Course Outline
EE 2093 – Theory of Electricity
B.Sc. Engineering Degree
Semester 3 – 2021
Credits:
Lecturer :
Lectures:
Duration:
2
Dr. V. Logeeshan
2 hours per week
14 weeks
Module Objectives
▪ To develop analysis tools in electrical engineering and to
analyze electrical circuits and waveforms using the tools.
• To Introduce the fundamental concepts and to develop
analytical skills for the understanding and application of
basic electrical principles.
• To apply DC and AC electrical principles to electrical
circuit networks.
• Basic network theorems and methods of analysis are
combined with complementary laboratory exercises to
provide a solid working foundation in electrical
fundamentals.
Outline Syllabus
1. Review of fundamentals (6hrs)
• Fundamentals of electric circuits
• Transient solutions of RLC circuits using differential
equations
• AC theory
Outline Syllabus
2. Coupled circuits and Dependent
sources (4hrs)
• Series and parallel resonance
• Mutual inductance
• Electromagnetic coupling in circuits
• Dependent sources
Outline Syllabus
3. Network theorems (8hrs)
•
•
•
•
•
•
•
Superposition theorem
Thevenin’s theorem
Norton’s theorem
Milliman’s theorem
Reciprocity theorem
Maximum power transfer theorem
Nodal-mesh transformation and compensation
theorems.
• Network topology
• Nodal and mesh analysis.
• Two-port theory: Impedance, admittance, hybrid and
ABCD parameters.
Outline Syllabus
4. Three-phase Analysis (6hrs)
• Analysis of three phase balanced circuits
• Analysis of three phase unbalanced circuits
• Symmetrical components Analysis
• Single line equivalent circuits.
Outline Syllabus
5. Non-sinusoidal waveforms (4hrs)
• Waveform parameters:
• Mean, rms, peak, rectified average etc.
• Power, power factor
• Harmonics
• Fourier analysis
• Laplace transform
• Transient analysis using the Laplace transform.
Learning Outcomes
After completing the module the student should be able to
1. solve coupled circuits involving mutual impedance and/or
resonance phenomena.
2. apply network theorems in solving circuits.
3. solve circuits containing three phase generators and loads.
4. analyze circuits with non-sinusoidal voltage and current
waveforms.
Continuous Assessment
40% of Overall mark for module
Components of Continuous Assessment
• unannounced in-class tests (around 5)
• attendance at lectures
• tutorials/assignments may be given
End of Semester Assessment
60% of Overall mark for module
Components of End of Semester Assessment
• closed book examination of duration 2 hrs
• will usually consist of 5 questions
• all questions must be answered
• questions not necessarily of equal
weightage
Recommended Texts
• Electric Circuits, E.A.Edminster, Schaum Outline
Series, McGraw Hill
• Theory and Problems of Basic Electrical Engineering, D
P Kothari, I J Kothari, Prentice Hall of India, New Delhi
• Electrical Engineering Fundamentals, Vincent Del Toro,
Prentice Hall of India, New Delhi
1. Review of Fundamentals
1.1 Fundamentals of Electric circuits
Charge (unit: coulomb, C;
letter symbol: q or Q)
• The electric charge is the most basic quantity in electrical
engineering and arises from the atomic particles of which
matter is made.
Positive Charge: Protons
Negative Charge: Electrons
Potential Difference (unit: volt, V;
letter symbol: v or V)
• The potential difference, also known as voltage, is the work
done (or energy required) to move a unit positive charge from
one point to another (across a circuit element). Thus the
change in work done 𝒅𝒘 when a charge 𝒅𝒒 moves through a
potential difference of v.
𝒅𝒘 = 𝒗. 𝒅𝒒
Current
(unit: ampere, A;
letter symbol: i or I)
• The electric current is the rate of charge flow in a circuit.
𝑖=
𝑑𝑞
𝑑𝑡
Energy
𝑞 = ‫𝑖 ׬‬. 𝑑𝑡
(unit: joule, J;
letter symbol: w or W)
• The Energy is the capacity to do work. Thus, in electrical
quantities this may be expressed as
න 𝑑𝑤 = න 𝑣. 𝑑𝑞 = න 𝑣. 𝑖. 𝑑𝑡
Power
(unit: watt, W; letter symbol: p or P)
• The electric power is the rate of change of energy.
