Electrical and Electronic Fundamental

4. Consider the circuit in Figure 2.22. Find the equivalent resistance at
terminals:

a. a-b

b. c-d

50Ω 15Ω 60Ω
a
c
100Ω 100Ω d

b
150Ω

Figure 2.22

Summary

In this chapter we have studied that:

1. A resistor is a passive element in which the voltage v across it is
directly proportional to the current i through it. That is, a resistor is a
device that obeys Ohm’s law, v = iR, where R is the resistance of the
resistor.

2. A short circuit is a resistor (a perfectly conducting wire) with zero
resistance (R=0). An open circuit is a resistor with infinite resistance
(R=∞).

3. The conductance G of a resistor is the reciprocal of its resistance,

G = 1
R

4. A branch is a single two-terminal element in an electric circuit. A
node is the point of connection between two or more branches. A
43

loop is a closed path in a circuit. The number of branches b, the
number of nodes n, and the number of independent loops l in a
network are related as b=1+n-1
5. Kirchhoff’s current law (KCL) states that the currents at any node
algebraically sum to zero. In other words, the sum of the currents
entering a node equals the sum of currents leaving the node.
6. Kirchhoff’s voltage law (KVL) states that the voltages around a closed
path algebraically sum to zero. In other words, the sum of voltage
rises equals the sum of voltage drops.
7. Two elements are in series when they are connected sequentially,
end to end. When elements are in series, the same current flows
through them. They are in parallel if they are connected to the same
two nodes. Elements in parallel always have the same voltage across
them.

44

CHAPTER 3

CIRCUIT ANALYSIS

Introduction
Having understood the fundamental laws of circuit theory (Ohm’s

law and Kirchhoff’s laws), we are now prepared to apply these laws to
develop two powerful techniques for circuit analysis: nodal analysis, which
is based on a systematic application of Kirchhoff’s current law (KCL), and
mesh analysis, which is based on a systematic application of Kirchhoff’s
voltage law (KVL).

Learning Objectives
The objectives of this chapter are to :
1. describe nodal and mesh analysis.
2. apply the concept of supernode and supermesh.

3.1 Nodal Analysis
A node is any point or line segment connecting circuit elements.

Nodal analysis is applying KCL at each non-reference node

45

i4
ac

4

-- + + v4 - -

i1 1 v1 i2 2 v2 i3 3 v3 i5 5 v5

++ - i6 +

6

b + v6 - d

Figure 3.1

Steps to the solution:

1. Identify nodes

2. Express currents as a function of node voltages

3. Kirchhoff’s Current Law at each node except for reference node

4. Solve the resulting n –1 simultaneous linear equations for the voltages,
where n equals the number of nodes (using Cramer’s rule or
simultaneous equation)For active or passive elements:

v1 +_ v2 v1 R v2

v1 - v2 +_

v1 - v2

(a) (b)

Figure 3.2: (a). Source element, (b). Resistive element
For source element we get the result

46

Vs = v1 - v2 (3.1)

For a resistive element we apply Ohm’s law and the passive convention to
get

i = v1 - v2 (3.2)
R

with the current flow from left to right.

Example 3.1
Calculate the node voltages in the circuit shown in Figure 3.3(a).

5A

4Ω 2
1

2Ω 6Ω 10A

(a)

Figure 3.3(a): Original circuit

Solution:
Consider Figure 3.3(b), where the circuit in Figure 3.3(a) has been

prepared for nodal analysis. Notice how the currents are selected for the
application of KCL. Except for the branches with current sources, the
labeling of the currents is arbitrary but consistent. (By consistent, we mean

47

that if, for example, we assume that 2 enters the 4 Ω resistor from the left-
hand side, 2 must leave the resistor from the right-hand side.) The reference
node is selected, and the node voltages 1 and 2 are now to be

determined.

i1 5A 10A

i2 i1
v1 4Ω i2

i3 v2
i5
2Ω 6Ω

(b)

