# Fall 2011_PHY301_3_SOL

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Past assignments, CS201, CS301, CS304, CS601, Pak301, Eng201, PHY301

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```							Assignment 3 (Fall 2011)
(Solution)
Circuit Theory (PHY301)
Marks: 30

Q.1:
Using the source transformation method, find Vo in the network given below. Label
and draw each step where it required. Write each step of the calculation to get maximum
marks and also mention the units of each derived value.

Solution:

As 4k resistance is in series with 12v, So
12
 3mA
4
3mA and 1mA are parallel to each other. So, it can be converted a current source having
the value:
3mA – 1mA = 2mA

As 2mA and 4k  are parallel to each other so it can be converted to the voltage source
having the value:
4k 2mA  8Volts
And the resistance will become in series to this source.

4k  and 2k  are in series so we can add them:
4k  +2k  = 6k 

Applying voltage division rule:
R1
Vo             V1
R1  R2
6
Vo     8
12
 4volts

Q.2:
Find VO in the network given below using Thévenin’s theorem.
Show each step of calculation otherwise you will lose your marks. Draw and label the
circuit diagram of each step and also mention the units of each derived value.

Sol.

Using Nodal analysis
At Node I:
V1/16 + (V1-V2)/4 -3=0
5V1 -4V2= 48 --------- (A)

At Node II:
V2 – V1/4 + (V2 -12)/5 =0
5V1 - 9V2= 48 --------- (B)
Solving A and B we have
VTh=V2= 19.2
THEVENIN’S EQUIVALENT:

Vo = 15X19.2/(15+6)
= 13.7V

SECOND METHOD
Q.NO.2

We want to calculate VO by using Thévenin’s theorem. We will follow the
steps given below:

First step: Removing RL
Here RL is 10 resistor at which we want to calculate the voltage VO

Second step: Calculating Vth

Applying KVL to the circuit
Here     I1  3 A

For loop 2
4I 2  5I 2  12  16( I 2  I1 )  0
4I 2  5I 2  12  16I 2  16I1 )  0
25I 2  16I1  12  0                      I1  3 A
25I 2  16(3 A)  12  0
25I 2  48  12  0
25I 2  36  0
25I 2  36
36
I2 
25
I 2  1.44 A

Now voltage across 5 resistor
V5  I 2  5
 1.44 A  5
 7.2v
So
Vth  V5  12v
 7.2v  12v
 19.2v

Third step: Calculating Rth

To calculate Rth short circuiting the voltage source and open circuiting the current
source. The modified circuit will be as:

16 is in series with 4  16  4  20

20 5 100
20 5                   4
20  5 25

4 is in series with 2 4  2  6

So Rth  6
Fourth step: Calculating the unknown quantity.
After calculating Vth and Rth, re-inserting the load resistance RL in the circuit in
series with Rth and considering the Vth as a battery in series with these two
resistances.

15
Vo             19.2
15  6
 13.7v

3rd method

Solution:
Thévenin’s theorem involves four steps. Details for these steps are here under.
Step 1:
Removing load resister RL of 15 . so we have the circuit as

Step 2:
Calculating the Vth using the source transformation method.
In the circuit diagram given above, there is a 3A current source connected parallel with
a 16 resister. So it can be change to a voltage source with a 16 resister in series as.
V  (3 A)(16)
V  48V .
So, the circuit can be re-drawn as given
In the circuit diagram given above, 16 resister and 4 resister are in series. Their
combine effect will be
16  4  20
So, the circuit can be re-drawn as given

There is a 48V voltage source connected series with a 20 resistor. So it can be change to
a current source as.
48V
I
20
I  2.4 A.
So, the circuit can be re-drawn as given

There is a 12V voltage source connected series with a 5 resistor. So it can be change to
a current source as.
12V
I
5
I  2.4 A.
So, the circuit can be re-drawn as given
In the circuit diagram given above the combine current source can be found as
I  2.4 A  2.4 A  4.8 A.
So, the circuit can be re-drawn as given

In the circuit diagram given above, 20 resister and 5 resister are parallel. Their
combine effect will be
(20 5) 100
20 5                           4
20  5      25
So, the circuit will be re-drawn as given

In the circuit diagram given above, there is a 4.8A current source connected parallel with
a 4 resister. So it can be change to a voltage source with a 4 resister in series as.
V  (4.8 A)(4)
V  19.2V .
So, the circuit can be re-drawn as given
In the circuit diagram given above, 4 resister and 2 resister are in series. Their
combine effect will be
4  2  6
So, the circuit can be re-drawn as given

In the circuit diagram given above we have
Voltage drop across 6 will be zero.
So,
Vth  19.2V
Thus step 2 has been completed.

Step 3:
Now we have to calculating the Rth by open circuiting all current sources and short
circuiting all voltage sources in the given circuit.

So, the circuit can be re-drawn as given
In the circuit given above 16 & 4 are in series.
So their sum will be
16  4  20
The circuit can be re-drawn as given

In the circuit given above 20 4
So,
(20 5) 100
20 5                          4
20  5     25
The circuit can be re-drawn as given

In the circuit given above 4 & 2 are in series.
So their sum will be
4  2  6
The circuit can be re-drawn as given
Step 4:
Replacing RL and calculating VO
The circuit can be re-drawn as given

By voltage division rule we have
15
VO            19.2V
6  15
15
VO       19.2V
21
VO  13.71V

Q.3:
Find IO in the network given below using Norton’s theorem.
Show each step of calculation otherwise you will lose your marks. Draw and label the
circuit diagram of each step and also mention the units of each derived value.
Solution:
We want to calculate I0 by using Norton’s theorem.we will follow these steps .
First step: Replacing RL with a short circuit to find IN

Here RL is 2k resistor
Finding IN
We have to find Io through 2kΩ so according to Norton theorem we follow these steps

Converting the current source to voltage source

KVL for loop1
8k I1 -6=0
8k I1 =6
I1 =6/8
I = 0.75mA
1
KVL for loop 2
8k I2 +8=0
8k I2 =-8
I2 =-8/8
I2= -1mA
Now we see
IN = I1 - I2 so
IN = 0.75-(-1)
IN =1.75mA
Third step: Calculating R
N
To find RN short the voltage source and open the current source

4k and 4k in series and in parallel to 8k

RN = 8k || (4k+4k)
RN = 4kΩ
Fourth Step:
After calculating IN and RN, re-inserting the load resistance RL in the
circuit in parallel RN and considering the IN current source parallel with these two
resistances.

Using current division rule, Current through RL
Io= (1.75m x4k)/2K+4K
I0= 1.16mA
………………………………………………………………..

```
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