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Module 2

This document contains solutions to multiple circuit analysis problems using nodal analysis techniques. The first problem involves determining the current i in a circuit using nodal analysis at two nodes. The second problem determines the value of k that results in the node voltage vx being equal to zero. The third problem determines all four nodal voltages in a given circuit. The fourth problem finds the Norton equivalent of a given circuit seen between two terminals. The fifth problem determines the Norton equivalent and selects a value for an external resistor Rout to deliver maximum power.

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TKC
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954 views15 pages

Module 2

This document contains solutions to multiple circuit analysis problems using nodal analysis techniques. The first problem involves determining the current i in a circuit using nodal analysis at two nodes. The second problem determines the value of k that results in the node voltage vx being equal to zero. The third problem determines all four nodal voltages in a given circuit. The fourth problem finds the Norton equivalent of a given circuit seen between two terminals. The fifth problem determines the Norton equivalent and selects a value for an external resistor Rout to deliver maximum power.

Uploaded by

TKC
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Chapter 4

Exercise 8 : In the circuit of Fig. 4.34, determine the current labeled i with the assistance of nodal
analysis techniques.

Solution:

Considering bottom line as reference node.


Applying nodal analysis at nodes V1 and V2 we get:

5=

V 1 V 1V 2
+
eq 1
1
5

V 1V 2 V 2
=
+4eq 2
5
2

Rearrrangingthe equations we get :

6 V 1V 2=25

2V 17 V 2=40
Solving these two equations we get :

V 1=

27
19
V V 2=
V
8
4

Thus i=

V 1V 2 13
= A=1.625 A (ans)
5
8

Figure 4.34
Exercise 26

Determine the value of k that will result in vx, being equal to zero in the circuit of Fig. 4.52.

Solution:

Applying nodal analysis at nodes VxVy we get :

2Vx VxVy Vx
=
+
1
4
1
VxVy
VykVy
=1+
4
3

Now assuming Vx=0, putting first equation we get :

20 0Vy 0
=
+
1
4
1
>Vy =8 V

Putting the value of Vx=0Vy=8 V second equation we get :


08
8k
=1+
4
3

>k =

17
(ans)
8

FIGURE 4.52

Exercise: 28 For the circuit of Fig. 4.54, determine all four nodal voltages.

FIGURE 4.54
Solution:

The central node istaken as ground :


So V 1=1 V

Now for others applying nodal analysis at each of the nodes V 2,V 3V 4 we get :
1V 2
V 2V 3
+3=
1
2
V 3V 4
V 2V 3
+2 ( V 3V 4 )=
1
2

V 3V 4 V 4 V 41
=
+
1
3
4

Rearranging these equations we get :

3 V 2V 3=8
V 27 V 3+6 V 4=0

12V 319V 4=3

Solving the simultaneous equations we get :

V 1=

253
=3.085V
82

V 2=

103
=1.256V
82

V 3=

( ans )

39
=0.9512 V
41

Chapter 5
Question: 40

Determine the Norton equivalent of the circuit drawn in Fig. 5.81 as seen by terminals a and b. (There
should be no dependent sources in your answer.)

Solution:
Taking above portion as reference and bottom as voltage V,
Also inserting a voltage source of 1V

Now we have nodal equation as

Ix +0.7=

V +1
2500

also Ix=

2 IxV
500

Putting Ix solving for V we get ,

V =290.5276 V
thus current flowing across1 V sorce is

I=

V +1
1
+
2500 1.5 k

0.117266 A
1
Now norton resistance= =8.5275ohms
I

Now for Norton current we short the terminals

Thus applying nodal analysis we have:

Ix +0.7=

V
2500

also Ix=

2 IxV
500

Solving we get :
Ix=

875
=0.58372 A
1495

So Ix+.7=0.116277 A
116.277 mA

So nortoncurrent=116.3 mA

Question: 48
Study the circuit of Fig. 5.89. (a) Determine the Norton equivalent connected to resistor Rout. (b) Select a
value for Rout such that maximum power will be delivered to it.

Solution:
a)

To determine the Norton resistor, we will short the voltage source and open circuit the current source.

Now the resistor across AB will be:

Rab=

1 k2 k
1 k +2 k

2
k ohms
3

2
b m aximum power will be delivered if Rout =Rab=Rnorton= k ohms( ans)
3

Question: 60
Determine the effective resistance Rin of the network exhibited in Fig. 5.99.

Solution:

Before doing this question it is important to learn star-delta and vice-versa


conversion.

Equations for the transformation from -load to Y-load

Equations for the transformation from Y-load to -load

Transferring the first part into star from delta we get:


R=( R*R+ R*R+ R*R)(R+R+R) = R/3

Which is:

Now above portion is in parallel. So,

Now the last is in parallel, which gives:


R=8/7 R (ans)

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