Figure 3.3 (b): Circuit for analysis

At node 1, applying KCL and Ohm’s law gives

i1 = i2 + i3

5= v1 - v2 + v1 - 0
4 2

Multiplying each term in the last equation by 4, we obtain

20 = v1 - v2 + 2v1

Or

3v1 - v2 = 20  (1)

At node 2, we do the same thing and get

i2 + i4 = i1 + i5

48

v1 - v2 + 10 = 5 + v2 - 0
4 6

Multiplying each term by 12 result in

3v1 - 3v2 + 120 = 60 + 2v2

Or

-3v1+5v2=60  (2)

Using the elimination technique, we add Equations (1) and (2).
4v2 = 80
80
v2 = 4 = 20V

Substituting v2=20V in Equation (1) gives
3v1 - 20 = 20
40
v1 = 3 = 13.33V

49

Example 3.2
Determine Ix in the circuit shown in Figure 3.4 using nodal analysis.

1kΩ 4kΩ

+ Ix
9V _
+
2kΩ _ 6V

Figure 3.4

Solution:

Let Vx be the voltage at the node between 1kΩ and 4kΩ resistors.

9 - Vx + 6 - Vx = Vx
1k 4k 2k

Vx = 6V

Ix = Vx = 3mA
2k

50

3.2 Node Analysis with Voltage Sources

When the voltage source is connected between ground and another
node.

a R2 b

+ R1 R3 is
vs _

Figure 3.5

Note that va = vs and apply KCL at node b to get

vb = R2R3is + Rsvs (3.3)
R2 + R3

3.2.1 Supernode+
_
When the voltage source is not connected to ground or connected
between two nodes.

supernode
vs

va vb

R1 R2 is

Figure 3.6
51

Apply KCL to nodes a and b

va + vb = is = vs + vb + vb (3.4)
R1 R2 R1 R2 (3.5)

Solving for

vb = R1R2is - R2vs
R1 + R2

Example 3.3
For the circuit shown in Figure 3.7, find the node voltages.

10Ω

2V+
v1 v2_

2A 2Ω 4Ω 7A

Figure 3.7

Solution:

The supernode contains the 2V source, nodes 1 and 2, and the 10 Ω
resistor. Applying KCL to the supernode as shown in Figure 3.8(a) gives

2 = i1 + i2 + 7

Expressing i1 and i2 in terms of the node voltages

2 = v1 - 0 + v2 - 0 + 7
2 4

52

8 = 2v1 + v2 + 28
Or

v2 = -20 - 2v1  (1)
To get the relationship between v1 and v2 , we apply KVL to the circuit in
Figure 3.8(b). Going around the loop, we obtain

-v1 - 2 + v2 = 0
v2 = v1 + 2  (2)

1 v1 2 v2 2V+
2A i1 i2 7A 12_

2A 2Ω 4Ω 7A ++

(a) v1 v2

--

(b)

Figure 3.8: Applying: (a) KCL to the supernode, (b) KVL to the loop.

From Equation (1) and (2), we write
v2 = v1 + 2 = -2 - 2v1

or
3v2 = -22
v1 = -7.333V and v2 = v1 + 2 = -5.333V

Note that the 10 Ω resistor does not make any difference because it is
connected across the supernode.

53

Example 3.4
Apply nodal analysis to find io and the power dissipated in each resistor in
the circuit of Figure 3.9.

2A

10V ++ 3S 4A
io __ 5S

6S

Solution: Figure 3.9

2A

10V 3S 3
1 2 4A
io
5S
6S

Figure 3.10

Nodes 1 and 2 form a supernode so that

v1 = v2 + 10  (1)

54

At the supernode,
2 + 6v1 + 5v2 = 3(v3 - v2)
2 + 6v1+ 8v2 = 3v3  (2)

At node 3,
2 + 4 = 3(v3 - v2)
v3 = v2 + 2  (3)

