Summary Notes of Electrical Technology

Delta-Star and Star-Delta Transformation

Delta Star

Circuit 01 01 Circuit

RA RB R1
RC
RA R1 RB R2 R3

R2 R3

RC

Formula 02 Method 1 Method 2 +0 2 F o rmula
+ +
= = + + = +

=
+ + = + + = + +

=
+ + = + + = + +


36

Example 8 : Delta-Star Transformation

.

ANS :
+ +
=

( )( )
= + + = . Ω

RB RA =
RC + +

( )( )
= + + = . Ω

=
+ +

( )( )
= + + = . Ω

37

Example 8 : (Continue) Delta-Star Transformation

.

= + = . + = . Ω
= + = . + = . Ω

= = . .
+ . + .

.
= . = . Ω

= +
= . + .

= . Ω

38

Example 9 : Star-Delta Transformation

Find the equivalent resistance between A & B in the given network.

.

ANS :

=
+ +
( )( )
= + + = = Ω

=
+ +
( )( )
= + + = = . Ω

=
+ +
( )( )
= + + = = Ω

39

Example 10 : Star-Delta Transformation

Find the equivalent resistance between A & B in the given network. Where P = 10kΩ, Q = 20kΩ and R = 30kΩ.
.
ANS :

= + +
( )( )
= + + = . Ω

= + +
( )( )
= + + = . Ω


= + +

( )( )
= + + = Ω

40

03
Network
Theorem

41

Kirchoff's Voltage Law (KVL)

2 Resistor 3 Resistor

01 01

Circuit Circuit

02 ෍ = ෍ ෍ = ෍ 02

Formula Formula

= + + + = + +

In any closed loop, the total Voltage Supply (Vs) equal to the total Voltage
Drop (VD) across conductors

42

Example 1 : Kirchoff’s Voltage Law(KVL)

Calculate the suitable value for E2.

ANS :

෍ = ෍

+ + = + + +

30 + (− 2) + 45 = 15 + 2 + 8 + 4
( − 2) + 75 = 29
− 2 = 29 − 75
− 2 = −46
=

43

Kirchoff’s Current Law (KCL)

01 06

At any junction in a

electric circuit, the total ෍ = ෍

02 current flowing towards 05
that junction is equal to
+ + = +
the total current

flowing away from the

03 junction. 04

44

Example 2 : Kirchoff’s Current Law (KCL)

Calculate I1, I2, I3 and I4 at each node.

ANS :

Applying KCL to each node :-

At node a; At node b;

෍ = ෍ ෍ = ෍

5 = 1 + 6 6 = 2 + 10
1 = 5 − 6 2 = 6 − 10
= − = −

At node c; At node d;

෍ = ෍ ෍ = ෍

1 = 3 + 9 2 + 3 = 45
−1 = 3 +9 2 + (−10) = 45

3 = −1 − 9 2 = 10 + 45
= − =
45

Mesh Analysis

Draw mesh currents in loops of

12 the circuit.

1 Mercury

2Mesh Analysis or Label resistor voltage drop
polarities based on assumed
Mesh Current Analysis is a directions of mesh current.
technique used to find the
Loop 1 :
3currents circulating around a Write KVL equations for each
3෍ = ෍ loop or mesh within any loop of the circuit.
closed path of a circuit.
= +
= + ( + )

Loop 2: ෍ = ෍ Solve for unknown mesh
= +
4 currents using equations
derived from the third step.

= + ( + )

46

Example 3 : Mesh Analysis VR1 VR2
R1=14Ω R2=20Ω
Using Mesh Analysis method, calculate the current flow
through resistor R3. + +

ANS : + R3=50Ω + VR3 +
Step 1 : Write Kirchoff’s Voltage law for each loop E1=5V
Loop 1: E2=15V

෍ = ෍ Loop 1 Loop 2

(I1) (I2)

= + Step 2 : Using Cramer’s rule, solve the unknown currents of simultaneous equations
= + ( + )
= + + V
= + +
=
= + - - - - eqn 1

Loop 2: ∆= , = = ሼ − ( )} =

෍ = ෍ ∆ −
∆ = , = = ሼ − ( )} = − ∴ = ∆ = = − .
= +
= + ( + )
= + + ∆ = , = =ሼ − ( )} = ∴ = ∆ = = .
= + + ∆

= + - - - - eqn 2 Current flow through the R3 resistor is; = + = − . + . = .

