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PHY301 – Circuit Theory
Virtual University
PHY 301
LECTURE 18
Example:
Solution:
Calculate the voltage V
o.
We want to calculate the voltage V
o.
The circuit can be redrawn as
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
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PHY301 – Circuit Theory
Example:
Solution:
Calculate the voltage V
a.
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
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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
Example:
= 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
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
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PHY301 – Circuit Theory
KVL equation for super mesh
1(I -I )+4I +3I +2+4I - 4I +6I =0 ------------ (2)
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
KVL
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
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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
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