Substituting (1) and (3) into (2),
2 + 6v2 + 60 + 8v2 = 3v2 + 6
56
v2 = - 11

Thus,
54

v1 = v2 + 10 = 11
io = 6v1 = 29.45A

Hence

P6S = v12 = v21G = �1541�2 × 6= 144.6W
R

P5S= v22G = �-1516�2 × 5 = 129.6W

P3S = (vL - v3)2G = (2)23 = 12W
55

3.3 Mesh Analysis

A loop is any closed path which does not pass the same node more
than once. Loop analysis is developed by applying KVL around loop in the
circuit

a I1 R1 b I2 R2
I3 i2
+ i1 R3 c
V1 _
+
_ V2

Figure 3.11: A circuit with two meshes

Steps of Mesh Analysis:
1. Assign a current to each mesh: i1, i2
2. Applying KVL around each loop to get an equation in terms of the

loop currents
V1 = i1R1 + (i1 - i2)R3 = (R1 + R3)i1 - R3i2
-V2 = (i2 - i1)R3 + i2R2 = -R3i1 + (R2 + R3)i2

3. Solve the resulting system of linear equations
Note: Use small letter i for mesh currents
Use block letter I for branch currents
I1 = i1, I2 = i2 , I3= i1 - i2

56

Example 3.5

For the circuit in Figure 3.16, find the branch currents I1, I2 and I3 using mesh
analysis.

I1 5Ω I2 6Ω

I3

+ 10Ω
15V _
i1 i2 4Ω

+
_ 10V

Figure 3.12

Solution:

We first obtain the mesh currents using KVL. For mesh 1,

-15 + 5i1 + 10(i1 - i2) + 10 = 0
3i1 - 2i2 = 1  (1)
For mesh 2,

6i2 + 4i2 + 10(i2 - i1) - 10 = 0
i1 = 2i2 - 1  (2)

Using the substitution method, we substitute equation (2) into equation (1),
and write

6i2 - 3 - 2i2 = 1
i2 = 1A

57

From equation (2), i1 = 2i2 - 1 = 2 - 1 = 1A. Thus,

I1 = i1 = 1A, I2 = i2 = 1A, i3 = i1 - i2 = 1 - 1 = 0A,

Example 3.6
Obtain voin the circuit of Figure 3.13.

+ + +
30V _ 20V _
vo 4kΩ
2kΩ 5kΩ
-

Figure 3.13

Solution:

+ + +
30V _ 20V _
vo 4kΩ
2kΩ i1
i2 -
5kΩ

Assume that 1 and 2 are in mA. We apply mesh analysis.

For mesh 1,

-30 + 20 + 7i1 - 5i2 = 0
7i1 - 5i2 = 10  (1)

58

For mesh 2,
-20 + 9i2 - 5i1 = 0
-5i1 + 9i2 = 20  (2)

Solving (1) and (2), we obtain,
i2 = 5A
vo = 4i2 = 20V

3.4 Mesh Analysis with Current Source
We can’t use KVL around the loop because we don’t know the voltage.
The 4mA current source sets i2:

i2 = -4mA
The 2mA current source sets a constraint on i1 and i3:

i1 - i3 = 2mA
We have two equations and three unknowns.

2kΩ

2mA i3

1kΩ

+ i1 2kΩ i2 4mA
12V _

Figure 3.14
59

Where is the third equation?

2kΩ

The supermesh 2mA i3 The supermesh
surrounds this does not
source 1kΩ
include this
+ i1 2kΩ i2 source
12V _
4mA

Figure 3.15

3.4.1 Supermesh
When there is a current source between two loops
KVL around the Supermesh:

-12 + i3(2k) + (i3 - i2)(1k) + (i1 - i2)(2k)=0
i3(2k) + (i3 - i2)(1k) + (i1 - i2)(2k) = 12

60

Example 3.7
For the circuit in Figure 3.16, find i1 to i4using mesh analysis.

2Ω

i1 4Ω 2Ω Io

i1 3Io 8Ω +
P i3 i4 _ 10V
i2 5A
6Ω

i2

i2 Q i3

Figure 3.16

Solution:

Note that meshes 1 and 2 form a supermesh since they have an
independent current source in common. Also, meshes 2 and 3 form another
supermesh because they have a dependent current source in common.
The two supermeshes intersect and form a larger supermesh as shown.