47

Example 4 : Mesh Analysis

Calculate the current flow through the 25 Ω resistor,
using Mesh Analysis.

ANS :

Step 1: Draw mesh currents in loops of the circuit

Step 2: Write Kirchoff’s Voltage law for each loop
Loop 1:

෍ = ෍
+ = + +
+ = + ( + ) +
+ = + + +
+ = + + +
= + - - - - eqn 1

48

Example 4 : (Continue) Mesh Analysis

Calculate the current flow through the 25 Ω resistor,

using Mesh Analysis.

ANS : Step 3: Solve the unknown currents of simultaneous equations for step 2

Step 2: Write Kirchoff’s Voltage law for each loop = + −− −
= + −− −
Loop 2:
෍ = ෍ 1, ℎ 45,
+ = + = + −− −
+ = + ( + )
+ = + + 2, ℎ 75,
+ = + + = + −− −
= + - - - - eqn 2
Solve for : :
∴ − = + − .

= − 1950 = − .
= + .

− = .
= − . =
= .
= = .

49

Nodal Analysis

Nodal analysis, node-voltage analysis, or the branch current method is a method of determining the
voltage (potential difference) between "nodes" (points where elements or branches connect) in an
electrical circuit in terms of the branch currents.

Determine the number of common nodes and
reference node within the network.

Assign current and its direction to each distinct branch
of the nodes in the network.

Apply KCL at each of the common nodes in the
network.

Solve the resulting simultaneous linear equation for the
nodal voltages and Determine the currents through
and voltages across each the elements in the network.

50

Example 5 : Nodal Analysis

By using Nodal Analysis, calculate the voltages V1 dan V2 and hence calculate the
current flowing through the 20 Ω resistor.

ANS :

Step 1 : Identify the principal node and Reference
node. Voltages V1 and V2 assigned.
Step 2 : Assume all the currents are leaving from the
node.

51

Example 5 : (Continue) Nodal Analysis

By using Nodal Analysis, calculate the voltages V1 dan V2 and hence calculate the
current flowing through the 20 Ω resistor.

ANS : Step 3 : Apply Kirchoff” Current Law at each nodes.

∶
1 − 10 1 − 0 1 − 2
8 + 20 + 5 =0

1 − 10 + 1 − 0 + 1 − 2 = 0
8 8 20 5 5
111 2 10
Step 3: Apply Kirchoff” Current Law 1 8 + 20 + 5 − 5 = 8
at each nodes.

0.38 1 − 0.2 2 = 1.25 −− − 1

∶
2 − 1 2 − 0
5 + 15 −2=0

2 − 1 + 2 − 0 − 2 = 0
5 5 15
1 11
− 5 + 2 5 + 15 =2

−0.2 1 + 0.27 2 = 2 −− − 2

52

Example 5 : (Continue) Nodal Analysis

By using Nodal Analysis, calculate the voltages V1 dan V2 and hence calculate the

current flowing through the 20 Ω resistor. 1 2 ∶

ANS : 3 + 4
0.063 2 = 1.01
Step 4 : Solve for V1 and V2 1.01
. − . = . −− − 2 = 0.063
− . + . = −− − = .

2 1

1 , ℎ 0.2 ∶ 0.38 1 − 0.2 16.032 = 1.25

0.38 1 − 3.21 = 1.25

0.076 1 − 0.04 2 = 0.25 −− − 3 0.38 1 = 1.25 + 3.21
2, ℎ 0.38 ∶ 0.38 1 = 4.46
4.46
1 = 0.38
= .
−0.076 1 + 0.103 2 = 0.76 −− − ℎ ℎ ℎ 20Ω is

20Ω = 1
20
11.74
20Ω = 20 = .

53

Superposition

Superposition theorem states that in any linear, bilateral network where more than one source is present,
the response across any element in the circuit, is the sum of the responses obtained from each source
considered separately while all other sources are replaced by their internal resistance.

Take only one independent source

of voltage or current and

deactivate the other sources

STEP 1 If the current obtained by

By activating one source and each branch is in the same
deactivating the other source
direction, then add them and if
find the current in each
it is in the opposite direction,
branch of the network. STEP 2
STEP 5 subtract them to obtain the
net current in each branch.

SUPERPOSOTION

If there is a voltage source Determine the net branch
than short circuit it and if
there is a current source current utilizing the
then just open circuit it.
superposition theorem, add the
.
STEP 3 currents obtained from each

STEP 4 individual source for each
branch.

