CHAPTER 5

CAPACITORS
INDUCTORS
5.1 INTRODUCTION
5.2 CAPACITORS
5.3 SERIES AND PARALLEL CAPACITORS
5.4 INDUCTORS
5.5 SERIES AND PARALLEL INDUCTORS
5.1 INTRODUCTION
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.
The application of resistive circuits is quite limited. With the
introduction of capacitors and inductors in this chapter, we will be
able to analyze more important and practical circuits.
We begin by introducing capacitors and describing how to
combine them in series or in parallel. As typical applications, we
explore how capacitors are combined with op-amps to form
integrators, differentiators, and analog computers.

2
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 is typically
constructed as depicted in Fig.
3
6.1
5.2 CAPACITORS
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.
When a voltage source 𝑣 is
connected to the capacitor, as in Fig.
6.2, the source deposits a positive
charge 𝑞 on one plate and a
negative charge −𝑞 on the other.
The capacitor is said to store the
electric charge.
𝑞 = 𝐶𝑣
where C, the constant of proportionality, is known as the
4
capacitance of the capacitor.
5.2 CAPACITORS
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).
1 farad (𝐹) = 1 coulomb 𝐶 /volt 𝑉 .
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. For
example, for the parallel-plate capacitor shown in Fig. 6.1, the
capacitance is given by
𝜖𝐴
𝐶=
𝑑
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.

5
5.2 CAPACITORS
To obtain the current-voltage relationship of the capacitor, we
take the derivative of both sides. Since
𝑑𝑞
𝑖=
𝑑𝑡
So that
𝑑𝑣
𝑖=𝐶
𝑑𝑡
The voltage-current relation of the capacitor can be obtained by
integrating both sides of above Equation. We get
1 𝑡
𝑣 𝑡 = න 𝑖 𝜏 𝑑𝜏
𝐶 −∞
Or
1 𝑡
𝑣 𝑡 = න 𝑖 𝜏 𝑑𝜏 + 𝑣 𝑡0
𝐶 𝑡0

6
5.2 CAPACITORS
The instantaneous power delivered to the capacitor is
𝑑𝑣
𝑝 = 𝑣𝑖 = 𝐶𝑣
𝑑𝑡
The energy stored in the capacitor is therefore
𝑡 𝑡 𝑣 𝑡 𝑣 𝑡
𝑑𝑣 1 2
𝜔 = න 𝑝 𝜏 𝑑𝜏 = 𝐶 න 𝑣 𝑑𝜏 = 𝐶 න 𝑣 𝑑𝑣 = 𝐶𝑣 ቤ
−∞ −∞ 𝑑𝜏 𝑣 −∞ 2 𝑣 −∞
We note that 𝑣 −∞ = 0, because the capacitor was uncharged
at 𝑡 = −∞. Thus,
1 2 𝑞2
𝜔 = 𝐶𝑣 =
2 2𝐶

1. A capacitor is an open circuit to dc.

2. The voltage on a capacitor cannot change abruptly.

7
5.2 CAPACITORS
Example 6.1
(a) Calculate the charge stored on a 3𝑝𝐹 capacitor with 20𝑉
across it.
(b) Find the energy stored in the capacitor.

Solution
(a) Since 𝑞 = 𝐶𝑣
𝑞 = 3 × 10−12 × 20 = 60𝑝𝐶
(b) The energy stored is
1 2 1
𝜔 = 𝐶𝑣 = × 3 × 10−12 × 202 = 600 × 10−12 𝐽 = 600𝑝𝐽
2 2

Problem 6.1
What is the voltage across a 4.5𝜇𝐹 capacitor if the charge on
one plate is 0.12 𝑚𝐶? How much energy is stored?

8
5.2 CAPACITORS
Example 6.2
The voltage across a 5𝑝𝐹 capacitor is
𝑣 𝑡 = 10𝑐𝑜𝑠 6000𝑡 𝑉
Calculate the current through it
Solution
By definition, the current is
𝑑𝑣 −6
𝑑
𝑖 𝑡 =𝐶 = 5 × 10 10𝑐𝑜𝑠 6000𝑡
𝑑𝑡 𝑑𝑡
= −5 × 10−6 × 6000 × 10 𝑠𝑖𝑛 6000𝑡 = −0,3 𝑠𝑖𝑛 6000𝑡 𝐴