𝑑𝑤
𝑝=
𝑑𝑡
Example:
Suppose that charge versus time for a given circuit element is given by
q(t) = 0 for t < 0
and
q(t) = 2 - 2e -100t C for t > 0
Sketch q(t) and i(t) to scale versus time.
𝑖(𝑡) =
𝑑𝑞(𝑡)
𝑑𝑡
= 0 for t < 0
= 200e -100tA for t > 0
Ohm’s Law
Current through a conductor between two
points is directly proportional to
the voltage across the two points.
𝑉∝𝐼
𝑉 = 𝐼𝑅
Kirchhoff’s Current Law
The net current entering a node is
zero.
Kirchhoff’s Voltage Law
The algebraic sum of the voltages
equals zero for any closed path
(loop) in an electrical circuit
Loop 1:
-va + vb + vc = 0
Node a: -i1 - i2 + i3 = 0
Loop 3:
va - vb + vd - ve= 0
Basic Elements
Active Elements
An active element is an electronic
component which supplies energy to a
circuit. Active elements have the
ability to electrically control electron
flow (i.e. the flow of charge). All
electronic circuits must contain at least
one active component.
• Voltage sources, Current sources
• Generators (such as alternators and
DC generators)
• All different types of transistors
Passive Elements
A passive element is an
electronic component which
can only receive energy,
which it can either dissipate,
absorb or store it in an
electric field or a magnetic
field.
• Resistors
• Capacitors
• Inductors
Passive Elements
Resistance (unit: ohm, Ω;
letter symbol: R , r )
𝑣(𝑡) = 𝑅 . 𝑖(𝑡)
𝑖(𝑡) = 𝐺 . 𝑣(𝑡), 𝐺 = 1/𝑅
𝑝(𝑡) = 𝑣(𝑡) . 𝑖(𝑡) = 𝑅 . 𝑖2(𝑡) = 𝐺 . 𝑣 2(𝑡)
𝑤(𝑡) = ‫ )𝑡(𝑣 ׬‬. 𝑖(𝑡) . 𝑑𝑡 = ‫ 𝑅 ׬‬. 𝑖2(𝑡) . 𝑑𝑡 = ‫ 𝐺 ׬‬. 𝑣 2(𝑡) . 𝑑𝑡
Series Resistance
Parallel Resistance
Capacitance (unit: farad, F;
letter symbol: C, c )
q=C.V
1
𝑣 = ‫𝑖 ׬‬. 𝑑𝑡
𝐶
𝑝 𝑡 = 𝑣 𝑡 .𝑖 𝑡
or
𝑖=𝐶
𝑑𝑣
𝑑𝑡
𝑑𝑣
𝑤 𝑡 = න 𝑝 𝑡 . 𝑑𝑡 = න 𝑣 𝑡 . 𝑖 𝑡 . 𝑑𝑡 = න 𝑣. 𝐶
. 𝑑𝑡
𝑑𝑡
1
= න 𝐶. 𝑣. 𝑑𝑣 = 𝐶. 𝑣 2
2
1
Stored energy w(t) = 𝐶. 𝑣 2
2
Voltage across capacitor cannot change instantaneously
𝑖 = 𝑖1 + 𝑖2 + 𝑖3
𝑑𝑣
𝑑𝑣
𝑑𝑣
𝑖 = 𝐶1
+ 𝐶2
+ 𝐶3
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑣
𝑖 = (𝐶1 +𝐶2 + 𝐶3 )
Capacitances in Parallel
𝑑𝑡
𝑑𝑣
𝑖 = (𝐶𝑒𝑞 )
𝑑𝑡
Capacitances in Series
𝑣 = 𝑣1 + 𝑣2 + 𝑣3
𝑣=
1
1
1
න 𝑖. 𝑑𝑡 + න 𝑖. 𝑑𝑡 + න 𝑖. 𝑑𝑡
𝐶1
𝐶2
𝐶3
1
1
1
𝑣 = ( + + ) න 𝑖. 𝑑𝑡
𝐶1 𝐶2 𝐶3
1
𝑑𝑣
𝑑𝑣
𝑖=
= (𝐶𝑒𝑞 )
1
1
1 𝑑𝑡
𝑑𝑡
( + + )
𝐶1 𝐶2 𝐶3
Example:
Suppose that the voltage v(t) shown in the figure below is applied
to a 1-µF capacitance. Plot the stored charge and the current
through the capacitance versus time.
