tutorials on lesson 3 - 5

7/28/2019 Tutorials on Lesson 3 - 5

http://slidepdf.com/reader/full/tutorials-on-lesson-3-5 1/10

BASIC ELECTRICAL ENGINEERING

Tutorials on Lessons 3-5

1. Determine the Thevenin’s equivalent circuit with respect to terminals AB of the

circuit shown in Figure 1.Use the result to find the current in two impedances

551

j Z , and

0

2010 Z , connected in turn to terminals AB.

Determine also the power delivered to them.

Solution

0

00

0105

050

555

050

j j Z

V I

The Thevenin’s equivalent voltage V’ is the voltage drop across the (5 + j5) Ω

impedance.i.e V j j I V V AB

0'457.70505055

The driving point impedance at terminals AB is

Z’ = -j5 in parallel with 5 + j5

i.e

55

555

555' j j j

j j Z

Thus, Thevenin’s equivalent is

To find the current when

(i) 551

j Z is connected across AB



A j Z Z

V I

0

0

1

'

'

1905

1010

457.70

W x R I P 125552

1

2

11

(ii) 0

2010 Z

A j Z Z

V I

0

0

2'

'

243.6347.4

515

457.70

W x R I P 2001047.42

2

2

22

2. Obtain the Thevenin’s equivalent circuit for the network shown in figure 2.

Solution

(i) Find VTH i.e V AB = I2 x 6 Ω

The mesh currents in matrix form are:



0

4.178.55

885

5550

2

1 V

I

I

j j

j j

885

555

05

4.178.55550

22

2

j j

j j

j

V j

I Z

A0

0

0

03.336.727.83

6.72279

V xV AB0

00.2063.33 (ii) Find ZTH when the source is shorted

5.25.2

55

55 j

j

xj Z

p

01.236.341.131.35.55.10

5.55.46 j

j j Z TH

Thus the Thevenin’s equivalent circuit is:



3. Determine the Norton’s equivalent circuit with respect to terminals AB of the

circuit shown in figure 1.

Use the result to find the current in the two impedances

551

j Z , and

0

2010 Z , connected in turn to terminals AB.

Determine also the power delivered to them.

Solution

To find the Norton’s equivalent circuit with respect to terminals AB of:

When a short circuit is applied to terminals AB, the short-current will be:

A j

V I

0

0

0

'9010

905

050

5

When the source is set to zero, the effective impedance is:

55

555

555' j

j j

j j Z

Thus, the Norton’s equivalent circuit is:

(i) Connect 551

j Z across terminals AB

The current that flows in Z1 will be:

A j

j

Z Z

Z I I

00

1

'

'

'

1905

1010

559010

Power, W x R I P 125552

1

2

11

(ii) Connect 0

2 010 Z

across terminals ABThe current that flows in Z2 will be:



A j

j

Z Z

Z I I 00

2

'

'

'

243.6347.4

515

559010

Power, W x R I P 2001047.42

2

2

22

Note that from Questions 1 and 3, it is seen that Thevenin’s and Norton’scircuits are equivalent to each other.

4. Replace the active network shown in figure 3 with

(i) The Thevenin’s equivalent circuit, and

(ii)The Norton’s equivalent circuit respectively, at terminals AB.

Solution

(i) To find the Thevenin’s equivalent circuit of Figure 3,

With AB open circuited, the current I1 flowing is given by:

A j

I 0

0

00

11.1747.1

1.176.13

020

4310

020

The voltage drop across the 10Ω resistor is

V x I V 0

1101.177.1410

The voltage across terminals AB will be the sum of the two voltage sources and

the voltage across the 10Ω resistor with polarity as shown.

That is:



V j

j j j

V AB

0

000

4.26444.1139.1112.1

32.405.14071.7071.7020

1.177.144510020

The equivalent impedance looking at terminals AB with all sources short-circuited is:

0

2.1526.816.297.7413

43105 j

j

j Z AB Therefore, Thervenin’s equivalent circuit is:

(ii) To find the Norton’s equivalent circuit of Figure 3,

Short-circuit terminals AB and determine the short-circuit current, I2

The mesh equation in matrix form is: V I Z

V

V

I

I j

0

0

2

1

451020

02

1510

10413

03.3236.112609510015413 j j Z

00

0

22

451002010

020413

j

20007.793.12413 j j



063.24773.15563.14319.6020063.14381.139 j j

A I Z

0

0

0

22

293.27939.1

3.32112

63.24773.155

The impedance looking at terminals AB is the same as obtained in Q 4(i).

Therefore, the Norton’s equivalent circuit is:

5. Write the Nodal Voltage equation for the circuit shown in figure 4 in the form

I V Y

.

Solution

At Node 1,

0

2

21

1

1

1

R

V V

R

V I

Or 12

2

1

21

111 I V

RV

R R

At Node 2,

03

2

2

21

2

R

V

R

V V I



Or 22

32

1

2

111 I V

R RV

R

The two equations can be written in matrix form as follows:

2

1

2

1

322

221

111

111

I

I

V

V

R R R

R R R

Note: It is seen that the LHS consist of currents flowing into or out of the Nodes

as a result of the Nodal Voltages, while the RHS are source currents.

In general, I V Y

6. In the network shown in figure 5, determine the voltages of Nodes 1 and 2 withrespect to the selected reference.

Trainees are to solve this Question and submit.



Solution

The Nodal equation may be written in matrix form by inspection as follows:

i.e I V Y

Take note that since Z

V I

At Node 1,

425

0502111

0V V

j

V V

0

21 010

44

1

2

1

5

1

V

jV

At Node 2,

22

9050

4

22

0

21

j

V V V V

0

2

19025

2

1

2

1

4

1

4

V

V

0

0

2

1

9025

010

2

1

2

1

4

1

4

1

4

1

4

1

2

1

5

1

V

V

j

j

095.15546.0

5.075.025.0

25.05.045.0

j

jY

0

11 3.565.135.075.0250

25.0010

j j

j



0

228.3735.18

25025.0

0105.045.0

j

j j

V V Y

0

0

011

125.727.24

95.15546.0

3.565.13

V V Y

0

0

0

22

275.536.33

95.15546.0

8.3735.18

tutorials on lesson 3 - 5

Documents