Problem 6.2
If a 10𝜇𝐹 capacitor is connected to a voltage source with
𝑣 𝑡 = 75 sin 2000𝑡 𝑉
Determine the current through the capacitor

9
5.2 CAPACITORS
Example 6.3
Determine the voltage across a 2𝜇𝐹 capacitor if the current
through it is
𝑖 𝑡 = 6𝑒 −3000𝑡 𝑚𝐴
Assume that the initial capacitor voltage is zero.
Solution
1 𝑡
Since 𝑣 = න 𝑖 𝑑𝑡 + 𝑣 0 , and 𝑣 0 = 0
𝐶 0
𝑡
1 −3000𝑡 . 10−3 . 𝑑𝑡 + 0 = 1 − 𝑒 −3000𝑡 𝑉
𝑣= න 6𝑒
2 × 10−6 0

Problem 6.3
The current through a 100𝑝𝐹 capacitor is 𝑖 𝑡 =
50 sin 120𝜋𝑡 𝑚𝐴 . Calculate the voltage across it at 𝑡 = 1𝑚𝑠 and
𝑡 = 5𝑚𝑠. Take 𝑣 0 = 0.
10
5.2 CAPACITORS
Example 6.4
Determine the current through a 200𝑝𝐹 capacitor whose voltage
is shown in below Figure.

11
5.2 CAPACITORS
Solution
The voltage waveform can be described mathematically as.
50𝑡 𝑉 ,0 < 𝑡 ≤ 1
100 − 50𝑡 𝑉 , 1 < 𝑡 ≤ 3
𝑣 𝑡 =
−200 + 50𝑡 𝑉 , 3 < 𝑡 ≤ 4
0 ,𝑡 ≥ 4
We know that:
50 , 0 < 𝑡 ≤ 1
𝑑𝑣 −6 −50 , 1 < 𝑡 ≤ 3
𝑖=𝐶 = 200 × 10 ×
𝑑𝑡 50 , 3 < 𝑡 ≤ 4
0 ,𝑡 ≥ 4
10 𝑚𝐴 , 0 < 𝑡 ≤ 1
−10𝑚𝐴 , 1 < 𝑡 ≤ 3
=
10𝑚𝐴 , 3 < 𝑡 ≤ 4
0 ,𝑡 ≥ 4 12
5.2 CAPACITORS
Solution
Thus the current waveform is as shown in below Figure

13
5.2 CAPACITORS
Problem 6.4
An initially uncharged 1-mF capacitor has the current shown in
Fig. 6.11 across it. Calculate the voltage across it at 𝑡 = 2𝑚𝑠 and
𝑡 = 5𝑚𝑠

14
5.2 CAPACITORS
Example 6.5
Obtain the energy stored in each capacitor in Fig. 6.12(a) under
dc conditions.

15
5.2 CAPACITORS
Solution
Under dc conditions, we replace each capacitor with an open
circuit, as shown in Fig. 6.12(b). The current through the series
combination of the 2𝑘Ω and 4𝑘Ω resistors is obtained by current
division as
3
𝑖= × 6 = 2(𝑚𝐴)
3+2+4
Hence, the voltages 𝑣1 and 𝑣2 across the capacitors are
𝑣1 = 2000𝑖 = 4 𝑉; 𝑣2 = 4000𝑖 = 8 𝑉
and the energies stored in them are
1 2
1
𝜔1 = 𝐶1 𝑣1 = 2 × 10−3 × 42 = 16 𝑚𝐽
2 2
1 2
1
𝜔2 = 𝐶2 𝑣2 = 4 × 10−3 × 82 = 128 𝑚𝐽
2 2

16
5.2 CAPACITORS
Problem 6.5
Under dc conditions, find the energy stored in the capacitors in
Fig. 6.13.

17
5.3 SERIES AND PARALLEL CAPACITORS

The equivalent capacitance of N parallel-connected capacitors is

the sum of the individual capacitances.