54

Example 6 : Superposition

By means of the Superposition theorem, calculate
the currents flowing through the 18 Ω resistor.

ANS :

Step 1 : Take the source 10 V alone and
open-circuit the 2 A source as shown in the
circuit.

Step 2 : Calculate the current across 18 Ω ( 18Ω′) due

to the one source, 10 V

Since 5Ω = 0 A, then by using Ohm’s law;

3Ω′ = 18Ω′ =

3Ω′ = 18Ω′ = 10
3 + 18
10
3Ω′ = 18Ω′ = 21

3Ω′ = 18Ω′= 0.48

55

Example 6 : (Continue) Superposition

By means of the Superposition theorem, calculate
the currents flowing through the 18 Ω resistor.

ANS :

Repeat step 1 and 2 for other sources and find the current

across R18 Ω caused by the individual sources acting alone.

Using CDR,

18Ω" = 3 ×
3 + 18

Since = 2 , 3 To find the total current flows through the 18 Ω resistor, perform
21 algebraic sum of individual currents considering their direction of
18Ω" = × 2 flow to the 18 Ω resistor

18Ω" = 0.28 18Ω = 18Ω′ + 18Ω"
18Ω =0.48+0.28

= .

56

Thevenin Theorem

Thevenin’s Theorem states that any linear electric network or a complex circuit

Mercurwith current and voltage sources can be replaced by an equivalent circuit
ycontaining a single independent voltage source VTH and a Series ResistanceMRaTHrs.

Earth ∶ =
+

57

Steps to Analyze an Electric Circuit using
Thevenin’s Theorem

Thevenin’s • Open / remove the load resistor.
Voltage, • Calculate / measure the open circuit

voltage.
• This is the Thevenin’s voltage .

THEVENIN’S Thevenin’s • Open / remove the load resistor.
THEOREM Resistance, • Remove the power source; voltage sources

shorted and current sources open.
• Calculate / measure the open circuit
Thevenin’s
equivalent resistance.
• This is the Thevenin’s Resistance .
circuit
• Redraw the circuit with measured open
circuit Voltage as voltage source and
measured open circuit Resistance as a
series resistance.

• Connect the load resistor which had
removed.

58

Example 7 : Thevenin Theorem

Calculate the current flow through Load Resistance, RL using Norton Theorem.

ANS :

Step 1 :
Remove the Load, RL. Calculate Thevenin’s Voltage

is the open circuit voltage at terminal A-B

= = 2 ;

Using the concept that the voltage

in parallel is same ;

2 = = 2 ×
1 + 2

= 5 × 10
4+5

= .

59

Example 7 : (Continue) Thevenin Theorem

Calculate the current flow through Load Resistance, RL using Norton Theorem.

Step 2 : Calculate Thevenin’s Resistance which is the Step 3 : Insert the load resistive load into the
resistance seen from terminal A-B. The power source Thevenin’s Equivalent Circuit. Calculate the current
is replaced by short circuit. (IL) that flows through the 5 Ω resistor.

= 1 × 2 =
1 + 2 +
5.56
4×5 = 2.22 + 2
= 4 + 5
20 5.56
= 9 = 4.22
= .
= .

60

Norton Theorem

Any linear circuit, no matter how complicated, can be reduced to an equivalent

Mceirrccuuirt with just one current source, IN and parallel resistance, RN coupled to a load,
RL, according to Norton's Theorem.
y Mars

IL

Earth ∶ =
+

61

Steps to Analyze an Electric Circuit using
Norton’s Theorem

Norton’s • Short the load resistor.
Current, • Calculate / measure the Short Circuit

Current.
• This is the Norton’s current .

NORTON’S Norton’s • Open / remove the load resistor.
THEOREM Resistance, • Remove the power source; voltage sources

shorted and current sources open.
• Calculate / measure the Open Circuit
Norton’s
equivalent Resistance.
• This is the Norton’s Resistance .
circuit
• Redraw the circuit with measured short
circuit Current (IN) as Current Source and
measured open circuit resistance (RN) as a
parallel resistance and connect the load
resistor.

• Connect the load resistor which had
removed.

62

Example 8 : Norton Theorem

Calculate the current flow through Load Resistance, RL using Thevenin Theorem.

ANS :

Step 1 : Remove the load resistance, RL and
shorted the A - B terminals to determine the
Norton current, IN

By applying Current Divider Rule ;

25
= 10 + 5 + 25 ×

25
= 40 × 2
= .