q(t) = Cv(t) = 10-6v(t)
𝑑𝑣
10𝑉
−6
𝑡 = 0 𝑡𝑜 2 𝜇𝑠; 𝑖 𝑡 = 𝐶
= 10
= 5𝐴
𝑑𝑡
2 × 10−6
𝑑𝑣
0𝑉
−6
𝑡 = 2 𝑡𝑜 4 𝜇𝑠; 𝑖 𝑡 = 𝐶
= 10
= 0𝐴
𝑑𝑡
2 × 10−6
𝑑𝑣
−10𝑉
−6
𝑡 = 4 𝑡𝑜 5 𝜇𝑠; 𝑖 𝑡 = 𝐶
= 10
= −10𝐴
𝑑𝑡
1 × 10−6
Example:
Consider the situation shown in Figure below. Prior to t = 0, the
capacitor C1 is charged to a voltage of v1 = 100 V and the other
capacitor has no charge (i.e., v2 = 0). At t = 0, the switch closes.
Compute the total energy stored by both capacitors before and
after the switch closes.
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑆𝑡𝑜𝑟𝑒𝑑 𝐸𝑛𝑒𝑟𝑔𝑦:
1 2 10−6 × 1002
𝑤1 = 𝐶𝑣1 =
= 5 𝑚𝐽
2
2
𝑤2 = 0 𝑚𝐽
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐶ℎ𝑎𝑟𝑔𝑒:
𝑞1 = 𝐶𝑉 = 10−6 × 100 = 100 𝜇𝐶
𝑞2 = 0
𝐴𝑓𝑡𝑒𝑟 𝑡ℎ𝑒 𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑠, 𝑡ℎ𝑒 𝑐ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡
𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒:
𝑞𝑒𝑞 = 𝑞1 + 𝑞2 = 100 𝜇𝐶
𝐴𝑓𝑡𝑒𝑟 𝑡ℎ𝑒 𝑠𝑤𝑖𝑡𝑐ℎ 𝑐𝑙𝑜𝑠𝑒𝑠, 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒:
𝐶𝑒𝑞 = 𝐶1 + 𝐶2 = 2 𝜇𝐹
𝑣𝑒𝑞
𝑞𝑒𝑞 100
=
=
= 50 𝑉 = 𝑣1 = 𝑣2
𝐶𝑒𝑞
2
1 2 10−6 × 502
𝑤1 = 𝐶𝑣1 =
= 1.25 𝑚𝐽
2
2
1 2 10−6 × 502
𝑤2 = 𝐶𝑣2 =
= 1.25 𝑚𝐽
2
2
𝑇ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑠𝑡𝑜𝑟𝑒𝑑 𝑒𝑛𝑒𝑟𝑔𝑦: 2.5 𝑚𝐽
Inductance (unit: henry, H;
letter symbol: L , l )
The basic equation governing the behavior of an
inductor is Faraday’s law of electromagnetism.