𝐶𝑒𝑞 = 𝐶1 + 𝐶2 + 𝐶3 + ⋯ + 𝐶𝑁

𝑄𝑒𝑞 = 𝑄1 + 𝑄2 + 𝑄3 + ⋯ + 𝑄𝑁

18
5.3 SERIES AND PARALLEL CAPACITORS
The equivalent capacitance of series-connected capacitors is the
reciprocal of the sum of the reciprocals of the individual
capacitances.
1 1 1
= +
𝐶𝑒𝑞 𝐶1 𝐶2

1 1 1 1 1
= + + + ⋯+
𝐶𝑒𝑞 𝐶1 𝐶2 𝐶3 𝐶𝑁
𝑄𝑒𝑞 = 𝑄1 + 𝑄2 + 𝑄3 + ⋯ + 𝑄𝑁 19
5.3 SERIES AND PARALLEL CAPACITORS
Example 6.6
Find the equivalent capacitance seen between terminals a and b
of the circuit in Fig. 6.16

20
5.3 SERIES AND PARALLEL CAPACITORS
Solution
The 20𝜇𝐹 and 5𝜇𝐹 capacitors are in series; their equivalent
capacitance is
20 × 5
= 4𝜇𝐹
20 + 5
This 4𝜇𝐹 capacitor is in parallel with the 6𝜇𝐹 and 20𝜇𝐹
capacitors; their combined capacitance is
4 + 6 + 20 = 30𝜇𝐹
This 30𝜇𝐹 capacitors is in series with the 60𝜇𝐹 capacitor. Hence,
the equivalent capacitance for the entire circuit is
30 × 60
𝐶𝑒𝑞 = = 20𝜇𝐹
30 + 60

21
5.3 SERIES AND PARALLEL CAPACITORS
Problem 6.6
Find the equivalent capacitance seen at the terminals of the
circuit in Fig. 6.17.

22
5.3 SERIES AND PARALLEL CAPACITORS
Example 6.7
For the circuit in Fig. 6.18, find the voltage across each
capacitor.

23
5.3 SERIES AND PARALLEL CAPACITORS
Solution
The two parallel capacitors in Fig. 6.18 can be combined to get
40 + 20 = 60𝑚𝐹. This 60𝑚𝐹 capacitor is in series with the 20𝑚𝐹
and 30𝑚𝐹 capacitors. Thus,
1
𝐶𝑒𝑞 = = 10𝑚𝐹
1 1 1
+ +
60 30 20
The total charge is
𝑞 = 𝐶𝑒𝑞 𝑣 = 10 × 10−3 × 30 = 0,3𝐶
This is the charge on the 20𝑚𝐹 and 30𝑚𝐹 capacitors, because
they are in series with the 30𝑉 source.
𝑞1 𝑞 0,3
𝑣1 = = = −3
= 15𝑉
𝐶1 𝐶1 20 × 10
𝑞2 𝑞 0,3
𝑣2 = = = −3
= 10𝑉
𝐶2 𝐶2 30 × 10

24
5.3 SERIES AND PARALLEL CAPACITORS
Solution
Having determined 𝑣1 and 𝑣2 , we now use KVL to determine 𝑣3
by
𝑣3 = 30 − 𝑣1 − 𝑣2 = 5𝑉
Alternatively, since the 40𝑚𝐹 and 20𝑚𝐹 capacitors are in
parallel, they have the same voltage and their combined
capacitance is This combined capacitance is in series with the
20𝑚𝐹 and 30𝑚𝐹 capacitors and consequently has the same
charge on it. Hence
𝑞34 𝑞 0,3
𝑣3 = = = −3
= 5𝑉
𝐶34 𝐶34 60 × 10

25
5.3 SERIES AND PARALLEL CAPACITORS
Problem 6.7
Find the voltage across each of the capacitors in Fig. 6.20.

26
5.4 INDUCTORS
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.
𝑑𝑖
𝑣=𝐿
𝑑𝑡
where L is the constant of proportionality called the inductance
of the inductor.
27
5.4 INDUCTORS
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 and can be found in standard electrical
engineering handbooks.
𝑁 2 𝜇𝐴
𝐿=
𝑙
where N is the number of turns, is the length, A is the cross-
sectional area, and 𝜇 is the permeability of the core.

