# Lecture 08 by naveedawan53

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PHY301 – Circuit Theory

Virtual University
PHY 301
LECTURE 8

Example:         Write KCL equations for all nodes.

Solution:

The KCL equations for nodes 1 through 4 follow
For node 1
I +I –I =0
1 2 5
For node 2
-I +I -50I = 0
2 3       2
For node 3
-I + 50I + I = 0
1      2 4
For node 4
I –I –I =0
5 3 4
Example         Write KCL equations for all nodes.

Solution:

For node A 10mA is entering the node and the source current I is leaving and   60mA is also entering
t
so,
For node A
I -60mA – 10mA= 0
t
For node B
60mA – 40mA – 20mA = 0

Example           Write KCL equation for node A.

Virtual University of Pakistan                                                                   Page 44
PHY301 – Circuit Theory

Solution:

By KCL the equation for node A
-12mA + 4mA + I =0

Example        Write KCL equation for node A and node B.

Solution:
The equation of currents at node A and B by KCL
For node A
-I +I + 3mA = 0
1 2
For node B
-12mA + 4mA +I =0
1
Example      Write KCL equation for node A .

Writing equation for node A .
For node A
-10 I + I +44 mA – 12 mA = 0
x x
Example         Write KCL equation for node A .

Solution:
For node A equation will be by KCL
I + 10I – 44mA = 0
x     x

Example         Calculate the values of I and I
1      2.

Solution:

Virtual University of Pakistan                                  Page 45
PHY301 – Circuit Theory

We want to calculate the values of I and I
1      2
For node A
4mA + 8mA- I = 0
1
I = 12mA
1
Now for I
2
For node B
-8mA + 2 mA + I =0
2
I = 6mA
2
Example          Calculate the values of I , I , and I
1 2          3.

Solution:
We want to calculate the values of I , I , and I so we will use node analysis.
1 2         3,

For node A
-I – I +8mA = 0
1 2
-I – I = -8mA
1 2
For node B
I + I +4mA = 0
2 3
I + I = - 4mA
2 3

For node C
-I + 2mA – 8mA = 0
3
I = - 6mA
3
Putting the value of I in equation of node B
3
I - 6mA = -4mA
2
I = 2mA
2
Putting the value of I in equation of node A
2
-I – 2mA = -8mA
1
I = 6mA
1
Example:         Find the KCL equtations for node A,node B, node C and node D.

Solution:

Virtual University of Pakistan                                                           Page 46
PHY301 – Circuit Theory

We want to write the equations for nodes A, B, C and D.
For node A
-5mA + 8mA + 4mA =0
For node B
I – I + 5mA =0
1 2
For node C
-I – 2mA + 3mA – 8mA =0
1
For node D
-4mA – 3mA + I = 00
3

Example         Calculate the current I .
o

Solution:
At node 1
(V1/12k) + ((V1 – V2)/10k) = 6mA
5V1 +6V1 – 6V2 = (60k)(6mA)
11V1– 6V2 = 360
At node 2
(V2/3k) +(V2/6k) + ((V2- V1)/10k) =0
10V2 +5V2+3V2-3V1=0
18V2 -3V1 =0

Equating equation of node1 and node 2

33V1 -18V2 =1080
-3V1 + 18V2 = 0

30V1 =1080
V1 = 36 volts
18V2 -3V1 =0
6V2 – V1=0
6V2 -36 =0

V2 = 36/6
V2 = 6volts
I =V2/6k
o
= 6/6k
= 1mA

Virtual University of Pakistan                                                Page 47

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