Applying KVL to the larger supermesh,

2i1 + 4i3 + 8(i3 - i4) + 6i2=0
i1 + 3i2 + 6i3 - 4i4 = 0  (1)

For the independent current source, we apply KCL to node P:
i2 = i1 + 5  (2)

61

For the dependent current source, we apply KCL to node Q:
i2 = i3 + 3io

But = − 4, hence,
i2 = i3 + 3i4  (3)

Applying KVL in mesh 4,
2i4 + 8(i4 - i3) + 10 = 0
5i4 - 4i3 = -5  (4)

From equations (1) to (4),
i1 = -7.5A, i2 = -2.5A, i3 = 3.93A, i4 = 2.143A

62

Example 3.8
Find current, i in the circuit in Figure 3.17.

4Ω 8Ω

2Ω 4A
6Ω
i 1Ω
3Ω
+
30V _

Figure 3.17

Solution: 8Ω
Redraw circuit in Figure 3.17

4Ω

i3 i4
2Ω 6Ω

+ i1 3Ω i2 1Ω
30V _

For loop 1,
30 = 5i1 - 3i2 - 2i3  (1)

63

For loop 2,
10i2 - 3i1 - 6i4 = 0  (2)

For the supermesh,
6i3 - 14i4 - 2i1 - 6i2 = 0  (3)

But i4-i3=4 which leads to
i4 = i3 + 4  (4)

Solving (1) to (4) by elimination gives
i = i1 = 8.561A

3.5 Nodal Versus Mesh Analysis
Nodal for voltage, mesh for current.
1. Nodal for network with current source and parallel elements, mesh

for network with voltage source and series elements.
2. Nodal for network having fewer nodes than meshes, mesh for

network having fewer mesh than nodes.
3. Nodal more suitable for computer program, and for circuits

containing op-amps.

64

Exercise

1. Determine Ix in the circuit shown in Figure 3.18 using nodal analysis.

1kΩ 4kΩ +
+ Ix _ 6V
9V _ 2kΩ

Figure 3.18

2. Find the currents i1 through i4 and the voltage vo in the circuit in Figure
3.19.

vo i3 i4
i1 i2 60Ω

10A 10Ω 20Ω 30Ω 2A

Figure 3.19

3. Calculate v1 and v2 in the circuit of Figure 3.20 using nodal analysis.

v1 2Ω 2V
v2
+
_

8Ω 4Ω 3A

Figure 3.20
65

4. Using nodal analysis, find vo in the circuit of Figure 3.21.

5A

2Ω 8Ω

1Ω + + 20V
_
+ vo
40V _
_

Figure 3.21

5. Apply mesh analysis to find io in Figure 3.22.

10Ω

i1 6V
2Ω

+
_

1Ω

4Ω i2 i3 5Ω

+

_ 8V

Figure 3.22

66

6. Use mesh analysis to obtain io in the circuit of Figure 3.23.

6V

+
_

2Ω 4Ω +
io 1Ω 3A _ 12V

5Ω

Figure 3.23

7. Use mesh analysis to find the current io in the circuit in Figure 3.24.

io

4Ω 2Ω 8Ω
10Ω 3io

+
60V _

Figure 3.24

67

Summary
In this chapter we have studied that:
1. Nodal analysis is the application of Kirchhoff’s current law at the non-

reference nodes. (It is applicable to both planar and non-planar
circuits). We express the result in terms of the node voltages. Solving
the simultaneous equations yields the node voltages.
2. A supernode consists of two non-reference nodes connected by a
(dependent or independent) voltage source.
3. Mesh analysis is the application of Kirchhoff’s voltage law around
meshes in a planar circuit. We express the result in terms of mesh
currents. Solving the simultaneous equation yields the mesh currents.
4. A supermesh consists of two meshes that have a (dependent or
independent) current source in common.
5. Nodal analysis is normally used when a circuit has fewer node
equations than mesh equations. Mesh analysis is normally used when
a circuit has fewer mesh equations than node equations.

