# Lecture 18 by naveedawan53

VIEWS: 6 PAGES: 5

• pg 1
```									PHY301 – Circuit Theory

Virtual University
PHY 301
LECTURE 18

Example:      Calculate the voltage V
o.

Solution:   We want to calculate the voltage V        The circuit can be redrawn as
o.

Now
Vx = (I - I ) 6k -------------(A)
2 3
I = Vx/2k ---------------(B)
1
Put Vx from A into B we have
I = (I – I )6k/2k
1    2 3
= 3(I – I )
2 3
Now
I = 2mA
2
I = 6mA - 3I
1           3
Now KVL equation for mesh 3
6kI +6k(I – I ) +2k(I -I ) =0
3     3 2          3 1
6kI + 6kI – 12 +2kI -2k(6mA -3I )=0
3     3         3            3
14kI – 12 -12 + 6kI = 0
3               3
20kI = 24
3
I = 1.2mA
3
V = 6kI
0      3
= 6k(1.2mA)
=7.2 volts

Virtual University of Pakistan                                                        Page 75
PHY301 – Circuit Theory

Example:     Calculate the voltage V
a.

Solution:   We want to calculate the voltage V The circuit can be redrawn as
a.

Va can also be found by super node technique but it will take a lengthy
calculation lets see how it becomes very easy with loop analysis.

Here
I =1A
4
KVL for mesh 1
1(I – I ) +3(I -3I )-5 =0
1 2        1 3
I – I +3I -3I -5 = 0
1 2       1 3
4I – 3I – I = 5 --------(A)
1    3 2

KVL for mesh 2
2I2 +3I2 -2Va+I2-I1 = 0
6I2 –I1 = 2Va------(B)
Va = 3I2
Put this value in (a)
6I2 –I1 = 3I2
I1 = 0
KVL for mesh 3

Virtual University of Pakistan                                                    Page 76
PHY301 – Circuit Theory

2Va + 5(I – I ) +3(I –I ) =0
3 4        3 1
Putting the values of I ,I and Va
1 4
6I + 3I -0 +5I – 5 = 0
2    3        3
8I + 6I = 5
3    2
Solving equation for mesh 1 and mesh 3
- 18I - 6I = 30
3    2
8I + 6I = 5
3    2

-10I     = 35
3
I = - 3.5A
3
Putting values of I and I in eq of mesh 1 i.e in (A) we have
3      1
10.5– I =5
2
I = 5.5 A
2
As                                  Va =3I
2
Therefore                Va =3 * 5.5
Va = 16.5 V
Example:       Find Current through all meshes.

Solution:

Incorrect approach
I =-5A
2
I =+5A
3
We drive all these
KVL equation for mesh 1
-5+I -I +2I =0 -------------------- (1)
1 3 1

Virtual University of Pakistan                                                   Page 77
PHY301 – Circuit Theory

KVL equation for super mesh
1(I -I )+4I +3I +2+4I - 4I +6I =0 ------------ (2)           KVL
3 1    2 2         3 4 3
equation for Mesh 4
4I – 4I +2I -3=0 ------------ (3)
4    3 4
Simplifying
Equation (1)                           3I - I = 5
1 3
Equation (2)                -I +7I +11I -4I =-2
1 2      3 4
Equation (3)                        - 4I + 6I =3
3      4
Also                       -I +I =5
2 3
Therefore          I =2.481A
1
I =-2.556A
2
I =2.444A
3
I =2.130A
4
Example:    Calculate current I .
0

Solution: We want to calculate the current I The circuit can be redrawn as
0.

Here
I = -1mA
1
I = -2mA
2
KVL equation for loop 3
1k( I + I ) + 2kI + 1k(I + I ) = 2 + 4
3 1         3       3 2
1kI + 1kI + 2kI + 1kI + 1kI = 6
3     1      3      3     2
4kI – 1 – 2 = 6
3
4kI – 3 = 6
3
I = 9/4 mA
3
I = 2.25 mA
3
I = I = 2.25 mA
3 0

Virtual University of Pakistan                                                   Page 78
PHY301 – Circuit Theory

Example:     Calculate current I .
0

Solution: We want to calculate the current I The circuit can be redrawn as
0.

Here
I = 2mA
1
I = - 4mA
2
KVL for loop 3
2kI + 2k(I + I ) = 12
3      3 2
2kI + 2kI + 2kI = 12
3     3     2
4kI – 8 =12
3
4kI = 20
3
I = 5mA
3
Now
I =I +I
0 3 1
I = 5mA + 2mA
0
= 7mA

Virtual University of Pakistan                                               Page 79

```
To top