Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
Voltage divider Rule :-
RT = R1 + R2
I
E
RT
 E 
E.R1
V1  I .R1   .R1 
RT
 RT 
E
E.R2
V2  I .R2   .R2 
RT
 R2 
Vn 
ERn
RT
Voltage divider rule
Vn = Voltage across Rn
E = The ( emf ) voltage across the series elements .
RT = The total resistance of the series circuits .
Example :- Using voltage divider rule , determine the voltage V1 , V2 , V3 and
V4 for the series circuit in figure below , given that ; R1 = 2KΩ , R2 = 5KΩ ,
R3 = 8KΩ , E = 45 V ?
Solution :-
V1 
R1 E 2 *10 3 * 45

 6V
RT
15 *10 3
-٢٩-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
R2 E 5 *103 * 45
V2 

 15V
RT
15 *103
R3 E 8 *103 * 45
V3 

 24V
RT
15 *103
V4
R  R2 E
 1
RT
To check:
7 * 103 * 45

 21V
15 * 103
or V4 = V1 + V2 = 21V
E – V1 – V2 – V3 = 0
E = V1 + V2 + V3  45 = 6 + 15 + 24
45 = 45
Active Potential :‫رﻣوز اﻻرﺿﻲ‬
Va = 14 V a
Vab
Vab = 6 V
is the voltage difference between the
point a and point b
Vab = Va – Vb = 14 – 8 = 6V
Vb = 8 V
Vba = Vb – Va = 8 – 14 = - 6V
Vab = - Vba
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b
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
Va = -7 V
Va = 10 V
Va = 0 V
E = 20 V
R1

R1
20 V
R2
R2
E = -12 V
R1
R1
-12 V

R2
R2
-٣١-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))


Example :- Find Va , Vb , Vc , Vab , Vac and Vbc for the following diagram .
Solution :-
-٣٢-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
RT = R1 + R2 + R3
= 2 + 5 + 3 = 10 Ω
I
E 10

 1A
RT 10
E – V2 – Va = 0
Va = E – V2 = 10 – (2 * 1) = 8 V
Vb = V5 = (1 * 5) = 5 V = Vbc
or
E – V2 – V3 – Vb = 0
; Vc = 0 V
 Vb = E – V2 – V3 = 10 – 2 – 3 = 5 V
Vab = Va – Vb = 8 – 5 = 3 V
Vac = Va – Vc = 8 – 0 = 8 V
Vbc = Vb – Vc = 5 – 0 = 5 V
Equivalence of actual sources :-
Open
Voc = E
Circuit
I=0
Short
I sc 
circuit
Voltage
Current
Source
Source
Voc  I o
E
Ro
1
Go
Isc = Io
V=0
Kirchoff's Current Law ( K.C.L. ) :The algebraic sum of ingoing currents is equal to the out going currents
at any point .
I
in
  I out
Or , At any point , the algebraic sum of entering and leaving current is zero .
-٣٣-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
I  0
I2
I3
I1
I4
I5
I1 + I2 + I4 = I3 + I5
Or
I1 + I2 + I4 - I3 - I5 = 0
a
b
At a
Or
I1 = I2 + I3
13 + 5 – I = 0
I1 - I2 - I3 = 0
18 – I = 0
 I = 18 A
At b
-I1 + I2 + I3 = 0
-٣٤-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
Example :- Find the current in each section in the cct. Shown ?
Solution :At node a
3 – 1 – Iab = 0
2 – Iab = 0  Iab = 2 A
At node b
Iab + 3 – Ibc = 0
2 + 3 – Ibc = 0  Ibc = 5 A
At node c
Ibc + 4 – Icd = 0
5 + 4 – Icd = 0  Icb = 9 A
At node d
Icd – 8 – Ide = 0
9 – 8 – Ide = 0  Ide = 1 A
At node e
2 – 3 + Ide = 0
2–3+1=0
0 = 0 check .
-٣٥-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
Example :- Find the magnitude and direction of the currents I3 , I4 , I6 , I7 in the
following cct. Diagram?
=
12
A
I5
I2
=
8A
Solution :-
I
enter
  I leave
 I1 = I7 = 10 A
At node a ; suppose I3 is entering
I1 + I3 – I2 = 0
10 + I3 – 12 = 0  I3 = 2 A
At node b;
I2 enter , I5 leave ,  I4 must be leaving
I2 = I5 + I4
12 = 8 + I4  I4 = 12 – 8 = 4 A
At node c;
I4 enter , I3 leave ,  I6 leave
I4 = I3 + I6
4 = 2 + I6  I6 = 2 A
At node d;
I5 and I6 enter , I7 leave
-٣٦-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
I7 = I5 + I6
10 = 8 + 2
10 = 10
Ok.
Resisters in Parallel :-
From K.V.L.
V = V1 = V2
From K.C.L.
I = I1 + I2
I
From Ω.L.
V1 V2