68

CHAPTER 4

CAPACITOR AND INDUCTOR

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. 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.

Learning Objectives
The objectives of this chapter are to:
1. Learn about capacitors and capacitance.
2. Understand how to combine capacitors in series and parallel.
3. Learn about inductors and inductance.
4. Understand how to combine inductors in series and parallel.

69

4.1 Capacitors
Capacitor C consists of two conducting plates of area A separated

by distance d using an insulator (or dielectric of permittivity e)

Dielectric with permittivity ε
Metal plates,
each with area, A

d

Figure 4.1: A typical capacitor

C= εA (4.1)
d

4.1.1 Capacitance
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).

q

Slope = C

V

Figure 4.2
70

q = CV (4.2)
(4.3)
Note: there is no linear relationship between V and I (current) (4.4)
(4.5)
Since, i = dq
Hence dt

i= d(CV)
dt

Or

dV
i = C dt

i
Slope = C

dV/dt

Figure 4.3

4.1.2 Voltage/Current Relationships
Energy is stored in the capacitor as more charge is added when a

voltage is maintained between the plates. However, such voltage cannot
change abruptly as it depends on time.

71

1 tt
C 1
v = � idt= C � idt + v(to) (4.6)
(4.7)
-∞ to

tt t dv
dt
W = � pdt = � vidt = � Cv dt

-∞ -∞ -∞

=C t = 1 t = q2
2 2C
� vdv Cv2�

-∞ t=-∞

4.1.3 Voltage Across Capacitor

The amount of charge stored, q(t) must be continuous (i.e. it cannot
change instantaneously). since q(t) = C v(t), it implies that the voltage
across a capacitor cannot change instantaneously. Hence, v(t) must be
continuous (no abrupt jump).

v(t) Since = ( )
1

∆t i=0 t<0
= �∆1 �
0 < t < ∆t

t(s) i=0 t > ∆t

Figure 4.4

72

4.1.4 Current Through Capacitor

A capacitor is an open circuit when subjected to d.c. voltage in the
steady-state ( no current flow through the capacitor when voltage
applied across it is constant). When the voltage across a capacitor is not
changing with time, the current through the capacitor is zero ( iC = C
dV/dt ) Charge is deposited and accumulated on opposite plates of the
capacitor, but no current flow through.

An ideal capacitor does not dissipate energy: it merely stores energy
when iC > 0 and it delivers energy when iC < 0. A real capacitor has a parallel
model leakage resistance of high resistive value (∼108 W).

iC iC

+v - +v -
(a) (b)

Figure 4.5: Circuit symbols for capacitors a) fixed capacitor; b) variable
capacitor

Example 4.1

In 5 s, the voltage across a 40-mF capacitor changes from 160 V to 220 V.
Calculate the average current through the capacitor.

Solution:

i = dv = 40 × 10-3 �220 - 160� = 480mA
C dt 5

73

Example 4.2
The voltage across a 4-µF capacitor is shown in Figure 4.6. Find the current
waveform.

v(V)
10

2468 t(ms)
-10

Figure 4.6

Solution:

5000t, 0 < t < 2ms
v= � 20-5000t, 2 < t < 6ms

-40+5000t, 6 < t < 8ms

i = dv = 4 × 10-6 � 5, 0 < t < 2ms 20mA, 0 < t < 2ms
C dt 10-3 -5, 2 < t < 6ms = � -20mA, 2 < t < 6ms
5, 6 < t < 8ms 20mA, 6 < t < 8ms

4.2 Series and Parallel Capacitors

4.2.1 Capacitors in Parallel

The equivalent capacitance of N parallel connected capacitors is
the sum of the individual capacitances

