# Lecture 26

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```					PHY301 – Circuit Theory

Virtual University
PHY 301
LECTURE 26
Example: Calculate the voltage V by using Thevenin’s theorem .
o

Solution:
We want to calculate V by using Thevenin’s theorem .we will follow the
0
following steps .

First step:   Removing R
L

Here R is 4k resistor at which we want to calculate the voltage V
L                                                           o.
Second step: Calculating V
th

We want to calculate V        . Apply KVL in two loops to calculate the individual loop
th
currents.

Here
I = 2mA
1
KVL for the super mesh
-6 +6kI +12k(I +I ) + 12k(I +I ) =0
2      1 2          1 2
-6 +6kI +12kI + 12kI +24kI =0
2     2       2      1
30kI – 6 +24k(2m)=0
2
30KI -6 +48 =0
2

I = -42/30mA
2
I = -1.4mA
2
Now we will calculate the voltage across 6k and 12 k to find the value of V th
Therefore,
V = 6kI
6k      2
= 6k(-1.4m)
V = -8.4volts
6k
V      = 12k(I1 +I2)
12k

Virtual University of Pakistan                                                                             Page 107
PHY301 – Circuit Theory

= 12k( 2m -1.4m)
V    = 7.2volts
12k
So,
V = V6k + V12k
th
= -8.4 +7.2
V = -1.2volts
th

Third step: Calculating R
th

6k is in series with 12k .The resultant of these two is in parallel with 12k.
(6k+12k)||12k = 18k x12k/(12k+18k)
R = 7.2k
th
Fourth step: Calculating the unknown quantity.
After calculating V and R , re-inserting the load resistance R in the circuit in series with
th      th                                      L
R and considering the V as a battery in series with these two resistances.
th                        th

V = -1.2 x4k/(7.2k + 4k)
0
V = - 0 .4285 volts
0

Example:     Calculate the value of RL and the maximum power dissipation across it by Thevenin’s
Theorem.

Solution:
We want to calculate R and the maximum power across it by using Thevenin’s theorem .we will follow
L
the steps given earlier.

First step: Removing R
L

To remove R to calculate V .In this case our R is our R .
L              th                  th         L
Second step: Calculating V
th
We want to calculate V . Apply KVL in two loops to calculate the individual loop
th
currents.

Virtual University of Pakistan                                                                      Page 108
PHY301 – Circuit Theory

Here
I = 2mA
1
Applying KVL to loop 2
6kI + 3k(I – I ) +3 =0
2       2 1
9kI -3kI + 3 =0
2    1
9kI – 6 +3 =0
2
I = 1/3 mA
2
I = 0.33mA
2
Now we will calculate the voltage across 4k and 6k to find the value of V th
Therefore,
V     = 4k(I )
AB         1
= 4k( 2m)
V     = 8 volts
AB
V     = 6k(I )
BC         2
= 6k x 0.33m
V     = 2 volts
BC
So
V =V         +V
th     AB     BC
= 8 +2
V = 10 volts
th
Third step: Calculating R
th

To calculate R       Short circuiting the voltage source and open circuiting the current source.
th .
3k is in parallel with 6k and 4k is series with these two so
3k||6k + 4k = 3k x 6k/(3k + 6k) +4k
= 2k +4k
= 6k = R =R
th   L
R =R = 6k
th    L
Fourth step: Calculating unknown quantity.
After calculating V and R , re-inserting the load resistance R in the circuit in series with
th       th                                    L
R and considering the V as a battery in series with these two resistances.
th                          th

For maximum power dissipation
R =R = 6k
L    th
2
P = I R ------------- (A)
L
From Ohm’s law V=IR and I = V/R
so              I =10/12 put in A we have
2
P = (10/12k) (6k)
L
2
= (0.83) X 6
P = 4.1 mW
L

Virtual University of Pakistan                                                                          Page 109
PHY301 – Circuit Theory

Example:     Calculate the value of RL and the maximum power dissipation across it by Thevenin’s
Theorem.