28
5.4 INDUCTORS
The current-voltage relationship is obtained as
1
𝑑𝑖 = 𝑣 𝑑𝑡
𝐿

Integrating gives
1 𝑡 1 𝑡
𝑖 = න 𝑣 𝜏 𝑑𝜏 = න 𝑣 𝜏 𝑑𝜏 + 𝑡 𝑡0
𝐿 −∞ 𝐿 𝑡0
29
5.4 INDUCTORS
The inductor is designed to store energy in its magnetic field.
The power delivered to the inductor is
𝑑𝑖
𝑝 = 𝑣𝑖 = 𝐿 𝑖
𝑑𝑡
The energy stored is
𝑡 𝑡
𝑑𝑖
𝜔 = න 𝑝 𝜏 𝑑𝜏 = 𝐿 න 𝑖𝑑𝜏
−∞ −∞ 𝑑𝜏
𝑡
1 2 1 2
= 𝐿 න 𝑖 𝑑𝑖 = 𝐿𝑖 𝑡 − 𝐿𝑖 −∞
−∞ 2 2
Since 𝑖 −∞ = 0,
1 2
𝜔 = 𝐿𝑖
2
An inductor acts like a short circuit to dc.
The current through an inductor cannot change instantaneously.

30
5.4 INDUCTORS
Example 6.8
The current through a 0.1-H inductor is 𝑖 𝑡 = 10𝑡 𝑒 −5𝑡 𝐴 . Find
the voltage across the inductor and the energy stored in it.

Solution
Since 𝑣 = 𝐿 𝑑𝑖/𝑑𝑡 and 𝐿 = 0,1𝐻.
𝑑
𝑣 = 0,1 10𝑡 𝑒 −5𝑡 = 𝑒 −5𝑡 + 𝑡 −5 𝑒 −5𝑡 = 𝑒 −5𝑡 1 − 5𝑡 𝑉
𝑑𝑡
The energy stored is
1 2 1
𝜔 = 𝐿𝑖 = 0,1 100𝑡 2 𝑒 −10𝑡 = 5𝑡 2 𝑒 −10𝑡 𝐽
2 2

Problem 6.8
If the current through a 1𝑚𝐻 inductor is 𝑖 𝑡 = 60 cos 100𝑡 𝑚𝐴 ,
find the terminal voltage and the energy stored.

31
5.4 INDUCTORS
Example 6.9
Find the current through a 5-H inductor if the voltage across it is
30𝑡 2 , 𝑡 ≥ 0
𝑣 𝑡 =ቊ
0 ,𝑡 < 0
Also, find the energy stored at 𝑡 = 5𝑠. Assume 𝑖 𝑣 > 0
Solution
1 𝑡
Since 𝑖 = න 𝑣 𝑡 𝑑𝑡 + 𝑖 𝑡0 and 𝐿 = 5𝐻,
𝐿 𝑡0
1 𝑡
𝑖 = න 30𝑡 2 𝑑𝑡 + 0 = 2𝑡 3
5 0
The power 𝑝 = 𝑣𝑖 = 60𝑡 5 , and the energy stored is then
5
5
𝑡6
𝜔 = න 𝑝 𝑑𝑡 = න 60𝑡 5 𝑑𝑡 = 60 อ = 156,25 𝑘𝐽
0 6
0

32
5.4 INDUCTORS
Problem 6.9
The terminal voltage of a 2-H inductor is 𝑣 = 10 1 − 𝑡 𝑉 . Find
the current flowing through it at 𝑡 = 4𝑠, and the energy stored in it
at 𝑡 = 4𝑠. Assume 𝑖 0 = 2𝐴.

33
5.4 INDUCTORS
Example 6.10
Consider the circuit in below Figure. Under dc conditions, find:
𝑎 𝑖, 𝑣𝐶 and 𝑖𝐿 , (𝑏) the energy stored in the capacitor and
inductor.

34
5.4 INDUCTORS
Solution
(a) Under dc conditions, we replace the capacitor with an open
circuit and the inductor with a short circuit, as in below Figure.
12
𝑖 = 𝑖𝐿 = = 2𝐴
1+5
The voltage 𝑣𝐶 is the same
as the voltage across the 5Ω
resistor. Hence,
𝑣𝐶 = 5𝑖 = 10𝑉
(b) The energy in the
capacitor is
1 2 1
𝜔𝐶 = 𝐶𝑣𝐶 = 1 102
2 2
= 50𝐽
And that in the inductor is
1 2 1
𝜔𝐿 = 𝐿𝑖𝐿 = 2 22 = 4 𝐽 35
2 2
5.4 INDUCTORS
Problem 6.10
Determine 𝑣𝐶 , 𝑖𝐿 and the energy stored in the capacitor and
inductor in the circuit of below Figure under dc conditions.

36
5.5 SERIES AND PARALLEL INDUCTORS
The equivalent inductance of series-connected inductors is the
sum of the individual inductances.

