# Lecture 28

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```					PHY301 – Circuit Theory

Virtual University
PHY 301
LECTURE 28
Example: Find Norton equivalent circuit.

Solution:
First step:     Removing R replace open circuit with short circuit
L

by ohm’s Law we have
I =12/~0 → A
N
Now we will insert the resistance R=~0 parallel to current source.

So we have the Norton’s equivalent circuit.
Example: Calculate Norton’s current through the circuit.

Solution:
Inject a 1A current source into the port and define Vx across 200 so that

For node 1
1=(V /100)+(V -V )/50
1        1 x
100=V +2V -2V
1    1   x
3V -2V =100 ----------------- (A)
1   x
For node 2
-0.1V =(V /200)+(V -V )/50
1  x         x 1
-20V =V +4V -4V
1 x     x   1
5V +16V =0 ----------------(B)
x      1

Virtual University of Pakistan                                                        Page 119
PHY301 – Circuit Theory

Solving simultaneously A and B we have
10Vx+32V1= 0
-10Vx+15V1= 500

47V1= 500
V1=10.64V
Also Rth= Rn =V1/1A =10.64
Since no independent source is involved in the circuit, hence I =0
N
Example: Calculate the voltage V by using Norton’s theorem , do not use loop and node methods. Use
o
super postion method.

Solution:
We want to calculate V by using Norton’s theorem.we will follow these four steps .
0
First step: Replacing R with a short circuit to find I
L                               N.

Here R is 6k resistor.
L
Second step:
We cant use loop or node methods, so lets use super position method to calculate I
N
To apply super position method we will remove all circuits one by one i.e. after removing
voltage source we will replace it with short circuit and current source with open circuit.
Hint: Don’t remove all circuits simultaneously.
Only current source is acting.

Due to short circuit all current will follow through the short circuit so
I    = 2mA --------------(A)
N1
Only voltage source is acting.

4k is in parallel with 2k resistor which in return in series with 2k resistor .So
total resistance.
R = (4k||2k) +2k
= 8/6 k    +2k
R = 3.33 k
So
I    = 6/3.33k
N2
I    = 1.80mA -------------- (B)
N2

Virtual University of Pakistan                                                                      Page 120
PHY301 – Circuit Theory

total I       from both sources so from equation A and B we have
N
I     =I   +I
N N1 N2
= 2mA + 1.80mA
= 3.80mA

Third step: Calculating R
N
To calculate R            we will short circuit all voltage sources and open circuit all current
N
sources.

4k is in parallel with 2k.The combined effect of these is in series with 2k.
4k||2k + 2k = 1.33 + 2k
=3.33k
=R
N

Fourth Step:
After calculating I           and R , re-inserting the load resistance R in the circuit in parallel R and
N          N                                     L                          N
considering the I            current source parallel with these two resistances. So our Norton
N
equivalent circuit will be.

by current divider rule we have
I = ( 3.80m)(3.33) x 1/9.33k
0
= 1.356 mA
By ohm’s Law we have
V = 6k x 1.35 = 8.143 volts
0

Example: Calculate the current I by using Norton’s theorem .
o

Solution:
We want to calculate I by using Norton’s theorem.we will follow these steps .
0
First step: Replacing R with a short circuit to find I Here R is 6k resistor.
L                             N.       L

Virtual University of Pakistan                                                                                     Page 121
PHY301 – Circuit Theory

Second step: Calculate current IN

From loop 2, we can write
I =2mA
2
For loop 1
4k(I -I )+2k(I -I )-6=0
1 N        1 2
4kI -4kI +2kI -2kI -6=0
1    N       1    2
6kI -4kI -2kI =6
1     N     2
Putting the value of I
2
6kI -4kI -2k(2m) = 6
1    N
6kI -4kI = 10
1    N
For Loop 3
4k(I -I )+4k(I -I )=0
N 2        N 1
4kI -4kI +4kI -4kI =0
N     2     N    1
-4kI +8kI =8
1    N
Solving equations for loops 1 and 3
12kI -8kI =20
1    N
-12kI +24kI =24
1       N

16kI =44
N
I = 2.75 mA
N
Third step: Calculating R
N
To calculate R         we will short circuit all voltage sources and open the current sources.
N

For R
N
2kII4k+4k =((2kx4k)/(2k+4k))+4k
R = 5.33k
N

Fourth Step:
After calculating I       and R , re-inserting the load resistance R in the circuit in parallel R and
N         N                                     L                          N
considering the I        current source parallel with these two resistances.
N

Virtual University of Pakistan                                                                              Page 122
PHY301 – Circuit Theory

I = (2.75m) (5.33k)x1/(6+5.33)k)
o
I =1.29mA
o
Linearity Principle
In this technique we assume the unknown quantity and analyze the circuit in reverse manner. Until we
reach the source which is producing all voltages or currents and calculate its value.
Now by comparing this value with the original value of the source we calculate the exact unknown
quantity of the circuit.

Example: Calculate V by linearity principle.
o

Solution:
We want to calculate V by linearity principle. Let us assume that V is 1 Volt.
o                                            o

Our assumption is V =1V
o
Therefore, V2=1V
Hence I2=1/2k=0.5mA
Also, V =(0.5m)(4k)=2V
4k
Now, V =V +V = 2+1= 3V
1 4k 2
Then, I =3/3k=1ma
1
Also, I =I +I =0.5m+1m=1.5mA
s 2 1
V2k=Is(2k)=2k(1.5m)=3V
Vs=V2k+V1=3+3=6V
When Vo is 1, source voltage is 6V, but original source voltage is 12 V, hence output voltage will be 2
Volts.

Virtual University of Pakistan                                                                         Page 123

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