63

Example 8 : (Continue) Norton Theorem

Calculate the current flow through Load Resistance, RL using Thevenin Theorem.

ANS :

Step 2 : Remove the load resistance, RL and Step 3 : Draw Norton’s Equivalent Circuit and insert
open current source. Calculate the open circuit back the load resistor at terminal A – B.
resistance at terminal A - B.

Seen from terminal A – B; By applying Current Divider Rule;

= 10 + 5 + 25 = ×
= +
40
= 40 + 4 × 1.25

66 = .

Difference between Thevenin

and Norton Theorems

Thevenin’s Norton’s

Source Uses a voltage source. Source Uses a current source.

Resistor Uses a resistor in series. Resistor Uses a resistor in
parallel.

Formula = Formula =
+ +

• Thevenin’s resistance and Norton’s resistance = equal in magnitude.
• Thevenin’s equivalent circuit and Norton’s equivalent circuit = can be easily interchanged.

65

Thevenin-Norton Equivalencies

=
=

=

=

=


= ℎ
= ℎ

66

04

Capacitor and
Capacitance

67

Capacitor and Capacitance

CAPACITOR CAPACITANCE

Capacitors are simple passive device that Capacitance is the ratio of the amount
can store an electrical charge on their of electric charge stored on a
plates when connected to a voltage conductor to a difference in electric
source. potential.

UNIT SI & SYMBOL FORMULA

Farad & ,
, ( )

= , ( )


=

68

Types of Capacitors

Non – polarized 01
capacitor

Polarized 02 Fixed
capacitor
capacitor

01 Variable capacitor

Variable
capacitor

02 Trimmer capacitor
03 Air-gap capacitor

69

An oxidised coating on the anode Types of Capacitor The dielectric material of a
paper capacitor is paper, which
of an electrolytic capacitor is also known as a fixed
capacitor. The paper capacitor
serves as a dielectric. The holds a set quantity of electric
charge. Two metal plates are
capacitance of these devices is Paper sandwiched between two sheets
polarised. The dielectric of Electrolytic of paper that serve as a
dielectric material.
electrolytic capacitors is used to
Various other names for a
classify them. polymer film capacitor include a
plastic film capacitor or a film
Natural minerals are referred to Mica Film dielectric capacitor. Film
as "Mica." The dielectric of a 70 capacitors have the advantages
silver mica capacitor is known as of being cheap and having an
mica. Silver mica capacitors and indefinite shelf life. In the film
damped mica capacitors are the capacitor, one side of the
two main varieties of these capacitor is made of metalized
capacitors. In lieu of clamped dielectric material. The film
mica, silver mica capacitors are capacitor is rolled into thin films
employed for their lower depending on the application.
characteristics. Mica capacitors These capacitors can withstand
are typically low-loss capacitors voltages ranging from 50 V all
that are utilised in high- the way up to 2 kV.
frequency applications where
their value doesn't vary
substantially over time.

Terms related to Capacitance

CAPACITANCE

Electric Flux Density, D Electric Field Strength, E

Electric flux density is the amount of flux passing Electric field strength is the ratio between
through a defined area that is perpendicular to the the potential difference or voltage and
direction of the flux. Electric flux is defined as the total thickness of the d2ielectric
number of electric lines force emanating from a
charged body. As the flux is total number of lines of
force emanated from the charge body, the unit of flux = /
is also taken as Coulomb.
= ℎ /
= / =

= / 2 = ℎ 4

= ℎ 71

= ℎ 2

Terms related to Capacitance

CAPACITANCE

Absolute Permittivity,

Absolute Permittivity is the ratio Absolute permittivity of free space or
between electric flux density to the vacuum is equal to approximately -- 8.85 x
electric field strength 10-12 Farads / metre (F/m). It is normally
symbolised by : ε02.

Relative permittivity is defined as the
= / permittivity of a given material relative to that
= / of the permittivity of a vacuum. It is normally
= / 2 symbolised by : εr.4

= ℎ /

72

Construction of Capacitor

➢ The conductive plates are normally made
of materials such as aluminum, brass, or
copper

➢ The insulating layer between a capacitors
plates is commonly called the dielectric. A
dielectric material such as air, paper, mica,
ceramic, etc.

➢ The factors affecting capacitance are :
• Plate area
• Plate spacing (distance between the

plates)
• Dielectric material

➢ Formula : =

• C = Capacitance (F)

• Q = Charge stored (Coulomb)

• V = Voltage (V)

73

Capacitors in Series and Parallel Circuit

Series Parallel

Capacitance Total Capacitance Total

= + +
= + +

74

Example 1 : Series and Parallel Circuit

Calculate the equivalent capacitance of the following circuit.