Solution:
We want to calculate R and the maximum power across it by using Thevenin’s theorem .we will follow
L
the following steps .
First step: Removing R
L
To remove R to calculate V .In this case our R is our R .
L             th                  th       L

Second step: Calculating V
th

Apply KVL to the circuit
For loop1
-12 +6I + 6(I - I ) = 0
1      1 2
-12 +12I – 6I = 0
1    2
2I – I = 2 --------------- (A)
1 2

For loop 2

12I – 6I +3 =0
2    1
4I – 2I +1 =0
2    1
2I = 4I
1     2
put in equation (A) we have
4kI +1 –1kI =2
2         2
3kI =1
2
I = 0.33mA
2
from equation (A)
I = (I2 +2)/2
1
= 0.33+2/2
I = 1.166mA
1
Now voltage across 6k resistor
V      = 6kI
R6k        1
= 6k(1.166m)
V      = 7volts
R6k
V = 7V +3V
th
V =10 volts
th

Virtual University of Pakistan                                                                  Page 110
PHY301 – Circuit Theory

Third step: Calculating R
th

We want to calculate R 6k is in parallel with 6k .The resultant is again parallel with third 6k resistor.
th
6k||6k = (6k x 6k)/6k + 6k
= 3k
3k||6k = 6k x 3k/(6k + 3k)
= 2k
So
R =R = 2k
th    L
Fourth step: Calculating unknown quantity.
After calculating V and R , re-inserting the load resistance R in the circuit in series with
th       th                                   L
R and considering the V as a battery in series with these resistances.
th                         th

V      = 10 x2k/4k
RL
= 5 volts
To calculate the power dissipation we have
2
P = V /R = 25/2k
L
P = 12.5mW
L

THEVENIN’S THEOREM AND DEPENDENT SOURCES:
Working with dependent sources is different from working with independent sources while applying
Thevenin’s theorem .
While calculating Rth we can simply open circuit current sources and short circuit voltage sources.
Because the voltage or current of the dependent sources is dependent on the voltage or current of these
independent spources.
While calculating R we will short circuit the open terminals of the Thevenin circuit and will calculate the I
th                                                                                        sc
and then divide V with I to calculate R .
th     sc                th
Example: Calculate the voltage V by using Thevenin’s theorem .
o

Solution:
We want to calculate V by using Thevenin’s theorem .we will follow the following steps .
0
First step:  Removing R
L

Here R is 6k resistor at which we want to calculate the voltage V
L                                                          o.

Virtual University of Pakistan                                                                           Page 111
PHY301 – Circuit Theory

Second step: Calculating V
th
Voltage across 4k resistor
V = (4k/6k) 12
4k
V = 8volts
4k
Voltage across 2k resistor
V = (2k/6k) 12
2k
V = V = 4volts
A      2k
V = 8 – 4V
th            A
= 8 – 4(4)
= 8 – 16
V = - 8 volts
th
Third step:
When we use dependent sources we will use the following technique .
In the above circuit output which is opened after removing the load resistance now we will
replace this open circuit with short circuit. As in the circuit below.

We will first find I        to calculate R .
sc                 th
Here
V = 2kI
A     1
Applying KVL to loop 1
-12 +2kI +(I – I )4k= 0
1 1 sc
-12 +2kI +4kI – 4kI = 0
1      1        sc
6kI -4kI = 12
1      sc
For other loop
(I –I )4k +4V = 0
sc 1         A
4kI -4I +4(2kI )=0
sc 1         1
4kI -4I +8I =0
sc 1      1
I = -I
1     sc
Putting in equation for loop1
3I -2I = 6
1 sc
3(-I ) -2I = 6
sc     sc
I = - 6/5 mA
sc
So
Rth = Vth/Isc
= -8/(-6/5 m)
Rth = 6.67k
Fourth step:
After calculating V and R , re-inserting the load resistance R in the circuit in series with
th        th                               L
R and considering the V as a battery in series with these resistances.
th                        th

So
V = (6k/6k +6.67k) x8
0
= 6k/12.67k x8
V = 3.78 volts
0

Virtual University of Pakistan                                                                       Page 112

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