑩
𝑰𝑨
=
𝑵𝑩𝑲𝑨𝜱𝑨
𝑰𝑨
=
𝟏𝟐𝟎𝟎𝟎×𝟎.𝟕×𝟎.𝟎𝟒×𝟏𝟎−𝟑
𝟒
= 𝟎. 𝟎𝟖𝟒 𝑯
iii) Coupling Coefficient 𝑲 =
𝑴
√𝑳𝑨𝑳𝑩
=
𝟎.𝟎𝟖𝟒
√𝟎.𝟏×𝟎.𝟐𝟒
= 𝟎. 𝟓𝟒𝟐𝟐
7.9 Lifting Power of Electromagnets
Force of attraction between the two magnetized surfaces forms the basis of operation of
devices like lifting magnets, solenoid valves, magnetically operated contactors, clutches etc.
Consider two poles of two magnetized surface N and S having an air gap of length 'l' m
between them and a cross-sectional area of 'a' m2
. Let P newtons be the force of attraction
between them. This is shown in the Fig. 7.7. The energy stored in a magnetic field per unit
volume is,
𝐄 =
𝟏
𝟐
𝐁𝟐
𝛍
𝐉/𝐦𝟑
The energy stored in a magnetic field in air per unit volume is,
𝐄 =
𝟏
𝟐
𝐁𝟐
𝛍𝟎
𝐉/𝐦𝟑
…. μr = 1
The force in newtons existing between two magnetized surfaces.

𝑷 =
𝑩𝟐𝒂
𝟐𝝁𝟎
𝒏𝒆𝒘𝒕𝒐𝒏𝒔
7.10 INDUCTORS
An inductor is a passive element designed to store energy in its magnetic field. Inductors
find numerous applications in electronic and power systems. They are used in power supplies,
transformers, radios, TVs, radars, and electric motors.
Any conductor of electric current has inductive properties and maybe regarded as an
inductor. But in order to enhance the inductive effect, a practical inductor is usually formed
into a cylindrical coil with many turns of conducting wire.
An inductor consists of a coil of conducting wire.
If current is allowed to pass through an inductor, it is found that the voltage across the
inductor is directly proportional to the time rate of change of the current. Using the passive sign
convention,
𝒗 = 𝑳
𝒅𝒊
𝒅𝒕
(7.4)
where L is the constant of proportionality called the inductance of the inductor. The unit of
inductance is the henry (H), named in honor of the American inventor Joseph Henry (1797–
1878). It is clear from Eq. (6.4) that 1 henry equals 1 volt-second per ampere.
Inductance is the property whereby an inductor exhibits opposition to the change of
current flowing through it, measured in henrys (H).
The inductance of an inductor depends on its physical dimension and construction.
Formulas for calculating the inductance of inductors of different shapes are derived from
electromagnetic theory as shown in section 7.7.
Typical practical inductors have inductance values ranging from a few microhenrys (μH),
as in communication systems, to tens of henrys (H) as in power systems. Inductors may be
fixed or variable. The core may be made of iron, steel, plastic, or air. The terms coil and choke
are also used for inductors. The circuit symbols for inductors are shown in Fig. 7.8.
Figure 7.8 Circuit symbols for inductors: (a) air-core, (b) iron-core, (c) variable iron-core.
The current-voltage relationship is obtained from Eq. (7.4) as
Fig. 7.7

