Basic Electrical Engineering

1. A short course on

2. DC Network Theorem: Definition of electric circuit, network, linear circuit, non-
linear circuit, bilateral circuit, unilateral circuit, Dependent source, Kirchhoff’s law,
Principle of superposition. Source equivalence and conversion, Thevenin’s theorem,
Norton Theorem, nodal analysis, mesh analysis, stardelta conversion. Maximum
power transfer theorem with proof. 7L
Electromagnetism: Biot-savart law, Ampere’s circuital law, field calculation using
Biot-savart & ampere’s circuital law. Magnetic circuits, Analogous quantities in
magnetic and electric circuits, Faraday’s law, Self and mutual inductance. Energy
stored in a magnetic field, B-H curve, Hysteretic and Eddy current losses, Lifting
power of Electromagnet. 5L
AC fundamental: Production of alternating voltage, waveforms, average and RMS
values, peak factor, form factor, phase and phase difference, phasor representation of
alternating quantities, phasor diagram, behavior of AC series , parallel and series
parallel circuits, Power factor, Power in AC circuit, Effect of frequency variation in
RLC series and parallel circuits, Resonance in RLC series and parallel circuit, Q
factor, band width of resonant circuit. 9L

3. Text books:
• Basic Electrical engineering, D.P Kothari & I.J Nagrath, TMH, Second Edition
• Fundamental of electrical Engineering, Rajendra Prasad, PHI, Edition 2005.
• Basic Electrical Engineering, V.N Mittle & Arvind Mittal, TMH, Second Edition
• Basic Electrical Engineering, J.P. Tewari, New age international publication
Reference books:
• Basic Electrical Engineering(TMH WBUT Series), Abhijit Chakrabarti & Sudipta
Nath, TMH
• Electrical Engineering Fundamental, Vincent.D.Toro, Pearson Education,
• Second Edition.
• Hughes Electrical & Electronics Technology, 8/e, Hughes, Pearson Education.
• Basic Electrical Engineering, T.K. Nagsarkar & M.S. Sukhija, Oxford
• Introduction to Electrical Engineering, M.S. Naidu & S, Kamakshaiah, TMH
• Basic Electrical Engineering, J.J. Cathey & S.A Nasar, TMH, Second Edition.

4. Ohm’s Law
The ratio of potential difference (V) between any two points on a conductor to
the current (I) flowing between them is constant, provided that the temperature of
the conductor does not change.
In other words, V/I =constant
R
I
V

where R is the resistant of the conductor between the two points considered.
V
I R

5. Some important definitions :
1. Active Elements  Active Elements is one which supplies energy to the circuit.
2. Passive Elements  Passive Elements is one which receives energy from the
circuit and them either converts it to heat or stores it in an electric or magnetic
field. The example of passive elements are resistance, inductance, capacitance.
3. Network  It refers to any arrangements of
passive or active circuit elements which
forms a closed path.
4. Node  A Node of a network is an
equipotential surface at which two or more
circuit elements are joined. In the above
figure points a, b, c, d are the Nodes.
5. Junction  A Junction is that point in a network, where three or more circuit
elements are joined. In the above figure points b and d are two Junctions.
So all the Junctions are Node, but all the Nodes are not Junctions.

6. 6. Branch  A Branch is that part of a network which lies between junction points.
According to the figure the number of branches are three.
7. Loop  A Loop is any closed path of the network. That is abda, bcdb, and abcda
are the three loops in the network.
8. Mesh  A Mesh is the most elementary part of a loop. It cannot be further divided
into other loops. abda and bcdb are the two meshes, but abcda is not a mesh.
So all the meshes are loop, but all the loops are not mesh.

7. Kirchhoff’s Law
a) Kirchhoff’s Current Law (KCL)  This law is applicable to a node of the network
which is a junction of two or more branches of that network. Kirchhoff’s current
law states that the sum of the current flowing towards the node is equal to the
sum of the current flowing away from the node. That is, in any network the
current in all the branches meeting at a node is zero.
i.e. ∑I =0
Sign Convention :-
Coming towards the node  Positive (+)
Going away from the node  Negative (-)
So at junction P
I1+I2-I3+I4-I5=0
 I1+I2+I4= I3+ I5
 Incoming current = Outgoing current

