# Lecture 16 by naveedawan53

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```									PHY301 – Circuit Theory

Virtual University
PHY 301
LECTURE 16

Example:        Calculate the voltage Vac .

Solution:          We want to calculate the voltage Vac. To solve this problem the circuit        can
be redrawn as

Let the current I is flowing through the circuit. The KVL equation will be
1
10kI + 20kI + 30KI -6 = 0
1         1       1
10kI + 20kI + 30KI = 6
1       1         1
60kI = 6
1
I = 6/60k
1
I = 0.1mA
1
Now Vac will be equal to
Vac = (10k + 20k)I
1
= 30k(0.1mA)
Vac = 3 Volts
Example:       Calculate the voltage Vac .

Virtual University of Pakistan                                                               Page 67
PHY301 – Circuit Theory

Solution:   We want to calculate the voltage V .First we will have to calculate the voltage across 40k resistor.
bd
Let the current I be flowing through the loop.
1
Applying KVL
10kI + 9 + 40kI + 10kI – 6 = 0
1         1        1
60kI = -3
1
I = - 3/60k
1
= - 0.05 mA
So, voltage across 40k resistor
V      = (-0.05m)(40k)
40k
= - 2 volts ---------- (A)
We want to want to calculate Vbd we will redraw the circuit as

Now to calculate V     we take the path bdcb
bd
Applying KVL
V -V        -9 =0
bd    40k
by putting the value of V     from equation (A) we have
40k
V – (–2) –- 9 =0
bd
V +2 –- 9 =0
bd
V = 7 Volts
bd

Example:        Calculate the current I .
0

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

Virtual University of Pakistan                                                                        Page 68
PHY301 – Circuit Theory

KVL for mesh 1
Here
I = 120 mA
1
Applying KVL to mesh 2
8kI + 4kI + 4k(I – I ) = 0
2     2      2 1
16kI – 4k I = 0
2      1
16kI = 480
2
I = 480/16k
2
I = 30 mA
2
So I will be
0
I =I –I
0   1 2
I = 120 – 30
0
I = 90 mA
0

Example:       Calculate the current I and the voltage V .
0                 0

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

Applying KVL to mesh 1
2kI +6k(I – I )= 12
1      1 2
8kI – 6kI = 12
1     2

Applying KVL at mesh 2
8kI + 4kI + 6k (I – I ) = 0
2     2       2 1
18kI - 6kI = 0
2      1

Virtual University of Pakistan                                                        Page 69
PHY301 – Circuit Theory

Solving equations of mesh 1 and mesh 2
24kI  - 18kI = 36
1       2
-24kI + 72kI = 0
1       2
54kI = 36
2
I = 36/54k
2
I = 0.67mA ------------(A)
2
Now V       will be equal to
0
V = 4k I
0       2
by putting the value of I2 from equation (A) we have
V = 2.66 Volts
0
Example:        Calculate the current I .
0

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

Here
I = - 2mA
1
Negative value of I is due to this reason because the direction of I is going    opposite to the
1                                                1
Independent current source.
I = 4mA
3
KVL for mesh 2
4kI + 6k (I – I ) + 2k(I – I ) = 12
2        2 3          2 1
12kI – 24 + 4 =12
2
I = 32/12k
2
= 8/3 mA
= 2.66 mA
I = I = 2.66mA
0 2
Example:       Calculate the current I .
0

Virtual University of Pakistan                                                                        Page 70
PHY301 – Circuit Theory

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

Here
I = - 2mA
1
I = 4mA
3
Apply KVL on the mesh 2
2kI + 1k(I – I ) = 12
2      2 1
3KI + 2 =12
2
I = 10/3 mA
2
I = 3.33mA
2
Now
I =I –I
0 1 2
= - 2 – 3.33
I = - 5.33 mA
0

Virtual University of Pakistan                                                      Page 71

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