𝒅𝒊 =
𝟏
𝑳
𝒗 𝒅𝒕
Integrating gives 𝒊 =
𝟏
𝑳
∫ 𝒗(𝒕)
𝒕
𝒕𝟎
𝒅𝒕 + 𝒊(𝒕𝟎) (7.5)
where i(t0) is the total current for −∞ < t < t0 and i(−∞) = 0.
The inductor is designed to store energy in its magnetic field. The energy stored can be
obtained from Eqs. (6.4) and (6.5). The power delivered to the inductor is
p = vi = (L
𝒅𝒊
𝒅𝒕
) i (7.6)
The energy stored is 𝒘 =
𝟏
𝟐
𝑳𝒊𝟐
(7.7)
We should note the following important properties of an inductor.
1. Note from Eq. (7.4) that the voltage across an inductor is zero when the current is constant.
Thus, An inductor acts like a short circuit to dc.
2. An important property of the inductor is its opposition to the change in current flowing
through it. The current through an inductor cannot change instantaneously.
However, the voltage across an inductor can change abruptly.
3. Like the ideal capacitor, the ideal inductor does not dissipate energy. The energy stored in it
can be retrieved at a later time.
4. A practical, nonideal inductor has a significant resistive
component, as shown in Fig. 7.9. This is due to the fact that the
inductor is made of a conducting material such as copper, which
has some resistance. This resistance is called the winding
resistance Rw, and it appears in series with the inductance of the
inductor. The presence of Rw makes it both an energy storage
device and an energy dissipation device. The nonideal inductor also has a winding capacitance
Cw due to the capacitive coupling between the conducting coils. Cw is very small and can be
ignored in most cases, except at high frequencies. We will assume ideal inductors in this book.
Example 7.6: The current through a 0.1-H inductor is i(t) = 10te−5t
A. Find the voltage across
the inductor and the energy stored in it.
Solution: Since v = Ldi/dt and L = 0.1 H,
𝐯 = 𝟎. 𝟏
𝒅
𝒅𝒕
(𝟏𝟎𝒕𝒆−𝟓𝒕
) = 𝒆−𝟓𝒕
+ 𝒕 (−𝟓)𝒆−𝟓𝒕
= 𝒆−𝟓𝒕
(𝟏 − 𝟓𝒕) 𝑽
The energy stored is 𝒘 =
𝟏
𝟐
𝑳𝒊𝟐
=
𝟏
𝟐
(𝟎. 𝟏)𝟏𝟎𝟎𝒕𝟐
𝒆−𝟏𝟎𝒕
= 𝟓𝒕𝟐
𝒆−𝟏𝟎𝒕
𝑱
Practice problem 7.1: If the current through a 1-mH inductor is i(t) = 20 cos 100t mA, find
the terminal voltage and the energy stored.
Figure 7.9 Circuit model
for a practical inductor.

Answer: −2 sin 100t mV, 0.2 cos2
100t μJ.
Example 7.7: Consider the circuit in Fig. 7.10(a). Under dc conditions, find:
(a) i, vC, and iL, (b) the energy stored in the capacitor and inductor.
Solution: (a) Under dc conditions, we replace the capacitor
with an open circuit and the inductor with a short circuit, as in
Fig. 7.10(b). It is evident from Fig. 7.10(b) that
i = iL =
𝟏𝟐
𝟏 + 𝟓
= 𝟐 𝑨
The voltage vC is the same as the voltage across the 5-Ω
resistor. Hence,
vC = 5×i = 10 V
(b) The energy in the capacitor is
𝒘𝑪 =
𝟏
𝟐
𝑪𝒗𝑪
𝟐
=
𝟏
𝟐
(𝟏)(𝟏𝟎𝟐
) = 𝟓𝟎 𝑱
and that in the inductor is
𝒘𝑳 =
𝟏
𝟐
𝑳𝒊𝑳
𝟐
= 𝟏𝟐(𝟐)(𝟐𝟐
) = 𝟒 𝑱
Practice problem 7.2: Determine vC, iL, and the energy stored in the capacitor and inductor in
the circuit of Figure below under dc conditions.
Answer: 3 V, 3 A, 9 J, 1.125 J.
7.11 SERIES AND PARALLEL INDUCTORS
Now that the inductor has been added to our list of passive elements, it is necessary to
extend the powerful tool of series-parallel combination. We need to know how to find the
equivalent inductance of a series-connected or parallel-connected set of inductors found in
practical circuits.
Consider a series connection of N inductors, as shown in Fig. 7.11(a), with the equivalent
circuit shown in Fig. 7.11(b). The inductors have the same current through them. Applying
KVL to the loop,
v = v1 + v2 + v3 +· · ·+ vN (7.8)
Substituting vk = Lk di/dt results in v = L1
𝒅𝒊
𝒅𝒕
+ L2
𝒅𝒊
𝒅𝒕
+ L3
𝒅𝒊
𝒅𝒕
+· · ·+LN
𝒅𝒊
𝒅𝒕
= (L1 + L2 + L3 +· · ·+LN)
𝒅𝒊
𝒅𝒕
= (∑ 𝑳𝒌
𝑵
𝒌=𝟏 )
𝒅𝒊
𝒅𝒕
= Leq
𝒅𝒊
𝒅𝒕
(7.9)
Figure7.10 For Example 7.7.

