Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (3))
Example :- For the following circuit diagram , find the potential difference
between Node ( A & D ) , and Node ( A & F ) ?
A
6V
D
12V
8V
B
E
5V
C
F
Solution : To find the potential difference between Node A & D , we will apply
K.V.L. on the closed loop BADEB
 +6 + V – 12 – 8 = 0
V = 20 – 6 = 14 volt
or
+6 – V1 – 12 – 8 = 0
–14 – V1 = 0
 V1 = –14 volt
‫ أﻋﻠـﻰ ﻣـن ﺟﻬـد‬D ‫* ﻣﻌﻧﻰ ذﻟك ان ﺟﻬد ﻧﻘطـﺔ‬
. ‫ ﻓوﻟت‬١٤ ‫ ﺑـ‬A ‫ﻧﻘطﺔ‬
Take the loop FEDAF to find the potential difference between Node C & F .
 –5 + 12 – V – V2 = 0
–5 + 12 – 14 – V2 = 0  –7–V2 = 0
 V2 = –7 volt .
Or Take the loop FEBAF  –5 –8 +6 – V2 = 0  V2 = –7 volt .
-٢١-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (3))
Example :- For the following circuit diagram , find the current ?
Solution :
A
2Ω
B
2Ω
2Ω
C
2Ω
2Ω
V
I
8V
15V
2Ω
2Ω
F
2Ω
E
D
Take the loop FABCDEF
+8 + V – 15 = 0  +V – 7 = 0  V = +7 volt
V = IR  I 
7
 3 .5 A
2
Definitions :Node :- Meeting point of 3 or more branches .
Branch :- Series of elements carrying the same current .
Loop :- Is any closed path in a circuit .
-٢٢-
3Ω
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (3))
Hence for the loop circuit, we can find :A
4 nodes and 6 branches
E1
and we can find : V1 , V2 , V3 and V4
V2
as follows :-
D
Take the loop BACB ; to find V1
E2 – V1 – E3 = 0  V1 = E2 – E3
E2
V3
B
E3
V4
Or, if we take BCAB ;
E3 – V1 – E2 = 0  V1 = E2 – E3
C
Take the loop ADBA ; to find V2
–E1 + V2 + E2 = 0  V2 = E1 – E2
Take the loop ABCDA ; to find V3
–E2 + E3 + V3 + E1 = 0
V3 = E2 – E3 – E1
V4 = –V3
Or;
–E2 + E3 – V4 + E1 = 0
V4 = E1 + E3 – E2
Example :- For the following circuit diagram , find ; RT , I , V1 , V2 , P4Ω , P6Ω
, PE , verify by K.V.L. ?
Solution :RT = R1 + R2 = 4 + 6 = 10
-٢٣-
V1
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (3))
E
20

 2A
RT 10
I
V1  IR1  2  4  8V
V2  IR2  2  6  12V
P4   I 2 R1  2   4  16W
V12 8

 16 w
R1
4
2
; or
P4  
P6   I R2  2   6  24W
; or
V22 12 
P6  

 24 w
R2
6
PE  IE  2  20  40W
; or PE  P4  P6  16  24  40W
2
2
2
2
To verify results by using K.V.L. ; then
N
V
i 1
i
0
E – V1 – V2 = 0
E = V1 + V2
20 = 8 + 12
20 = 20
checks
Internal Resistance :Every practical voltage or current source has an internal resistance that
adversely affects the operation of the source.
In a practical voltage source the internal resistance represent as a resistor in series
with an ideal voltage source.
In a practical current source the internal resistance represent as a resistor in
parallel with an ideal current source, as shown in the following figures.
-٢٤-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (3))
Where
Ro = Internal resistance
RL = load resistance
According to K.V.L.
E – Vo – V = 0
E – IRo – V = 0
 V = E – IRo
Note that an ideal sources have Ro = 0
We can representing a load as a group of parallel resistances.
Hence as the load will increase the current will be increase ( because the
resistance will decrease ) and the voltage will decrease .
This is because the drop voltage due to the internal resistance , as shown in the
following figure :-٢٥-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (3))
As seen
from the above
figure , if Ro2 > Ro1 , then V2 < V1 and the drop voltage will be ( E – V2 ) ,
which is greater than ( E –V1 ) .
Example :- For the following circuit diagram , calculate I and VL for the
following cases :a) Ro = 0 Ω
b) Ro = 8 Ω
c) Ro = 16 Ω
Solution :a.) By apply K.V.L.
E – Vo – V = 0
120 – IRo – IRL = 0
 120 – 0 – 22I = 0
120 = 22I  I = 5.46 A
VL = I RL = 5.46 * 22 = 120 V
b.) E – Vo – V = 0
120 – 8I – 22I = 0
 120 – 30I = 0
-٢٦-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (3))
120 = 30I  I = 4 A
VL = I RL = 4 * 22 = 88 V
c.) E – Vo – V = 0
120 – 16I – 22I = 0
 120 – 38I = 0
120 = 38I  I = 3.16 A
VL = I RL = 3.16 * 22 = 69.5 V
Then we can conclude that as Ro increase the total current and load voltage will
decrease.
Example :- A circuit have load one with 20 Ω and 4A , and load two with 10 Ω
& 6A . Find the current for load three which have 30 Ω ?
I
Solution :1) 20 Ω & 4A
Ro
2) 10 Ω & 6A
E
3) 30 Ω & I = ?
From K.V.L. , then
E – Vo – VL = 0
VL = E – IRo
IRL = E – IRo
4 * 20 = E – 4Ro
80 = E – 4Ro
-----------------
(1)
-----------------
(2)
Also
6 * 10 = E – 6 Ro
60 = E – 6Ro
-٢٧-
Vo
VL
RL
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (3))
From eq. (1) & (2) , we have
20 = ( 6 – 4 ) Ro   Ro = 10 Ω ; sub. this result it in eq. (1) , then
80 = E – 4 * 10   E = 120 V
Now , we Apply K.V.L. for load 3 ;
V3 = E – IRo  30I = 120 – 10I
40I = 120   I = 3 A for load three.
See from this example that the current will increase as the load will decrease
with constant E & Ro .
Example :- A circuit have Voc = 25 v and Isc = 50 A , find its current and RL
when VL = 15 V ?
Solution :E = Voc = 25 V
Ro 
E
25

 0.5
I sc 50
From K.V.L.
E – Vo – VL = 0
E – 0.5I – 15 = 0
25 – 0.5I – 15 = 0
0.5I = 25 – 15
I 
RL 
10
 20 A
0 .5
VL 15

 0.75
I
20
-٢٨-