8. Kirchhoff’s Voltage Law (KVL)  This law is applicable to a close mesh of the
circuit, which may consist of a number of branches, having resistances only or a
branch in addition having a source of emf. KVL states that the algebraic sum of the
product of current and resistance of various branches of closed mesh of a circuit
plus the algebraic sum of the emf in that closed mesh is equal to zero.
i.e. ∑IR + ∑E=0
In the figure, the application of Kirchhoff’s voltage law results V - IR1 - IR2 = 0

9. in this figure,
+v1-R1i-R2i+v2-R3i=0
Application of KVL

10. Maxwell’s Mesh Current’s Equation
Mesh 1
Mesh current I1
Branch current I1 of R1, (I1+I2) of R2
Mesh 2
Mesh current I2
Branch current I2 of R3, (I1+I2) of R2
The number of independent mesh equation to the number of branches and the
junction point of the network, it relates as follows,
m=b-(j-1)
=b-j+1
Where m = number of independent mesh equation
b = number of branches
j = number of junction point
in the above circuit, number of mesh equation (m)
= b-j+1 =3-2+1 =2

11. Application of Maxwell’s Mesh Current’s Equation
Mesh 1
E1 – R1I1 – R2 (I1 + I2) =0
Mesh 2
E2 – R3I2 – R2 (I2 + I1) =0

12. Nodal Analysis with voltage source
For node 1, the following current equation can be written with the help of KCL
I1=I4+I2--- --- --- (i)
Now, I1R1=E1-VA

1
A
1
1
R
V
E
I


Obviously I4=VA/R4 and I2R2=VA-VB (since VA >VB)
A B
2
2
V V
I
R

 

13. Substitute the value of I1, I2 and I4 in equation (i) and simplifying the equation we
get
0
)
1
1
1
(
1
1
2
4
2
1





R
E
R
V
R
R
R
V B
A
Similarly for node 2,
0
)
1
1
1
(
3
2
2
5
3
2





R
E
R
V
R
R
R
V A
B

14. Star / Delta Transformation
In solving networks (having
considerable number of branches) by
the application of Kirchhoff’s Laws,
one sometime experience great
difficulty due to a large number of
simultaneous equations that have to be
solved. However such complicated
networks can be simplified by
successively replacing delta meshes by
equivalent star system and vice-versa.
Suppose we are giving three resistances R12, R23 and R31 connected in delta fashion
between terminals 1, 2 and 3 as shown in the figure (a). So far as the respective
terminals are concerned, these three giving resistances can be replaced by the three
resistances R1, R2 and R3 connected in star as shown in figure (b).

15. These two arrangements will be electrically equivalent if the resistances as
measured between any pair of terminals are the same in both the arrangements.
Let us find the condition.
First take Delta connection :
Resistance between terminals 1 and 2 =  
23
31
12
23
31
12 )
(
R
R
R
R
R
R



Now take star connection :
Resistance between terminals 1 and 2 = R1+R2
As terminals resistance have to be same
R1+R2 =
 
23
31
12
23
31
12 )
(
R
R
R
R
R
R



--- --- --- (i)
Similarly for terminals 2 and 3 and terminals 3 and 1 we get,
R2+R3 =
 
23
31
12
12
31
23 )
(
R
R
R
R
R
R



--- --- --- (ii)

16. And R3+R1 =
 
23
31
12
23
12
31 )
(
R
R
R
R
R
R



--- --- --- (iii)
Now subtracting (ii) from (i) and adding the result to (iii) we get
31
23
12
31
12
1
R
R
R
R
R
R



31
23
12
12
23
2
R
R
R
R
R
R



31
23
12
23
31
3
R
R
R
R
R
R



Star/Delta Transformation 
This transformation can be easily done by using equation (i), (ii) and (iii)
giving above. Multiplying (i) and (ii), (ii) and (iii), (iii) and (i) and adding them
together and then simplify them, we get
2
3
1
3
1
2
1
3
3
2
2
1
12
1
3
2
3
2
1
1
3
3
2
2
1
23
3
2
1
2
1
3
1
3
3
2
2
1
12
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R
R



















17. Example:
Find the resistance between the terminals a-b of the bridge circuit using Star-Delta
Transformation.

18. Source Transformation
Open circuit voltage = V
Short circuit current = V/R
Here also,
Open circuit voltage = V
Short circuit current = V/R

19. So a voltage source with series resistance can be converted to its equivalent
current source with same parallel resistance

20. Superposition Theorem In a linear bilateral network, containing two or more
voltage sources, the current through any element (resistor or conductor) may be
determine by adding together algebraically the current produced by each source
acting alone when all other voltage sources are replaced by their internal
resistances. If a voltage source has no internal resistance, the terminals to which it
was connected are joined together. If there are current sources present, they are
removed and the terminals where they ware connected are left open.
+
V2
R2
+
V1
R5
R4
R3
R1
Consider the following circuit:
The circuit contains two voltage sources
(V1 and V2) and five resistors of
resistances R1, R2, R3, R4 and R5. Using
superposition theorem you have to find
the current flowing through the resistor
R3.