Where Leq = L1 + L2 + L3 +· · ·+LN (7.10)
Figure 7.11 (a) A series connection of N inductors, (b) equivalent circuit for the series inductors.
The equivalent inductance of series-connected inductors is the sum of the
individual inductances.
Inductors in series are combined in exactly the same way as resistors in series.
We now consider a parallel connection of N inductors, as shown in Fig. 7.12(a), with the
equivalent circuit in Fig. 7.12(b). The inductors have the same voltage across them. Using
KCL,
i = i1 + i2 + i3 +· · ·+iN (7.11)
But ik =
𝟏
𝑳𝒌
∫ 𝒗𝒅𝒕 + 𝒊𝒌
𝒕
𝒕𝟎
(𝒕𝟎) ; hence,
Figure 7.12 (a) A parallel connection of N inductors, (b) equivalent circuit for the parallel inductors.
i =
𝟏
𝑳𝟏
∫ 𝒗𝒅𝒕 + 𝒊𝟏
𝒕
𝒕𝟎
(𝒕𝟎) +
𝟏
𝑳𝟐
∫ 𝒗𝒅𝒕 + 𝒊𝟐
𝒕
𝒕𝟎
(𝒕𝟎) + · · ·+
𝟏
𝑳𝑵
∫ 𝒗𝒅𝒕 + 𝒊𝑵
𝒕
𝒕𝟎
(𝒕𝟎)
= (
1
𝐿1
+
1
𝐿2
+ ⋯ +
1
𝐿𝑁
) ∫ 𝒗𝒅𝒕 + 𝒊𝟏
𝒕
𝒕𝟎
(𝒕𝟎) + 𝒊𝟐(𝒕𝟎) + ⋯ + 𝒊𝑵(𝒕𝟎) =
𝟏
𝑳𝒆𝒒
∫ 𝒗𝒅𝒕 + 𝒊(𝒕𝟎)
𝒕
𝒕𝟎
(7.12)
Where
𝟏
𝑳𝒆𝒒
=
𝟏
𝑳𝟏
+
𝟏
𝑳𝟐
+ ⋯ +
𝟏
𝑳𝑵
(7.13)
The initial current i(t0) through Leq at t = t0 is expected by KCL to be the sum of the inductor
currents at t0. Thus, according to Eq. (7.12),
i (t0) = i1(t0) + i2(t0)+· · ·+ iN(t0)
According to Eq. (7.13),
The equivalent inductance of parallel inductors is the reciprocal of the sum of the
reciprocals of the individual inductances.
Note that the inductors in parallel are combined in the same way as resistors in parallel. For
two inductors in parallel (N = 2), Eq. (7.13) becomes

