Chapter 2

Introduction to Circuit Element

An element is the basic building block of a circuit. An electric circuit is simply an interconnection
of the elements. Circuit analysis is the process of determining voltages across (or the currents
through) the elements of the circuit. There are two types of elements found in electric circuits:
passive elements and active elements. An active element is capable of generating energy while a
passive element is not. Examples of passive elements are resistors, capacitors, and inductors.
Typical active elements include generators, batteries, and operational amplifiers.
2.1. Resistance and Resistors
Resistors are materials which in general have a characteristic behavior of resisting the flow of
electric charge. This physical property, or ability to resist current, is known as resistance and is
represented by the symbol R. The unit of measurement of resistance is the ohm, for which the
symbol is Ω (omega).
The resistance of an electrical conductor depends on four factors, these being: (a) the length of the
conductor, (b) the cross-sectional area of the conductor, (c) the type of material and (d) the
temperature of the material.
At a fixed temperature of 20°C (room temperature), the resistance is related to three factors by

Where ρ (Greek letter rho) is known as the resistivity of the material in ohm-meters, l is the length of
the sample in meters, A is the cross-sectional area of the sample in square of meters, and R is resistance in
ohm. The circuit symbol of resistance is

Example: a copper has a length of 100m and cross sectional area of 2.5mm2 how much will be the
resistance value? Resistivity of Cu=1.72 x10-8 Ω.m

Solution

1
 100𝑚
R = 1.72*10-8Ω.m = 0.688Ω
2.5∗10−6 𝑚.𝑚

2.2. Capacitance and Capacitor

A capacitor is a passive element designed to store energy in its electric field. A circuit element that
is composed of two conducting plates or surfaces separated by a dielectric (non-conducting) materials. If a
voltage source (v) is connected to the capacitor, positive charge will be transferred to one plate while
negative charge will be transferred to the other plate.

Figure 2.1 capacitor with applied voltage v.

The capacitor is said to store the electric charge. The amount of charge stored, represented by q,
is directly proportional to the applied voltage v so that
𝑞 = 𝐶𝑣
Where C, the constant of proportionality, is known as the capacitance of the capacitor. 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
𝐶=
𝑣
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 It depends on the physical dimensions of the capacitor. the
capacitance is given by

Where 𝜀0 = 8.85 X10-12F/m, 𝜀𝑟 = relative permittivity, A is the surface area of each plate, d is the
distance between the plates.
In general, three factors determine the value of the capacitance:

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1. The surface area of the plates—the larger the area, the greater the capacitance.
2. The spacing between the plates—the smaller the spacing, the greater the capacitance.
3. The permittivity of the material—the higher the permittivity, the greater the capacitance.
The circuit symbol for capacitor is

To obtain the current-voltage relationship of the capacitor,

𝑑𝑞(𝑡)
𝑖(𝑡) =
𝑑𝑡
Then

The voltage-current relation of the capacitor

dvc (t)
𝑖𝑐 (t) = C
dt
𝑖𝑐 (t)dt = Cdvc (t)
1
dvc (t) = i (t)dt
c c
τ=t
1
vc (t) = vc (t o ) + ∫ ic (τ)dτ
c
τ=to

Where to is the initial time

The capacitor is a passive element and follows the passive sign convention

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Examples

2. If the charge accumulated on two parallel conductors charged to 12 V is 600 pC, what is the
capacitance of the parallel conductors?

2.3 Inductance and Inductors

An inductor is a passive electrical component that can stores magnetic energy. It is measured in
henries. An inductor is usually constructed as a coil of conducting material.
If a current is flowing in the inductor, it produce a magnetic field, Φ.
Φc (t) = Li(t)
As the current increases or decreases, the magnetic field spreads or collapse. The change in
magnetic field induces a voltage across the inductor.
dΦ(t)
vL (t) =
dt
diL (t)
vL (t) = L
dt
Current in inductors
diL (t)
vL (t) = L
dt
1
diL (t) = v (t)dt
𝐿 L
Integrate both sides
t
1
iL (t) = iL (t o ) + ∫ vL (t) dt
𝐿
to

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There are four basic factors of inductor construction determining the amount of inductance.
1. Number of turns of coil: the greater number of turns results greater inductance; the fewer
turns of wire results less inductance.
2. Coil area: The greater coil area results greater inductance; the less coil area results less
inductance.
3. Coil length: the longer the coil's length, the less inductance; the shorter the coil's length,
the greater the inductance.
4. Core material: the greater the inductance; the less the permeability of the core, the less
the inductance.

Example

2.4 Electric sources

1. Independent voltage source:
It is a 2-terminal sources that maintains a specific voltage across its terminals regardless of the
current through it. The circuit symbol of independent voltage sources is given below,

2. Dependent voltage source:

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It is a 2-terminal sources that generates a voltage that is determined by a voltage or current at a
specified location in the circuit. The circuit symbol of dependent voltage sources is,

3. Independent current source:

It is a 2-terminal sources that maintains a specific current through it regardless of the voltage across
it terminals. The circuit symbol of independent current sources is given below,

4. Dependent current source:

It is a 2-terminal sources that generates a current that is determined by voltage or current at a
specified location in the circuit. The circuit symbol of dependent current sources is given below,

Example - Compute the power that is absorbed or supplied by each of the elements in the
following circuit

1IX
IX=4A R1
+
_

+ 12 V -
IR2 IR3=
+ 2A +
Vs=36 V 24 V 28 V
R2 R3
- -

Solution
𝑃𝑉𝑠 = 𝑉𝑠 𝐼𝑋 = (36)(−4) = −144 𝑊 (𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑠)
𝑃𝑅1 = 𝑉𝑅1 𝐼𝑋 = (12)(4) = 48 𝑊 (𝑎𝑏𝑠𝑜𝑟𝑏𝑠)
𝑃𝑅2 = 𝑉𝑅2 𝐼𝑅2 = 𝑉𝑅2 (𝐼𝑋 − 𝐼𝑅3 ) = (24)(4 − 2) = 48 𝑊 (𝑎𝑏𝑠𝑜𝑟𝑏𝑠)
𝑃𝐷𝑠 = 𝑉𝐷𝑠 𝐼𝑅3 = (1𝐼𝑋 )(𝐼𝑅3 ) = (4)(−2) = −8 𝑊 (𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑠)
𝑃𝑅3 = 𝑉𝑅3 𝐼𝑅3 = (28)(2) = 56 𝑊 (𝑎𝑏𝑠𝑜𝑟𝑏𝑠)