# Lecture 20

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```					PHY301 – Circuit Theory

Virtual University
PHY 301
LECTURE 20

Example:      Calculate the voltage V
o.

Solution:   we want to calculate the voltage V        The circuit can be redrawn as
0.

Here
I = - 2mA
1
KVL equation for loop 2
2kI + 4kI + 2k(I + I ) = 12
2     2       2 1
6kI + 2kI + 2kI = 12
2      2     1
8kI – 4 = 12
2
8kI = 16
2
I = 2mA
2
Now
V = 2k(I )
o     2
= 2k (2m)
V = 4 volts
o
Example:      Calculate the voltage V
o.

Solution:   We want to find the voltage V The circuit can be redrawn as
0.

Here
I =3mA
1
I = 1mA
2

Virtual University of Pakistan                                                        Page 83
PHY301 – Circuit Theory

KVL for loop 3
4kI + 2k(I +I ) + 4k(I –I ) +2k(I – I ) =6
3      3 2         3 1        3 1
4kI +2kI +2kI +4kI – 4kI +2kI -2kI =6
3    3      2    3      1     3     1
12kI +2 -12 – 6 =12
3
12kI =22
3
I =11/6mA
3
I =1.833mA
3
V = 4kI
0      3
= 4k x 1.833m
V = 7.33 volts
0
Example:       Calculate the current I
o.

Solution:    We want to find the current I        The circuit can be redrawn as
o.

Here
I =-2mA
1
I = 1mA
3
Now KVL for mesh 2
4k(I +I ) +2k(I – I ) +2k(I – I ) =0
2 4          2 3       2 1
8kI +4kI – 2 +4 =0
2    4
4kI +4kI = -1
2     4
Now KVL for loop 4
4k(I +I ) +2k(I +I ) +6kI -6 =0
4 2       4 3        4
12kI +4kI +2 -6 =0
4    2
12kI +4kI -4 =0
4    2
3kI +1kI =1
4     2
Multiplying equation of mesh2 by 3 and equation of loop 4 by 2 and        subtracting

12kI +6kI = -3
2     4
-2kI -6kI = -2
2    4

10kI =-5
2
I = -o.5mA
2
Now
I =I –I
0 1 2
I = -2 +0.5
0
so
I       = -1.5mA
0

Virtual University of Pakistan                                                                  Page 84
PHY301 – Circuit Theory

Example:         Find the currents I ,I and I .
1 2      3

Solution:   We want to find the currents I ,I and I . The circuit can be redrawn as
1 2      3.

For mesh 1
4I -4I +1 = 0 ...............(1)
1 2

For Super mesh (mesh 2 and mesh 3)
3I + 2I + 4I - 4I = 0 ...............(2)
2    3    2    1
Coupling Equation
I -I =2V
3 2       X
Here
V = 3I
X      2

Solving equation 1
4I - 4I = -1
1    2
Solving equation 2
- 4I + 7I + 2I = -1
1    2    3

For Mesh 3
-7I + I = 0 --------------(3)
2 3
Solving equation 1,2 and 3
I = - 308.8 mA
1
I = - 58.82 mA
2
I = - 411.8 mA
3

Virtual University of Pakistan                                                        Page 85

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