R1 R2
= V1G1 + V2G2
= V1 ( G1 + G2 )
or
I = V ( G1 + G2 )
I = VGT
Where
Hence
or
GT = G1 + G2
R  R2
1
1
1


 1
RT
R1
R2
R 1 .R 2
RT 
R1.R2
R1  R2
In the same minner , if we have three resistors
in parallel , then:
1
1
1
1



RT
R1
R2
R3
-٣٧-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
1
R .R  R1 .R 3  R1 .R 2
 2 3
RT
R1 .R 2 .R 3
RT 
R1 .R 2 .R 3
R 2 .R 3  R1 .R 3  R1 .R 2
And , if we have N of parallel resistance , then
1
1
1
1
1




RT
R1 R 2 R 3
RN
Also
PT = P1 + P2 + P3
P1  V1 I 1  I 12 R1 
V12
R1
Source power Ps  EI T  I T2 R T 
E T2
RT
Example :- For the following cct. Find RT , PT , IT , Ib?
Solution :RT 
IT 
‫ﻓﻲ ﺣﺎﻟﺔ ﻛون ﻗﯾم اﻟﻣﻘﺎوﻣﺎت ﻣﺗﺳﺎوﯾﺔ‬
R 8
  2
N 4
E 16

 8A
RT
2
-٣٨-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
I branch 
E 16

 2A
R1
8
PT  IT2 RT  8 .2   128W
2
or
PT = E.IT = 16 * 8 = 128W
or
PT = P1 + P2 + P3 + P4
 2  * 8  2  * 8  2  * 8  2  * 8
2
2
2
2
 32  32  32  32  128W
Example :- For the parallel network in fig. below , find :a) R3 , b) E , c) IT , I2 , d) P2 ; given that RT = 4 Ω ?
Solution :a)
1
1
1
1



RT R1 R2 R3
1 1
1
1



4 10 20 R3
0.25  0.1  0.05 
1
R3
0.25  0.1  0.05 
1
R3
0 .1 
b)
1
1
 R3 
 10
R3
0 .1
E = V1 = I1R1 = 4 * 10 = 40 V
-٣٩-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
c)
IT 
E 40

 10 A
RT
4
I2 
V2
E 40


 2A
R2 R2 20
d)
P2  I 22 R2  2 .20   80W
or
V22
P2 
R2
2
, or P2 = I2V2
Current division Rule :I
R .R
V I 1 2
R1  R2
V
I1 

R1
 I1  I
I
V
R1 .R2
R1  R2
R1
R2
R1  R2
In the same miner
I2  I
R1
R1  R2
R2
I
R
R R
Also 1  1 2  2
I 2 I R1
R1
R1  R2
I

I1 R2 G1


I 2 R1 G2
-٤٠-
I1
I2
R1
R2
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
Example :- For the following circut. , find V , I1 and I2?
5A
I1
V
100Ω
I2
0.1Ω
Solution :RT 
R1.R2
100 * 0.1
10


 0.0999
R1  R2 100  0.1 100.1
V = I . RT = 5 * 0.0999 = 0.4995 V
I1 
V
 0.004995 A
100
I2 
V
 4.995 A
0 .1
To check
I = I 1 + I2
5 = 0.004995 + 4.995
5 = 5 Ok.
Example :- Determine the resistance R1 in the figure below?
Solution :I = I 1 + I2
-٤١-
Dr. Sameir Abd Alkhalik Aziez
University of Technology Lecture (4))
or
I2 = I – I1 = 27 – 21 = 6 mA
V2 = I2R2 = 6 * 10-3 * 7 = 42 mV
V1 = V2 = 42 mV
V1 42 * 10 3
R1 

 2
I 1 21 * 10 3
or
I1  I
R2
27 * 103 * 7
 21 * 10 3 
R1  R2
R1  7
R1  2
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