𝟏
𝑳𝒆𝒒
=
𝟏
𝑳𝟏
+
𝟏
𝑳𝟐
or 𝑳𝒆𝒒 =
𝑳𝟏𝑳𝟐
𝑳𝟏 + 𝑳𝟐
(7.14)
Example 7.8: Find the equivalent inductance of the circuit shown in Fig. 7.13.
Solution:
The 10-H, 12-H, and 20-H inductors are in series;
thus, combining them gives a 42-H inductance.
This 42-H inductor is in parallel with the 7-H
inductor so that they are combined, to give
𝟕 × 𝟒𝟐
𝟕 + 𝟒𝟐
= 𝟔 𝑯
This 6-H inductor is in series with the 4-H and 8-H inductors. Hence,
Leq = 4 + 6 + 8 = 18 H
Practice problem 7.3: Calculate the equivalent inductance for the inductive ladder network in
Figure below.
Answer: 25 mH.
7.12 THE SOURCE-FREE RL CIRCUIT
Consider the series connection of a resistor and an inductor, as
shown in Fig. 7.14. Our goal is to determine the circuit response,
which we will assume to be the current i(t) through the inductor. We
select the inductor current as the response in order to take advantage
of the idea that the inductor current cannot change instantaneously. At
t = 0, we assume that the inductor has an initial current I0, or
i(0) = I0 (7.15)
with the corresponding energy stored in the inductor as
𝒘(𝟎) =
𝟏
𝟐
𝑳𝑰𝟎
𝟐
(7.16)
Applying KVL around the loop in Fig. 7.14,
vL + vR = 0 (7.17)
But vL = L
𝒅𝒊
𝒅𝒕
and vR = i R. Thus, L
𝒅𝒊
𝒅𝒕
+ Ri = 0
Rearranging terms and integrating gives
i(t) = I0e−Rt/L
(7.18)
Figure 7.14 A source-free
RL circuit.

This shows that the natural response of the RL circuit is an exponential decay of the initial
current. The current response is shown in Fig. 7.15. The time constant for the RL circuit is
𝝉 =
𝑳
𝑹
(7.19)
with τ again having the unit of seconds. Thus, Eq. (7.18) may be written as
i(t) = I0e−t/τ
(7.20)
With the current in Eq. (7.20), we can find the voltage across the resistor as
vR(t) = i×R = I0Re−t/τ
(7.21)
The power dissipated in the resistor is
p = vR×i = 𝑰𝟎
𝟐
R e−2t/τ
(7.22)
The energy absorbed by the resistor is
wR(t) =
𝟏
𝟐
𝑳𝑰𝟎
𝟐
(1 − e−2t/τ
) (7.23)
Note that as t →∞, wR(∞) →
𝟏
𝟐
𝑳𝑰𝟎
𝟐
, which is the same as
wL(0), the initial energy stored in the inductor as in Eq. (7.16).
Again, the energy initially stored in the inductor is eventually
dissipated in the resistor.
The Key to Working with a Source - free RL Circuit is to Find :
1. The initial current i(0) = I0 through the inductor.
2. The time constant τ of the circuit.
Example 7.9: Assuming that i(0) = 10 A, calculate i(t) and ix (t) in the circuit in Fig. 7.16.
Solution: There are two ways we can solve this problem. One
way is to obtain the equivalent resistance at the inductor
terminals and then use Eq. (7.20). The other way is to start from
scratch by using Kirchhoff’s voltage law.
The equivalent resistance is the same as the Thevenin resistance
at the inductor terminals. Because of the dependent source, we
insert a voltage source with vo = 1 V at the inductor terminals a-b, as in Fig. 7.17(a).
Applying KVL to the two loops results in
2(i1 − i2) + 1 = 0 ⇒ i1 − i2 = −1/2 (7.9.1)
6i2 − 2i1 − 3i1 = 0 ⇒ i2 = (5/6)i1 (7.9.2)
Substituting Eq. (7.9.2) into Eq. (7.9.1) gives
i1 = −3 A, io = −i1 = 3 A
Hence, Req = RTh = vo/io =(1/3)Ω
Figure 7.15 The current Response of the RL circuit.