𝑑∅
𝑒=−
𝑑𝑡
𝑑∅
When there are N turns in a coil 𝐸 = 𝑁
𝑑𝑡
∅ ∝ 𝑖,
𝑑𝑖
𝑣∝
𝑑𝑡
𝑑𝑖
𝑣=𝐿
𝑑𝑡
1
𝑖 = න 𝑣. 𝑑𝑡
𝐿
𝑝 𝑡 = 𝑣 𝑡 . 𝑖(𝑡)
𝑑𝑖
1 2
𝑤 𝑡 = න 𝑣 𝑡 . 𝑖 𝑡 . 𝑑𝑡 = න 𝐿. . 𝑖. 𝑑𝑡 = න 𝐿. 𝑖. 𝑑𝑖 = 𝐿. 𝑖
𝑑𝑡
2
Stored energy w(t)
1
= 𝐿. 𝑖 2
2
Current through inductor cannot
change instantaneously
𝑣 = 𝑣1 + 𝑣2 + 𝑣3
𝑑𝑖
𝑑𝑖
𝑑𝑖
𝑣 = 𝐿1 + 𝐿2 + 𝐿3
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑖
𝑣 = (𝐿1 +𝐿2 + 𝐿3 )
Inductances in Series
𝑑𝑡
𝑑𝑖
𝑣 = (𝐿𝑒𝑞 )
𝑑𝑡
Inductances in Parallel
𝑖 = 𝑖1 + 𝑖2 + 𝑖3
𝑖=
1
1
1
න 𝑣. 𝑑𝑡 + න 𝑣. 𝑑𝑡 + න 𝑣. 𝑑𝑡
𝐿1
𝐿2
𝐿3
1
1
1
𝑖 = ( + + ) න 𝑣. 𝑑𝑡
𝐿1 𝐿2 𝐿3
1
𝑑𝑖
𝑑𝑖
𝑣=
= (𝐿𝑒𝑞 )
1
1
1 𝑑𝑡
𝑑𝑡
( + + )
𝐿1 𝐿2 𝐿3
Example:
The voltage across a 150- µH inductance is shown in Figure below.
The initial current is i(0) = 0. Find and plot the current i(t) to scale
versus time.
1
1
15𝑡
1
15𝑡 2
𝑡 = 0 𝑡𝑜 2 𝜇𝑠; 𝑖 𝑡 = න 𝑣 𝑡 . 𝑑(𝑡) =
∗න
. 𝑑𝑡 =
×
𝐿
150 × 10−6
2 × 10−6
150 × 10−6 4 × 10−6
1
1
𝑡 = 2 𝑡𝑜 4 𝜇𝑠; 𝑖 𝑡 = න 𝑣 𝑡 . 𝑑(𝑡) =
∗ න 0. 𝑑𝑡 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝐿
150 × 10−6
1
1
1
𝑡 = 4 𝑡𝑜 5 𝜇𝑠; 𝑖 𝑡 = 𝐿 ‫ 𝑡 𝑣 ׬‬. 𝑑(𝑡) = 150×10−6 ∗ ‫ ׬‬−15. 𝑑𝑡 = 150×10−6 × −15𝑡
Summary
For a resistor,
𝑣 = 𝑅𝑖,
1
‫𝑖 ׬‬. 𝑑𝑡 ,
𝐶
𝑖 = 𝐺𝑣
𝑑𝑣
𝐶
𝑑𝑡
For a capacitor,
𝑣=
𝑖=
Voltage across a capacitor will never change suddenly.
𝑑𝑖
𝐿 ,
𝑑𝑡
1
‫𝑣 ׬‬. 𝑑𝑡
𝐿
For an inductor,
𝑣=
𝑖=
Current through an inductor will never change suddenly.
Impedance and Admittance
𝑣 = 𝑍 𝑝 .𝑖
𝑖 = 𝑌 𝑝 .𝑣
1
𝑍 𝑝 =
𝑌 𝑝
where Z(p) – impedance operator,
Y(p) – admittance operator.
Impedances and Admittances
• either linear or non-linear
Linear Systems
Non-Linear Systems
Active Elements
(Component capable of producing energy)
Sources of energy (or sources)
Independent source
Dependent source
For an independent voltage source (or
current source), the terminal voltage (or
current) would depend only on the loading
and the internal source quantity, but not on
any other circuit variable.
A dependent voltage source (or current
source) would have its terminal voltage (or
current) depend on another circuit quantity
such as a voltage or current. Thus four
possibilities exist.
o Voltage Source
o
o
o
o
- +
o Current Source
Voltage controlled voltage source
Current controlled voltage source
Voltage controlled current source
Current controlled current source
Example:
For the circuit shown in figure below, determine the current I and Vo.
Applying Kirchhoff's voltage law, gives
7.5 = 4 I + 4 Vo + 5 – Vo
from Ohm’s law
1I = – Vo
Thus by substitution, we have
I = 2.5 A and Vo = – 2.5 V
Example:
For the circuit shown in figure below, determine the current I
E = R1 I + R2 (1+α) I
Ro αI = – Vo