𝐿𝑒𝑞 = 𝐿1 + 𝐿2 + 𝐿3 + ⋯ + 𝐿𝑁
𝑣 = 𝑣1 + 𝑣2 + 𝑣3 + ⋯ + 𝑣𝑁

37
5.5 SERIES AND PARALLEL INDUCTORS
The equivalent inductance of parallel inductors is the reciprocal
of the sum of the reciprocals of the individual inductances

1 1 1 1 1
= + + + ⋯+
𝐿𝑒𝑞 𝐿1 𝐿2 𝐿3 𝐿𝑁
𝑖 = 𝑖1 + 𝑖2 + 𝑖3 + ⋯ + 𝑖𝑁

38
5.5 SERIES AND PARALLEL INDUCTORS
Example 6.11
Find the equivalent inductance of the circuit shown in Fig. 6.31.

Solution
The 10𝐻, 12𝐻 and 20𝐻 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
= 6𝐻
7 + 42
This 6-H inductor is in series with the 4-H and 8-H inductors.
Hence,
𝐿𝑒𝑞 = 4 + 6 + 8 = 18𝐻 39
5.5 SERIES AND PARALLEL INDUCTORS
Problem 6.11
Calculate the equivalent inductance for the inductive ladder
network in Fig. 6.32.

40
5.5 SERIES AND PARALLEL INDUCTORS
Example 6.12
For the circuit in Fig. 6.33, 𝑖 𝑡 = 4 2 − 𝑒 −10𝑡 𝑚𝐴. If 𝑖2 0 =
− 1𝑚𝐴, find 𝑎 𝑖1 0 ; 𝑏 𝑣 𝑡 , 𝑣1 𝑡 and 𝑣2 𝑡 ; 𝑐 𝑖1 𝑡 and 𝑖2 𝑡

41
5.5 SERIES AND PARALLEL INDUCTORS
Solution
(a) From 𝑖 𝑡 = 4 2 − 𝑒 −10𝑡 𝑚𝐴, 𝑖 0 = 4 2 − 1 = 4𝑚𝐴. Since
𝑖 = 𝑖1 + 𝑖2
𝑖1 0 − 𝑖 0 − 𝑖2 0 = 4 − −1 = 5𝑚𝐴
(b) The equivalent inductance is
𝐿𝑒𝑞 = 2 + 4||12 = 2 + 3 = 5𝐻
Thus,
𝑑𝑖
𝑣 𝑡 = 𝐿𝑒𝑞 = 5 −4 −10 𝑒 −10𝑡 𝑚𝑉 = 200𝑒 −10𝑡 𝑚𝑉
𝑑𝑡
And
𝑑𝑖
𝑣1 𝑡 = 2 = 2 −4 −10 𝑒 −10𝑡 = 80𝑒 −10𝑡 𝑚𝑉
𝑑𝑡
Since 𝑣 = 𝑣1 + 𝑣2
𝑣2 𝑡 = 𝑣 𝑡 − 𝑣1 𝑡 = 120𝑒 −10𝑡 𝑚𝑉

42
5.5 SERIES AND PARALLEL INDUCTORS
Solution
(c) The current 𝑖1 is obtained as
1 1 120 𝑡 −10𝑡
𝑖1 𝑡 = න 𝑣2 𝑑𝑡 + 𝑖1 0 = න𝑒 𝑑𝑡 + 5 𝑚𝐴
4 0 4 0
𝑡
= −3𝑒 −10𝑡 ቚ + 5𝑚𝐴 = 8 − 3𝑒 −10𝑡 𝑚𝐴
0
Similarly,
1 𝑡 120 𝑡 −10𝑡
𝑖2 𝑡 = න 𝑣2 𝑑𝑡 + 𝑖2 0 = න𝑒 𝑑𝑡 − 1 𝑚𝐴
12 0 12 0
𝑡
= −𝑒 −10𝑡 ቚ − 1𝑚𝐴 = −𝑒 −10𝑡 𝑚𝐴
0

43
5.5 SERIES AND PARALLEL INDUCTORS
Problem 6.12
In the circuit of Fig. 6.34, 𝑖1 𝑡 = 0,6𝑒 −2𝑡 𝐴. If 𝑖 0 = 1,4 𝐴, find:
(a) 𝑖2 0
(b) 𝑖2 𝑡 and 𝑖 𝑡
(c) 𝑣1 𝑡 , 𝑣2 𝑡 and 𝑣 𝑡

44