# Lecture 15 by naveedawan53

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```									PHY301 – Circuit Theory

Virtual University
PHY 301
LECTURE 15

Example:      Calculate the voltages Vad , Vac and Vbd.

Solution:         We want to calculate V        and V        . The circuit can be redrawn as

As seen form the figure there is only a voltage source between points a and d
so the voltage V will be 12 volts. There is no physical closed path between             point a and c we can
consider two closed paths acba and acda.

KVL equation for the path acba
V -4–6=0
ac
V = 10 Volts
ac
Second path in which only unknown value is V                is acda. KVL equation for
ac
this will be
V + 2 – 12 =0
ac
V = 10 Volts
ac
Note:                    Calculate Vbd yourself.
Example:      Calculate the voltages Vac and Vdb.

Solution:   We want to calculate Vac and Vdb. The circuit can be redrawn as

Virtual University of Pakistan                                                                            Page 65
PHY301 – Circuit Theory

To calculate V  , the two paths will be acda and acba.
ac
KVL equation for the path acda
V + 8 – 12 =0
ac
V = 4 volts
ac

KVL equation for the path acba will be
V + 6 – 10 = 0
ac
V = 4 volts
ac
Two paths for V    will be bdcb and bdab.
bd
KVL equation for bdcb
V -8 +6 = 0
bd
V = 2 volts
bd
V     = - 2 volts
db
KVL equation for the path bdab
V – 12 +10 =0
bd
V = 2 volts
bd
V    = - 2 volts
db

Example:     Calculate the voltage V .
0

Solution:  We want to calculate V0 .
Here
I = 2 mA
2
KVL for mesh 1
3kI + 6k (I – I ) -12 = 0
1         1 2
9kI – 6kI = 12
1      2
9kI – 12 =12
1
I = 8/3 mA
1
Voltage across 6k resistor
V = 6k( I – I )
1 2
= 6k(8/3 – 2
= 4 volts
As current through 2k resistor is
I = 2mA
2
Voltage across 2k = 2k x 2mA
= 4 volts
V = V–V
0             2k
=4–4
V = 0 volts
0

Virtual University of Pakistan                                            Page 66

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