The time constant is τ =L/Req =
𝟏/𝟐
𝟏/𝟑
= (3/2)s
Thus, the current through the inductor is
i(t) = i(0)e−t/τ
= 10e−(2/3)t
A, t>0
The voltage across the inductor is
𝐯 = 𝐋
𝐝𝐢
𝐝𝐭
= 𝟎. 𝟓(𝟏𝟎)(
−𝟐
𝟑
)𝐞−(𝟐/𝟑)𝐭
= −
𝟏𝟎
𝟑
𝐞−(𝟐/𝟑)𝐭
𝐕
Since the inductor and the 2-_ resistor are in parallel,
ix (t) = v/2 = −1.667e−(2/3)t
A, t>0
Practice problem 7.4: Find i and vx in the circuit in Figure below. Let i(0) = 5 A.
Answer: i= 5e−53t
A, vx= −15e−53t
V.
7.13 STEP RESPONSE OF AN RL CIRCUIT
Consider the RL circuit in Fig. 7.18(a), which may be replaced by the circuit in Fig. 7.18(b).
Again, our goal is to find the inductor current i as the circuit response. Rather than apply
Kirchhoff’s laws. Let the response be the sum of the natural current and the forced current,
i = in + if (7.24)
We know tht the natural response is always a decaying exponential, that is,
in = Ae−t/τ
, τ= L/R (7.25)
where A is a constant to be determined.
The forced response is the value of the current a long time after the switch in Fig. 7.18(a) is
closed. We know that the natural response essentially dies out after five time constants. At that
time, the inductor becomes a short circuit, and the voltage across it is zero. The entire source
voltage Vs appears across R. Thus, the forced response is
if =Vs/R (7.26)
Substituting, Eqs. (7.24) and (7.25) into Eq. (7.23) gives
i = Ae−t/τ
+Vs/R (7.27)
Figure 7.17 Solving the
circuit in Fig. 7.9.

We now determine the constant A from the initial value of i. Let I0 be the initial current
through the inductor, which may come from a source other than Vs. Since the current through
the inductor cannot change instantaneously,
i(0+
) = i(0−
) = I0 (7.28)
Thus at t = 0, Eq. (7.27) becomes
I0 = A +VsR
From this, we obtain A as
A = I0 –Vs/R
Substituting for A in Eq. (7.27), we get
i(t) =Vs/R + (I0 –Vs /R )e−t/τ
(7.29)
This is the complete response of the RL circuit. It is illustrated in Fig.7.19. The response in Eq.
(7.29) may be written as
i(t) = i(∞) + [i(0) − i(∞)]e−t/τ
(7.30)
where i(0) and i(∞) are the initial and final values of i. Thus, to find the step response of an RL
circuit requires three things:
1. The initial inductor current i(0) at t = 0+
.
2. The final inductor current i(∞).
3. The time constant τ.
We obtain item 1 from the given circuit for t < 0 and items 2 and 3 from the circuit for t > 0.
Once these items are determined, we obtain the response using Eq. (7.30). Keep in mind that
this technique applies only for step responses.
Figure 7.19 Total response of the RL circuit with initial inductor current I0.
Figure 7.18 An RL circuit with a step input voltage.

Again, if the switching takes place at time t = t0 instead of t = 0, Eq. (7.30) becomes
i(t) = i(∞) + [i(t0) − i(∞)]e−(t−t0)/τ
(7.31)
If I0 = 0, then
𝒊(𝒕) = {
𝟎 , 𝒕 < 0
𝑽𝒔
𝑹
(𝟏 − 𝒆−𝒕/𝝉
), 𝒕 > 0
(7.32a)
or i(t) =
𝑽𝒔
𝑹
(1 − e−t/τ
)u(t) (7.32b)
This is the step response of the RL circuit. The voltage across the inductor is obtained from Eq.
(7.32) using v = Ldi/dt. We get
v(t) = L
𝒅𝒊
𝒅𝒕
= Vs
𝑳
𝝉𝑹
e−t/τ
, τ =
𝑳
𝑹
t>0
or v(t) = Vs e−t/τ
u(t) (7.33)
Figure 7.20 shows the step responses in Eqs. (7.32) and (7.33).
Figure 7.20 Step responses of an RL circuit with no initial inductor current: (a) current response, (b) voltage response.
Example 7.12: Find i(t) in the circuit in Fig. 7.21 for t > 0. Assume that the switch has been
closed for a long time.
Solution: When t < 0, the 3-Ω resistor is short-circuited, and the inductor acts like a short
circuit. The current through the inductor at t = 0− (i.e., just before t = 0) is
i(0−
) =10 / 2 = 5 A
Since the inductor current cannot change instantaneously,
i(0) = i(0+
) = i(0−
) = 5 A
When t > 0, the switch is open. The 2-Ω and 3-Ω resistors are in series, so that

