Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Voltage Regulation :VR % 
Voltage Regulation
VNL  VFL
 100%
VFL
Where
VNL = No load voltage
VFL = Full load voltage
Also we can write
VR % 
R int .
 100 %
RL
Where
Rint. = Internal resistor .
RL = load resistor .
Example :- Find the voltage VL and power lost to the internal resistance , if the
applied load is 13 Ω , also find the voltage regulation ?
Solution :IL 
E
30

 2A
Rint  RL 2  13
VL  E  I L Rint .  30  2 * 2  26V
Ploss  I L2 Rint .  2 .2  8W
2
or
VR % 
VNL  VFL
30  26
 100% 
 100%  15.385%
VFL
26
VR % 
Rint .
2
 100%   100%  15.385%
RL
13
-٤٢-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Example :- Find the current I1 , for the network shown:
I = 42 mA
I1
R2 =24Ω
R1 = 6Ω
R3 =24Ω
Solution :- All resistance in parallel , so if we define that R = R2 // R3 then :R
R2 R3
24 * 24

 12
R2  R3 24  24
Hence
I1  I


R
12
 42 * 10  3
 28mA
R  R1
12  6
Example :- Calculate I & V for the network shown
Solution :- We have a short circuit on R2 resistance , hence no current through
R2 , hence the above cct. Can redrawn as fellows:
-٤٣-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
E
E
18


 3.6mA
RT R1 5 * 103
I
V  I .R1  E  18V
Example :- For the following cct. Network , find RT , IA , IB , IC , VA , VB , I1 ,
I2 ?
Solution :-
RA 
R1R2
9*6

 3 .6 
R1  R2 9  6
RB = R3 + R4 // R5  4 
9*3
 6
63
RC = 3 Ω
RT = RA + RB // RC
 3 .6 
 IA 
6*3
 5 .6 
63
E 16.8

 3A
RT
5 .6
-٤٤-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Apply C.D.R.
IB 
I A RC
3*3

 1A
RB  RC 3  6
By K.C.L.
IC = IA – IB = 3 – 1 = 2 A
VA = IARA = 3 * 3.6 = 10.8 V
VB = IBRB = 1 * 6 = 6 V = VC
I1 
I A R2
6*3

 1 .2 A
R1  R2 6  9
I2 = IA – I1 = 3 – 1.2 = 1.8 A
To check
E – VA – VB = 0
16.8 – 10.8 – 6 = 0
0=0
Ok.
-٤٥-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Example :- Find the resistor required to connect in parallel with the ammeter to
flow 1.2 A , if you know that the fsd ( full scale deflection ) of ammeter is 120
mA , and the resistance of ammeter is 2.7 Ω ?
Rsh
Solution :-
Ish
From K.C.L.
I sh  1.2  0.12  1.08 A
I sh  I
RA
R A  Rsh
 2.7
1.08  1.2
 2.7  Rsh
1.08 
I = 1.2A



3.24
2.7  Rsh
3.24  1.082.7  Rsh 
3.24  2.916  1.08 Rsh
0.324  1.08 Rsh
 Rsh  0.3
* Another Solution:
V A  0.12  2.7  0.324 V
I sh  1.2  0.12  1.08 A
Rsh 
V A 0.324

 0.3
I sh
1.08
-٤٦-
0.12A
A
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Example :- for the following cct. Network , Given that (V= 24 v), Find E ?
4Ω
6Ω
Solution :I1 
24
 2A
1 .2
V
12Ω
16Ω
The same voltage ( V = 24 V ) on the
4Ω
resistor Ra = 4 +16 + 4 = 24 Ω
Hence
6Ω
I2 
24
 1A
24
8Ω
24Ω
 I 3  I1  I 2  3 A
Also from K.C.L. I4 = I3 = 3A
Take the closed loop ABCDA , from
8Ω
K.V.L.
V1 – 6I4 – V – 6I3 = 0
I3
6Ω
I2
I1
 V1  60V
V
12Ω
60
 2 .5 A
24
16Ω
D
I4
 I 6  I 5  I 4  2.5  3  5.5 A
4Ω
Take the closed loop CBC
6Ω
V1
– V1 – V2 + E – V3 = 0
E = V1 + V2 + V3
4Ω
A
V1 = 6 * 3 + 24 + 6 * 3
 I5 
E
B
24Ω
C
E = 60 + 5.5 * 8 + 5.5 * 8
V2
= 60 + 44 + 44
E = 148 V
8Ω
-٤٧-
8Ω
I5
V3
E
I6
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Current Source :Example :- Find the voltage ( Vs ) for the circuit below:
Solution :Vs = IRL = 10 * 2 = 20 V
if RL = 2 Ω
Vs = IRL = 10 * 5 = 50 V
if RL = 5 Ω
Example :- Calculate V1 , V2 , Vs for the following cct.:
Solution :V1 = IR1 = 5 * 2 = 20 V
V2 = IR2 = 5 * 3 = 15 V
Vs = V1 + V2 = 10 + 15 = 25 V
Source Conversions :A voltage source with voltage E and series resistor Rs can be replaced by
a current source with a current I and parallel resistor Rs as shown :-
I
E
Rs

Current source to voltage source
Voltage source to current sourc
-٤٨-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Example :- Convert the voltage source in the cct. Below to a current source,
then calculate the current through the load for each source:
Solution :-
IL 
E
6

 1A
Rs  RL 2  4

 For the current source cct.
IL  I
Rs
 2 
 3
  1A
Rs  RL
2 4
‫ ﻣﺗﺳﺎوي ﻓﻲ اﻟﺣﺎﻟﺗﯾن و ﻫذا‬IL ‫ﻻﺣظ ان‬
I
. ‫ﺻﺣﯾﺢ‬

E
Rs
6
 3A
2
Example :- Convert the current source in the cct. Shown below to a voltage
source and determine IL for each cct.:
Solution : For the current cct.
IL  I

Rs
 9 * 10 3
Rs  RL
 3 *103 *106 *10
3

3
3



 I L  3mA
 For the voltage source cct.
IL 
E
E
27


RT Rs  R L 3  6 *10 3
 I L  3mA
-٤٩-

Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Current source in parallel :-

Is = 10 – 6 = 4 A &
Rs = 3 Ω // 6 Ω = 2 Ω
Example :-

Is = 7 – 3 + 4 = 8 A
-٥٠-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (5))
Example :- Find the load current in the following cct.:
Solution :-


I1 
E 32

 4A
R1
8
Is = I1 + I2 = 4 + 6 = 10 A
Rs = R1 // R2 
8 * 24
Rs
10 * 6
 6  I L  I s

 3A
8  24
Rs  RL 6  14
-٥١-