Since i = i1 + i2 +….+ iN and ik = Ck dV (4.8)
dt

74

∴i = C1 dV + C1 dV + ….+CN dV = �∑kN=1 Ck� dV = Ceq dV
dt dt dt dt dt

Where Ceq = C1 + C2 +…+ CN

4.2.2 Current Through Capacitor in Parallel

Applying KCL to Figure 4.7,

is = i1 + i2
Where

i1 = C1 × is (4.9)
C1 + C2 (4.10)

i2 = C2 × is
C1 + C2

i1 i2
is C1 C2

Figure 4.7

75

From Figure 4.7,

v1 = v2 ⇒ Q1 = Q2 ⇒ Q2= C2 Q1
C2 C2 C1

is = i1 + i2 = dQ1 = dQ2 = d �Q1+ C2 Q1 �
dt dt dt C1

= d �C1C+1C2� Q1 = �C1C+1C2� dQ1 = �C1C+1C2� i1
dt dt

i1 = C1 is (proven)
C1 + C2

Similarly,

i2 = C2 is
C1 + C2

4.2.3 Capacitors in Series

The equivalent capacitance of series connected capacitors is the
reciprocal of the sum of the reciprocals of the individual capacitances.

Since v = v1 + v2 + v3 + ... + vN and

1 t
Ck
vk = � i(t) dt

t0

ttt t
1 1 1 1
∴v= C1 � i(t) dt + C2 � i(t) dt + C3 � i(t) dt + ... + CN � i(t) dt

t0 t0 t0 t0

�C11 1 1 tt
C2 C3 C1N� 1
= + + + . . . + � i(t) dt = C � i(t) dt

t0 t0

1111 1
where Ceq = C1 + C2 + C3 + . . . + CN (4.11)

76

4.2.4 Voltage Across Capacitor in Series

Applying KVL to the loop in Figure 4.8,

vs = v1 + v2
Where

v1 = C2 × vs (4.12)
C1+C2 (4.13)

v2 = C1 × vs
C1+C2

C1

+ + v1 - + C2
vs - v2
-

Figure 4.8

From Figure 4.8,

v1= 1 � idt , v2= 1 � idt
C1 C2

∴ C1v1 = C2v2 ⇒ v2= C1 v1
C2

vs= v1 + v2= v1 + C1 v1 = C1 + C2 v1
C2 C2

∴ v1 = C2 vs (proven)
C1 + C2

Similarly,

v2 = C1 vs
C1+C2

77

Example 4.3
The equivalent capacitance at terminals a-b in the circuit in Figure

4.9 is 30 µF. Calculate the value of C.
a
C
14µF
80µF
b
Figure 4.9

Solution:
C × 80

Ceq = 14 + C + 80 = 30
C = 20μF

Example 4.4
Find Ceq in the circuit of Figure 4.10 if all capacitors are 4 µF.

Figure 4.10
78

Solution:
4 µF in parallel with 4 µF = 8µF
4 µF in series with 4 µF = 2 µF
2 µF in parallel with 4 µF = 6 µF

Hence, the circuit is reduced to that shown below.
8µF

6µF 6µF

1 111 ⇒ Ceq=2.1818μF
Ceq = 6 + 6 + 8 =0.4583

4.3 Inductors

4.3.1 Inductance

The ability of a coil to oppose any change in current is a measure of
the self-inductance L of the coil, measured in henries (H).

N2μA N2μA (4.14)
L = l = μT l = μrLo

Inductance is the property whereby an inductor which consists of a
coil of conducting wire exhibits opposition to the change of current through
it.