i(∞) =
𝟏𝟎
𝟐 + 𝟑
= 2 A
The Thevenin resistance across the inductor terminals is
RTh = 2 + 3 = 5 Ω
For the time constant,
τ =
𝑳
𝑹𝑻𝒉
=
𝟏/𝟑
𝟓
=1 / 15 s
Thus,
i(t) = i(∞) + [i(0) − i(∞)]e−t/τ
= 2 + (5 − 2)e−15t
= 2 + 3e−15t
A, t>0
Practice problem 7.5: The switch in Figure below has been closed for a long time. It opens at t
= 0. Find i(t) for t > 0.
Answer: (2 + e−10t
) A, t > 0.

Chapter eight
Sinusoidal Alternating Waveforms
8.1 Introduction
The analysis thus far has been limited to dc networks, networks in which the currents or
voltages are fixed in magnitude except for transient effects. We will now turn our attention
to the analysis of networks in which the magnitude of the source varies in a set manner. Of
particular interest is the time-varying voltage that is commercially available in large
quantities and is commonly called the ac voltage, (The letters ac are an abbreviation for
alternating current.)
Each waveform of Fig. 8.1 is an alternating waveform available from commercial
supplies. The term alternating indicates only that the waveform alternates between two
prescribed levels in a set time sequence (Fig. 8.1).
Fig. 8.1 Alternating waveforms.
To beabsolutely correct, the term sinusoidal, square wave, or triangular must also be
applied. The pattern of particular interest is the sinusoidal ac waveform for voltage of Fig.
8.1. Since this type of signal is encountered in the vast majority of instances, the abbreviated
phrases ac voltage and ac current are commonly applied without confusion.
8.2 Sinusoidal ac Voltage characteristics and definitions
Generation Sinusoidal ac voltages are available from a variety of sources. The most
common source is the typical home outlet, which provides an ac voltage that originates at a
power plant; such a power plant is most commonly fueled by water power, oil, gas, or
nuclear fusion. In each case an ac generator (also called an alternator).
The power to the shaft developed by one of the energy sources listed will turn a rotor
(constructed of alternating magnetic poles) inside a set of windings housed in the stator (the
stationary part of the dynamo) and will induce a voltage across the windings of the stator, as
defined by Faraday’s law,
𝒆 = 𝑵
𝒅𝜱
𝒅𝒕
Definitions

The sinusoidal waveform of Fig.8.2 with its additional notation will now be used as a model
in defining a few basic terms. These terms, however, can be applied to any alternating
waveform. It is important to remember as you proceed through the various definitions that
the vertical scaling is in volts or amperes and the horizontal scaling is always in units of
time.
FIG. 8.2 Important parameters for a sinusoidal voltage.
Waveform: The path traced by a quantity, such as the voltage in Fig. 8.2, plotted as a
function of some variable such as time (as above), position, degrees, radians, temperature,
and so on.
Instantaneous value: The magnitude of a waveform at any instant of time; denoted by
lowercase letters (e1, e2).
Peak amplitude: The maximum value of a waveform as measured from its average, or
mean, value, denoted by uppercase letters (such as Em for sources of voltage and Vm for the
voltage drop across a load).
Peak value: The maximum instantaneous value of a function as measured from the zero-
volt level. For the waveform of Fig. 8.2, the peak amplitude and peak value are the same,
since the average value of the function is zero volts.
Peak-to-peak value: Denoted by Ep-por Vp-p, the full voltage between positive and negative
peaks of the waveform, that is, the sum of the magnitude of the positive and negative peaks.
Periodic waveform: A waveform that continually repeats itself after the same time interval.
The waveform of Fig. 8.2 is a periodic waveform.
Period (T): The time interval between successive repetitions of a periodic waveform (the
period T1 = T2 = T3 in Fig. 8.2).
Cycle: The portion of a waveform contained in one period of time. The cycles within T 1 ,
T 2 , and T 3 of Fig. 8.2 may appear different in Fig. 8.3.

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