INTRODUCTION TO ELECTRIC CIRCUIT

Delta-Star (Example)

4Ω 12Ω

Rc

x Ra 6Ω y

Rb 10Ω

8Ω

Convert --- Y

Ra = 4x8 = 1.78Ω Rb = 8x6 = 2.67Ω
4+8+6 4+8+6

Rc = 4 x 6 = 1.33Ω

4+8+6

Delta-Star (Example)

Rd 12Ω y

1.33Ω

x 1.78Ω

2.67Ω

Re 10Ω

Rd = 1.33 + 12= 13.33 Ω
Re = 2.67 + 10= 12.67 Ω

Delta-Star (Example) y

13.33Ω

Rd 12Ω

1.33Ω

x 1.78Ω

2.67Ω

Re 10Ω

12.67Ω

Rd = 1.33 + 12= 13.33 Ω

Re = 2.67 + 10= 12.67 Ω

Delta-Star (Example) y

Rf 13.33Ω

x 1.78Ω

12.67Ω

Rf = 13.33∗12.67 = 6.5Ω
13.33+12.67

Delta-Star (Example)

13.33Ω

x 1.78Ω 6.5Ω y

12.67Ω

Rf = 1133..3333+∗1122..6R677x=y 6=.51Ω.78 + 6.5 = 8.28Ω

Electrical Power & Energy

• ELECTRICAL POWER is defined as the rate at
which electrical energy is transferred by an
electric circuit.

• The SI unit of power is Watt.

• Equation:

Power, P = VI Equ. 1

V – voltage measured in Volts (V)
I – current measured in Ampere (A)

Electrical Power & Energy

• From Ohm’s Law;
I = V/R and V = I*R

V2 Equ.2
Equ.3
Hence Power, P =
P = I2R

Electrical Power & Energy

• ENERGY can be defined as capacity to do
work

• The unit of energy is Joule
• Equation :

Energy/Work Done, W = Pt

P – power measured in Watt (W)
t – time measured in seconds (s)

Electrical Power & Energy

Example 1.16
By referring to the circuit below, calculate:
i) Power that’s supplied by the battery
ii) Power that’s absorbed by 25Ω resistor
iii) Energy supplied by the battery after 30s
iv) Energy absorbed by the 15Ω resistor after 2 hours

15Ω

20V 25Ω

Electrical Power & Energy

RT 15Ω i) Power that’s supplied by the battery, Ps

IT RT = 15 + 25 = 40Ω

20V 25Ω

IT = = 20 = 0.5A
40

Use Equ. 1:

Power, Ps = V*I = 20 x 0.5 = 10W

Electrical Power & Energy

15Ω ii) Power that’s absorbed by 25Ω resistor PL

0.5A

20V 25Ω

Use Equ. 3:

Power, PL = I2*R = 0.52 x 25 = 6.25W

Electrical Power & Energy

15Ω iii) Energy supplied by the battery after 30s

0.5A

20V 25Ω

Energy, W = P*t = 10 x 30 = 300 J

Electrical Power & Energy

15Ω iv) Energy absorbed by the 15Ω resistor after 2
hours

0.5A

20V 25Ω

Energy,
W = P*t = I2*R*t = 0.52 x 15 x 2 x 60 x 60 = 27 kJ

SELF-EXERCISE

i) Diagrams below show a delta connection circuit
with its equivalent star connection circuit. If
R1=20kΩ, R2=40kΩ and R3 =80kΩ, calculate
Ra, Rb and Rc

11A.N4S3WkΩER

AN5.S7W1EkRΩ 2A2N.8SW6kEΩR

SELF-EXERCISE

ii) With refer to the diagram as below, calculate
power that supplied by the battery and power
dissipation at resistor 40kΩ.

PsAN=S4W.8EmR W

PL4A0N=SW1.E6RmW

RECAP

• Cell and battery are sources of DC type of
electricity.

• Voltage, current and resistance are recognized
as three basic elements of electrical circuit
which contribute in Ohm’s Law.

• Electrical circuit can be constructed in series,
parallel and combination of series-parallel
connection.

• Star-Delta transformation technique is required
to analyze network that involve Star/Delta
connection.

• Power and Energy is the product of voltage and
current elements of a circuit.

REFERENCES

Main:
John Bird (2010). Electrical Circuit Theory & Technology. Fourth

Edition. Newness. (ISBN: 978-0-08-089056-2)

Additional:
1. Allan R. Hambley (2011). Electrical Engineering, Principles

and Applications, Fifth Edition. Prentice Hall. (ISBN-13: 978-
0-13-213006-6)
2. B.L. Theraja (2010).Textbook of Electrical Technology .S
Chand & Co Ltd. (ISBN: 978-8121924900)

REFERENCES

3. Darren Ashby (2011). Electrical Engineering 101, (3rd Ed )
[Paperback] Elsevier Inc. (ISBN: 978-0123860019)

4. John Bird. (2010). Electrical And Electronic Principles And
Technology. Fourth Edition. Newness. (ISBN: 978-1-85617-
770-2)

5. Meizhong Wang. (2010). Understandable Electric Circuits
First edition © 2005 Higher Education Press, China, English
translation ©2010 The Institution of Engineering and
Technology. (ISBN 978-0-86341-952-2)

6. V. K. Mehta (2010). Principles of Electrical Engineering and
Electrical [Paperback] S Chand & Co Ltd. (ISBN: 978-
8121927291)