79

Length, l

Cross-sectional area, A

Core material
Number of turns, N

Figure 4.11: Typical form of an inductor

Magnetic flux f linking such a coil of inductance L is proportional to
the current I through the coil, where

L=proportionality constant i.e. Φ=Li (4.15)

φ

Slope = L
i

Figure 4.12

Voltage across an inductor is directly proportional to the time rate of
change of current through it,

di (4.16)
VL = L dt

80

Ndφ/dt

Slope = L
di/dt

Figure 4.13

4.3.2 Voltage / Current Relationships

An inductor stores its energy in its magnetic field as more flux is added
when a current is flowing through the coil. However, such current cannot
change abruptly as it involves time.

tt (4.17)
1 1 (4.18)
i = L � vdt = L � vdt +i(to)

-∞ to

tt t di
Li dt dt
W = � pdt = � vidt = �

-∞ -∞ -∞

t = 1 t = 1 Li2(t) - 1 Li2(-∞)
2 2 2
= L � idi Li2�

-∞ t=-∞

Since i(∞) = 0,

W = 1 Li2 (4.19)
2

81

4.3.3 Current Through Inductor

Φ(t) must be continuous (i.e. it cannot change instantaneously).
Since Φ(t) = L i(t), it implies that the current through an inductor cannot
change instantaneously. Hence, i(t) must be continuous (no abrupt jump).

i(t) Since v(t)=L di
1 dt

∆t v = 0 t <0

v(t)=L �1� 0 < t <∆t

∆t

v = 0 t > ∆t

t(s)

Figure 4.14

4.3.4 Voltage Across Inductor
An inductor is a short circuit when subjected to d.c. voltage in the

steady-state ( no voltage difference as current is constant through the
inductor). When the current through an inductor is not changing with time,
the voltage across the inductor is zero ( VL = L di/dt ).

An ideal inductor does not dissipate energy: it merely stores energy
when VL > 0 and it delivers energy when VL < 0. A real inductor has a parallel
winding capacitance of low capacitive value and a small winding
resistance in series with the main inductance

82

Figure 4.15: Circuit symbols for inductors

Example 4.5
The current through a 12-mH inductor is 4 sin 100t A. Find the voltage,

and also the energy stored in the inductor for 0 < t < π/200 s.

Solution:

v = di = 12 × 10-3 × 4(100) cos 100t
L dt

= 4.8 cos 100t V

p = vi = 4.8 × sin100t cos100t = 9.6 sin 200t

t 11/200

w = � pdt = � 9.6 cos200t dt

00

= - 9.6 cos 11/200 J
200
200t�

0

= -48(cosπ - 1) mJ = 96 mJ

83

Example 4.6
The current through a 5mH inductor is shown in Figure 4.16. Determine

the voltage across the inductor at t=1,3, and 5ms.

i(A)
10

2468 t(ms)

Figure 4.16

Solution:

5t, 0<t<2ms
i = �10, 2<t<4ms

30-5t, 4<t<6ms

v = di = 5 × 10-3 5, 0<t<2ms
L dt 10-3 � 0, 2<t<4ms
-25, 4<t<6ms

At t=1ms, v=25 V
At t=3ms, v=0 V
At t=5ms, v=-25 V

4.4 Series and Parallel Inductors

4.4.1 Inductors in Parallel

The equivalent inductance of parallel inductors is the reciprocal of
the sum of the reciprocals of the individual inductances.

84

1 t
Lk
Since i = i1 + i2 + i3 + ... + iN and ik = � v(t)dt

to

ttt t
1 1 1 1
∴ = L1 � v(t)dt + L2 � v(t)dt + L3 � v(t)dt + . . . + L � v(t)dt

to to to to

�L11 1 1 tt
L2 1 LN 1
= + + L3 + ... + � � v(t)dt = Leq � v(t)dt

to to

where 1 = �L11 1 1 + ... + L1N� (4.20)
Leq + L2 + L3

4.4.2 Current Flow in Inductor in Parallel

Applying KCL to Figure 4.17,

is = i1 + i2
Where

i1 = L2 × is (4.21)
L1 + L2 (4.22)

i2 = L1 × is
L1 + L2

i1 i2
is L1 L2

Figure 4.17
85

From Figure 4.17,

is = i1+ i2 = �L11 + L12� � vdt = L1 + L2 � vdt
L1L2

∴ � vdt = L1L2 is
L1 + L2

∴ i1 = 1 � vdt = 1 L1L2 is = L1 L2 L2 is (proven)
L1 L1 L1 + L2 +

Similarly, i2 = L1 L2 L2 is
+

4.4.3 Inductors in Series

The equivalent inductance of series connected inductors is the sum
of the individual inductances.