21. 1. Replacing all other independent voltage sources with a short circuit (thereby
eliminating difference of potential. i.e. V=0, internal impedance of ideal voltage
source is ZERO (short circuit)).
2. Replacing all other independent current sources with an open circuit (thereby
eliminating current. i.e. I=0, internal impedance of ideal current source is infinite
(open circuit).
R1
R2
R3
R4
+
V1
R5
 
4 5
1
1
4 5 3 3 4 5
1 2
3 4 5
( )
R R
V
I amp
R R R R R R
R R
R R R

 
  
 
 

22. +
V2
R1
R2
R3
R4
R5
 
2 1 2
2
1 2 3 1 2 3
4 5
1 2 3
( )
V R R
I amp
R R R R R R
R R
R R R

 
  
 
 
Total current IT =I1 + I2
   
4 5
1 2 1 2
4 5 3 1 2 3
3 4 5 1 2 3
1 2 4 5
3 4 5 1 2 3
( )
R R
V V R R
amp
R R R R R R
R R R R R R
R R R R
R R R R R R
   
   
 
   
  
   
 
   
   
   
   
   

23. Example: Determine the current I in the network shown in the figure by the principle of
superposition.
1. Only 24 V voltage source
acting alone in the circuit,
Fig(a).
1
24
2
6 6
I A
 


24. 2. Only 5 A current source acting alone in the circuit, Fig (b).
2
6 5
5 2.5
6 6 2
I A
  

By superposition theorem, the total current through Rab is
1 2 2 2.5 4.5
I I I A
    

25. Thevenin’s Theorem This theorem states that in any bilateral network the current
through any resistance connecting across any two points of an active network, can
be obtained by dividing the potential difference between the two points with the
load resistance disconnected (equivalent Thevenin’s source Vth) by the sum of load
resistance and the resistance of the network measured between these points with
load resistance disconnected and source of emf replaced by their internal
resistances (equivalent Thevenin’s resistance).
Thevenin’s Equivalent Circuit

26. +
V
R1
R2
R3 RL
R4
The circuit contains one voltage source (V) and five
resistors of resistances R1, R2, R3, R4 and RL. Using
Thevenin’s theorem you have to find the current
flowing through the load resistor RL.
open
circuit
+
V
R1
R2
R3
R4
Step 1 At first disconnect (remove) the load
resistance. Next find the open circuit voltage
across the load terminals. This voltage is
designated as “Thevenin’s Equivalent Voltage
Source (VTh).
VTh = voltage across the resistance R3 (since no
current will flow through the resistance R4)
3
1 2 3
Th
V
V R
R R R
 
 

27. open
circuit
short
circuited
voltage
sources
R1
R2
R3
R4
Step 2 Next remove all the sources from the circuit leaving their internal
resistances (if any). Look backward into the circuit from the open circuited load
terminals to find the Thevenin’s resistance (RTh).
 
1 2 3
4
1 2 3
Th
R R R
R R
R R R

 
 
Step 3 Now draw the Thevenin’s Equivalent Circuit and connect the load
resistance to the circuit again. Find the load current (IL) using the formula
Load current = Th
L
Th L
V
I
R R



28. Example:
Using Thevenin’s theorem to determine the current through and the voltage across the 25 Ω
resistor given in the figure below.
Vth=open circuit voltage across a-b=Vab=10 V
Vab = voltage drop across R3=R3I1=R3 X
S
1 3
V 20
=10× 10V
R +R 10+10

Vth=open circuit voltage across a-b=Vab=10 V

29. Rth=resistance of the network of figure (c) measured at the terminals a-b
1 3
2 1 3 2
1 3
R R 10 10
R (R || R ) R 10 15
R R 10 10

      
 
The 25 Ω resistor is now connected between a-b of the Thevenin’s equivalent as
shown in figure (e). let I be the current through 25Ω resistor.
10
0.25
25 15 25
th
th
V
I A
R
  
 
=25×I=25×0.25=6.25V
Voltage across 25Ω resistor

30. Norton’s Theorem Any two terminal active network containing voltage sources and
resistances when viewed from its output terminals, is equivalent to a constant
current sources and a parallel resistance. The constant current is equal to the current
which would flow if a short circuit is placed across the load terminals and parallel
resistance is the resistance of the network when viewed from the open circuited load
terminals after all voltage and current sources have been removed and replaced by
their internal resistances.