Series Parallel

22 × 30 = 55 + 40
= 22 + 30 =

6.6−10 = 15 + 33 + 20
= 5.2−5 =
= .

1111
= 6 + 8 + 7

1
= 0.43
= .

75

Example 2 : Series-Parallel Circuit

Calculate the equivalent capacitance of the following circuit.

Solve the series capacitance;

= 2 × 3 = 4 × 6 = . = 1 + 1 + 4 + 2
2 + 3 4 + 6 = 3 + 2.4 + 8 + 6.47

5 × 6 23 × 9 = .
5 + 6 23 + 9
= = = .

76

Example 3 : Series-Parallel Circuit

Calculate the equivalent capacitance of the following circuit.

= 4 × 5 = 8 × 2 = = 2 × 2 = 4 × 6.4 = .
4 + 5 8 + 2 2 + 2 4 + 6.4

2 = 3 + 1 = 3 + 1
2 = 2.4 + 4 = 2.46 + 3

= . = .

77

Example 4

1. Find the charge on a 5 µF capacitor when it is connected across a 120 V source.

ANS :

=CV

= 5 × 120
= × −

2. Two parallel rectangular plate measuring 30 cm by 40 cm carry an electric charge of 0.3 µC.
Calculate the Electric Flux Density.

ANS :
ℎ , = 0.3

, = 30 × 40 = 1200 2 = 1200 × 10−4 2


, =
0.3

= 1200 × 10−4
= . /

78

Circuit with Resistive and Capacitive Load

Charging Process : , ;

Initial current, Io −

= −

The initial current, which is = ,
the maximum current, will =
01 = 03
be;
= = =

02 ℎ τ = CR 04
∶
= , = Energy stored in the
capacitor The capacitor is fully
charged when the
time reaches 5

, ( , , ℎ
ℎ , =
ℎ
= ∶ )
ℎ

0.632
ℎ ℎ

0.371 .

79

Circuit with Resistive and Capacitive Load

Discharging Process : , , ℎ

01 The current will increase ℎ 03

− ℎ

by = − 0.632
ℎ ℎ
0.371 .

=

ℎ The capacitor is fully 04
charged

when the time
reaches 5
02 ℎ ℎ ;

−

=
= ,
=
=
=

80

Exponential graph of capacitor during
charging and discharging process

Charging process Capacitor Discharging process

81

Example 5 : Charging Process

A capacitor of 16 µF is connected in series to a 1 MΩ resistor across a DC voltage supply of

400 V. Determine the voltage across capacitor after 5 s of charging and energy stored in the

capacitor.

ANS : Energy stored in the capacitor.

The time constant; = 1 2
= 2
= 1 16 µ × 4002
= 16 µ × 1 2
=
= .
The voltage across the capacitor after 5 s;
Vmax = 400 V

−t

Vc t = Vmax 1 − e τ

−5

Vc t = 400 1 − e16
Vc t = 400 1 − e−0.3125
Vc t = 400 1 − 0.7316

Vc t = 400 0.2684
= .

82

Example 6 : Charging Process

A capacitor of 20 µF is connected in series to a 200 kΩ resistor across a DC voltage supply

of 200 V. Determine the time constant, initial charge current, initial voltage across capacitor,

the voltage across the capacitor after 6 s, and energy stored in the capacitor.

a R C ANS :
200kΩ 20µf
The time constant;
b =

= 20µ × 200
=

I(t) E Initial charge current;
200V

= =


= =
= = × − A

83

Example 6 : (Continue) Charging Process

A capacitor of 20 µF is connected in series to a 200 kΩ resistor across a DC voltage supply

of 200 V. Determine the time constant, initial charge current, initial voltage across capacitor,

the voltage across the capacitor after 6s, and energy stored in the capacitor.

ANS :

Initial voltage across capacitor; Energy stored in the capacitor.
,
= 1 2
ℎ , 2
= = 1 20 µ × 2002
2
The voltage across the capacitor after 6 s;
Vmax =200 V = .

−t

Vc t = Vmax 1 − e τ

−6

Vc t = 200 1 − e 4
Vc t = 200 1 − e−1.5
Vc t = 200 1 − 0.223

Vc t = 200 0.777
= .

84

05
Inductor and
Inductance

85