di
Since vs = v1 + v2 + v3 + ... +vN and vk = Lk dt

∴ vs = di di di di
L1 dt + L2 dt + L3 dt + . . . + LN dt

= ( L1+ L2+ L3+ ... + LN ) di = di
dt Leq dt

where Leq =( L1+ L2+ L3+ ... + LN) (4.23)

86

4.4.4 Voltage Across Inductor in Series

Applying KVL to the loop in Figure 4.18,

vs = v1 + v2.
Where

v1 = L1 × vs (4.24)
L1 + L2 (4.25)

v2 = L2 × vs
L1 + L2

L1
+ v1 -

+ +
vs - v2 L2

-

Figure 4.18

From Figure 4.18,

vs = v1+ v2 = (L1 + L2) di
dt

∴ di = vs
dt L1 + L2

v1 = di = L1 L2 vs (proven)
L1 dt L1 +

Similarly, v1 = L2 vs
L1 + L2

87

Example 5.7:

Find the equivalent inductance of the circuit in Figure 4.19. Assume
all inductors are 10 mH.

Figure 4.19

Solution:
Converting the wye-subnetwork to its equivalent delta gives the circuit
below.

30mH

30mH

5mH
30mH

30 // 0 = 0
30 × 5

30//5 = 30 + 5 = 4.286 mH
30 × 4.286

Leq = 30//4.286 = 30 + 4.286 = 3.75 mH

88

Example 4.7
Determine Leq at terminals a-b of the circuit in Figure 4.20.

10mH

60mH

a 25mH 20mH

b

30mH

Figure 4.20

Solution:

1 = �610 + 1 + 310� = 1
L 20 10

L =10 mH

Thus,

Leq = 10 // (25+10)= 10×35 = 7.78 mH
10+35

89

Exercise
1. The voltage waveform in Figure 4.21 is applied across a 30µF

capacitor. Draw the current waveform through it.

v(t) (V)
10

2 4 6 8 10 12 t(ms)
-10

Figure 4.21

2. Find the equivalent capacitance between terminals a and b in the
circuit of Figure 4.22. All capacitances are in µF.

80

12 40
a

50 20
12 30 10

b
60

Figure 4.22

3. The voltage across a 2-H inductor is 20(1 - e-2t) V. If the initial current
through the inductor is 0.3 A, find the current and the energy stored
in the inductor at t = 1 s.

90

4. Find Leq at the terminals of the circuit in Figure 4.23.

6mH 8mH
a 5mH 12mH

8mH 6mH 4mH
b 10mH 8mH

Figure 4.23

Summary

In this chapter we have studied that:

1. The current through a capacitor is directly proportional to the time
rate of change of the voltage across it.

dv
i = C dt

The current through a capacitor is zero unless the voltage is
changing. Thus, a capacitor acts like an open circuit to a d.c. source.

2. The voltage across a capacitor is directly proportional to the time
integral of the current through it.

v = 1 t = 1 t + v0(t0)
C C
� idt � idt

-∞ t0

The voltage across a capacitor cannot change instantly.

91

3. Capacitors in series and in parallel are combined in the same way as
conductances.

4. The voltage across an inductor is directly proportional to the time rate
of change of the current through it.

v = L di
dt

The voltage across the inductor is zero unless the current is changing.
Thus, an inductor acts like a short circuit to a d.c. source.

5. The current through an inductor is directly proportional to the time
integral of the voltage across it.

i = 1 t = 1 t + i(t0)
L L
� vdt � vdt

-∞ t0

The current through an inductor cannot change instantly.

6. Inductors in series and in parallel are combined in the same way
resistors in series and in parallel are combined.

7. At any given time t, the energy stored in a capacitor is 1 Cv2, while
2

the energy stored in an inductor is 1 Li2.
2

92