Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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Voltage divider Rule :-

TR

RER

R

ERIV 2

22

22

...

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 :-

RT = R1 + R2

TR

EI

TT R

RER

R

ERIV 1

111

...

T

nn R

ERV

VR

ERV

T

610*15

45*10*23

31

1

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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VR

ERV

T

1510*15

45*10*53

32

2

VR

ERV

T

2410*15

45*10*83

33

3

V

R

ERRV

T

2110*15

45*10*73

321

4

or V4 = V1 + V2 = 21V

To check: E – V1 – V2 – V3 = 0

E = V1 + V2 + V3 45 = 6 + 15 + 24

45 = 45

Active Potential :-

رموز االرضي

a

b

Va = 14 V

Vb = 8 V

Vab = 6 VVab is the voltage difference between the

point a and point b

Vab = Va – Vb = 14 – 8 = 6V

Vba = Vb – Va = 8 – 14 = - 6V

Vab = - Vba

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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R1

R2

R1

R2

20 V

E = 20 V

R1

R2

R1

R2

-12 V

E = -12 V

Va = -7 V

Va = 10 V

Va = 0 V

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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Example :- Find Va , Vb , Vc , Vab , Vac and Vbc for the following diagram .

Solution :-

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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E – V2 – Va = 0

Va = E – V2 = 10 – (2 * 1) = 8 V

Vb = V5 = (1 * 5) = 5 V = Vbc ; Vc = 0 V

or E – V2 – V3 – Vb = 0 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 :-

Voltage

Source

Current

Source

Open

Circuit

Voc = E

I = 0 oooc G

IV1

Short

circuit osc R

EI

V = 0Isc = Io

Kirchoff's Current Law ( K.C.L. ) :-

The algebraic sum of ingoing currents is equal to the out going currents

at any point .

outin II

Or , At any point , the algebraic sum of entering and leaving current is zero .

RT = R1 + R2 + R3

= 2 + 5 + 3 = 10 Ω

AR

EI

T

110

10

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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0I

I1

I5

I2

I3

I4

I1 + I2 + I4 = I3 + I5

Or I1 + I2 + I4 - I3 - I5 = 0

At a

I1 = I2 + I3 13 + 5 – I = 0

Or I1 - I2 - I3 = 0 18 – I = 0

At b I = 18 A

-I1 + I2 + I3 = 0

b

a

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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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 .

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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Example :- Find the magnitude and direction of the currents I3 , I4 , I6 , I7 in the

following cct. Diagram?

I 2= 12

AI5 = 8A

Solution :-

leaveenter II

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

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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I7 = I5 + I6

10 = 8 + 2

10 = 10 Ok.

Resisters in Parallel :-

or I = V ( G1 + G2 )

I = VGT

Where GT = G1 + G2

Hence21

21

21 .

111

RR

RR

RRR T

or

In the same minner , if we have three resistors

in parallel , then:

321

1111

RRRR T

From K.V.L. V = V1 = V2

From K.C.L. I = I1 + I2

From Ω.L.2

2

1

1

R

V

R

VI

= V1G1 + V2G2

= V1 ( G1 + G2 )

21

21.

RR

RRRT

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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321

213132

..

...1

RRR

RRRRRR

R T

213132

321

...

..

RRRRRR

RRRR T

And , if we have N of parallel resistance , then

NT RRRRR

11111

321

Also

PT = P1 + P2 + P3

1

21

12

1111 R

VRIIVP

Source powerT

TTTTs R

ERIEIP

22

Example :- For the following cct. Find RT , PT , IT , Ib?

Solution :-

في حالة كون قیم المقاومات متساویة 24

8

N

RRT

AR

EI

TT 8

2

16

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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AR

EIbranch 2

8

16

1

WRIP TTT 1282.8 22

or PT = E.IT = 16 * 8 = 128W

or PT = P1 + P2 + P3 + P4

W12832323232

8*28*28*28*2 2222

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)

101.0

111.0

105.01.025.0

105.01.025.0

1

20

1

10

1

4

1

1111

33

3

3

3

321

RR

R

R

R

RRRRT

b) E = V1 = I1R1 = 4 * 10 = 40 V

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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c)

d) WRIP 8020.2 22

222

or2

22

2 R

VP , or P2 = I2V2

Current division Rule :-

R1 R2

I2I1

I

V

In the same miner

21

12 RR

RII

Also1

2

21

1

21

2

2

1

R

R

RR

RI

RR

RI

I

I

2

1

1

2

2

1

G

G

R

R

I

I

AR

E

R

VI

AR

EI

TT

220

40

104

40

22

22

21

21.

RR

RRIV

1

21

21

11

.

R

RR

RRI

R

VI

21

21 RR

RII

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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Example :- For the following circut. , find V , I1 and I2?

100Ω 0.1Ω

I2I1

5A

V

Solution :-

0999.01.100

10

1.0100

1.0*100.

21

21

RR

RRRT

V = I . RT = 5 * 0.0999 = 0.4995 V

AV

I 004995.01001

AV

I 995.41.02

To check I = I1 + I2

5 = 0.004995 + 4.995

5 = 5 Ok.

Example :- Determine the resistance R1 in the figure below?

Solution :-

I = I1 + I2

Dr. Sameir Abd Alkhalik AziezUniversity of Technology Lecture (4))

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or I2 = I – I1 = 27 – 21 = 6 mA

V2 = I2R2 = 6 * 10-3 * 7 = 42 mV

V1 = V2 = 42 mV

210*21

10*423

3

1

11 I

VR

or

2

7

7*10*2710*21

1

1

33

21

21

R

RRR

RII