31. load
ohm
RL
R3
R2
R1
100
+
V
The circuit contains one voltage source (V) and four resistors of resistances viz R1, R2,
R3 and RL. Using Norton’s theorem you have to find the current flowing through the
load resistor RL.
Step 1 At first short circuited the
load resistance. Next find the short
circuit current through the load
terminals. This current is designated as
“Norton’s Equivalent Current Source
(IN).
IN = Current through the resistance R2
1
1 2 1 2
1 2
100
N
R
V
I
R R R R
R R
 




32. Step 2 Next remove all the sources from the circuit leaving their internal resistances
(if any). Look backward into the circuit from the open circuited load terminals to
find the Norton’s resistance (RN).
ohm
Short circuited
voltage source
terminals
open circuited
load terminals
100
R1
R2
R3
 
1
2 3
1
1 2 3
1
2 3
1
100
100
( ||100) ||
100
100
N
R
R R
R
R R R R
R
R R
R
 

 

 
  
 
 
 

 

33. Step 3 Now draw the Norton’s Equivalent Circuit and connect the load
resistance to the circuit again. Find the load current (IL) using the formula
Load current = N
L N
N L
R
I I
R R
 


34. Example:
Determine the current IL through the 15 Ω resistor given in figure (a) below by Norton’s
Theorem.
Solution:
Step 1: Since we are interested in determining the current through the 15 Ω resistor,
our first step is to remove the 15 Ω resistor from the original network of figure (a) and
to replaced by a short circuit. When the terminals a-b are short circuited, the original
circuit is reduced to that of figure (b).
N SC
totalvoltage 10
I =I = 1
totalresistance 10
A
 
Step 2: Determination of Norton current IN

35. Step 3: Determination of Norton resistance RN
1 1 1 1 1
, 5
10 20 20 5
N
N
R
R
     
Step 4: Norton Equivalent network
Step 5: Determination of current through 15Ω resistor
5
1 0.25
5 15
L
I A
  


37. Maximum Power Transform Theorem As applicable to dc networks, this theorem
may be stated as follows.
A resistive load will abstract maximum power from a network when the
load resistance is equal to the resistance of the network as viewed from the output
terminals, when all energy sources removed leaving behind their internal
resistances.
In the figure, a load resistance of RL is connected across
the terminals A and B of a network which consist of a
generator of emf E and internal resistance Rg and a series
resistance R which, in fact, represent the lumped
resistance of the connecting wires. Let Ri=Rg+R= internal
resistance of the network as viewed from A and B.
According to this theorem, RL will extract
maximum power from this network when RL=Ri

38. Proof :
Circuit current
i
L R
R
E
I


Power consumed by the load is --- (i)
2
i
L
2
2
)
R
(R 

 L
L
L
R
E
R
I
P
For PL to be maximum, 0

L
L
dR
dP
Differentiating equation (i) above, we have
       
   
i
2
L
2
2
L
L
2
Max
L
i
L
i
L
L
3
i
L
L
2
i
L
2
3
i
L
L
2
i
L
2
3
i
L
L
2
i
L
2
L
L
4R
E
4R
E
4R
R
E
P
is
Power
Max
R
R
or,
R
R
2R
or,
]
R
R
2R
R
R
1
[
E
0
]
R
R
2R
R
R
1
[
E
)]
R
R
2
(
R
R
R
1
[
E
dR
dP






















39. Example: In the figure below what is the value of the resistance R for maximum power
transfer and what is the value of the maximum power?
Solution: We know for maximum power
transfer,
Load resistance = Source resistance
Here, load resistance=12||R=12×R/(12+R)
Source resistance =4Ω
So for maximum power transfer
12R
4
12 R
12R 48 4R
8R 48
48
R 6
8


 

  

40. Network resistance looking from the
battery terminals
=4 + (12║6) = 8Ω
Circuit current=24V/8Ω=3 A
Therefore maximum power transfer to the
load
=3 × (12║6) = 12 Watt.

41. Thank you for listeni