Chapter 3a Soon

8/11/2019 Chapter 3a Soon

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1

By:

DR SOON CHIN FHONG

Department of Electronic Engineering

Faculty of Electrical and Electronic Engineering

Universiti Tun Hussein Onn Malaysia

ELECTRICAL AND ELECTRONIC

TECHNOLOGY

(BEX17003)



2

Chapter 3(a):

Direct Current CircuitsAnalysis



3

Lecture Contents

Basic Law

3.1 Nodes, Branches, and Loops

3.2 Kirchoff’s Laws

3.3 Series Resistors and Voltage Division

3.4 Parallel Resistors and Current Division

3.5 Series and

Parallel Resistors

3.6 Wye

-

Delta Transformations



4

3.1 Nodes, Branches,

And Loops



5

Nodes, Branches and Loops

A branch:

a single element such as a voltage

source or a resistor

represents any two-terminal element



6

A: 5 branches

How many branches exist in this circuit?



7

Cont…

A node:

the point of connection between two or

more branches. usually indicated by a dot in a circuit.



8

A: 3 nodes

How many nodes exist in this circuit?



9

Cont… A loop:

any closed path in circuit



10

3.2 Kirchoff’s

Laws



11

Kirchoff’s Laws

Kirchoff’s Current Law (KCL) states that the algebraicsum of currents entering a node (or a closed loop

boundary) is zero.

The sum of the currents entering a node is equal to the

sum of the currents leaving the node. Mathematically, KCL implies that

0i N

1nn

where,in= nth current entering (or leaving) the node

N= number of branches connected to the node

*Hukum Arus Kirchoff

……..( Eq. 3.1)



12

I6I1

I2

I3

I4

I5

I1 - I2 - I3 + I4 - I5 + 16=0 @ I1 + I4 + 16 = I2 + I3 +I5

current entering current leaving

leavingentering II ……..( Eq. 3.2 )

Kirchoff’s Current Law (KCL)



13

Cont… Kirchoff’s Voltage Law (KVL) states that the

algebraic sum of all voltages around a closedpath (or loop) is zero.

N

1n

n 0V

Where,

Vn= the nth voltageN= number of voltages in the loop

Sum of voltage across active element

= Sum of voltage across passive element

……..( Eq. 3.3 )



14

+ V1 -

- V3 +

V2

+

-E2

E1

R1

R2

R3

E1 – V1 – V2 – E2 – V3 = 0 @ E1 – E2 = V1 + V2 + V3

Kirchoff’s Voltage Law (KVL)



15

Example 3.1a

Write the KVL equation for the following circuit.





17

Example 3.1b

Write the KVL equation for the following circuit.



18

Solution 3.1b

-V5 + V4 – V3 – V2 + V1 = 0

v1 + v4 = v2 + v3 + v5



19

Example 3.2

For the following circuit, find voltages v1 and v2.

i

+

-

- + 1 v

2 v

W 2

W 3 V 20



20

Solution 3.2

From Ohm’s Law, v1 = iR = i x 2 = 2iv2 = iR = i x 3 = 3i

From KVL , 20 = v1 +v2

20 = 2i +3i

20 = 5i

i = 4 A

So, v1 = iR = 4 x 2 = 8 V

v2 = iR = 4 x 3 =12 V

Check! KVL: 20 = 8+12 =20

i

+

-

- +

1 v

2 v

W 2

W 3 V 20



21

Example 3.3

Find the power consumed by resistor Rx.



22

Solution 3.3



23

Example 3.4

Given that the power dissipated byResistor R1 is 4W. Find the value of Rx.



24

Solution 3.4



25

3.3 Series Resistors

andVoltage Division



26

Series circuit

A series circuit is a circuit which provides onlyone path for current to flow between two points

in a circuit so that the current is the same

through each series component.

Total Resistance

The total resistance of a series circuit

= the sum of the resistances of each

individual resistor .



27

Series Resistors and Voltage Division



28

Figure (a) Series circuit (b) Equivalent circuit

+ V -

Vs

R series

+ V1 - + V2 -

Vs

R 1 R 2

+ V3 - + VN -

R 3 R N

(a) (b)

)( R ...R R R R R N 32 1equivalent T W++++

(A)R

V(I)Current,

T

S

Voltage, v1 = iR1, v2 = iR2

…..( Eq. 3.4)

……..( Eq. 3.5)

……..( Eq. 3.6)



29

Voltage Divider Rules Voltage drop across any of the

resistors can be written as follows:

V2

IVs

+

-

V1

V4

V3

+

-

+

-

+

-

+

-

R 1

R 4

R 2

R 3

Substituting V S /R T for I results in

Rearranging terms yields

Voltage – divider formula

4,3,2,1

x

IR xx

V

x

T

S R

R

V )(x V

S

T

xV

R

R )(x

V

……..( Eq. 3.7)

……..( Eq. 3.8)

……..( Eq. 3.9)



30

Example 3.5

3 kΩ

V=24V

+ V1 -

+ V2

-

9kΩ

Find v1 and v2 in the circuit below.



31

Solution 3.5

Based on KVL,v = v1 +v2

= i(R1 +R2)

24 = i ((3 + 9)k)

= (12k) iSo, i = 24/(12k) = 2 (mA)

By using i value,

v1 = iR1 =2m x 3k = 6 V

v2 = iR2 =2m x 9k = 18 VCheck! KVL: 24V =( 6+18 )V =24 V

V=24V

+ V1 -

3 kΩ

+ V2

-

9kΩ

i



32

Example 3.6

Calculate the voltage drop across each resistor in thevoltage divider in figure below .

IVs

100V

+

-

V2

V1

V3

+

-

+

-

+

-

R 1=100Ω

R 2=220Ω

R 3=680Ω



33

Solution 3.6

IVs

100V

+

-

V2

V1

V3

+-+

-+

-

R 1=100Ω

R 2=220Ω

R 3=680Ω

V V R

RV

V V

R

RV

V V R

RV

S

T

S

T

S

T

68100)1000

680(

22100)1000

220(

10100)1000

100(

33

22

11

Check! KVL: 100V =( 10+22+68 )V =100V



34

3.4 Parallel

Resistorsand Current Division



35

Parallel Resistors and Current Division

Figure (a) Parallel Circuit (b) Equivalent Circuit

Vs

+ V -

Vs

R parallel

(a) (b)

I

+ V1

-

R 1+

V2

-

R 2+

VN

-

R N

I

I1 I2 IN



36

(V)V,...,VVVV N321

(A)I ...I I I (I)Current, N 32 1 ++++

(Ω)R

1...

R

1

R

1

R

1R

(Ω)R

1...

R

1

R

1

R

1

R

1

R

1

1

N321

T

N321equivalentT

-

++++

++++

…..( Eq. 3.10)

…..( Eq. 3.11)

…..( Eq. 3.13)

…..( Eq. 3.12)



37

Example 3.7

For the following circuit, find;

(a) Equivalent resistance

(b) Total current

(c) Currents through

each resistor

(d) Power for each element

and total circuit power(e) Power supplied by the source

30 V 10 Ω

+

V1-

5 Ω

+

V2-

I1 I2

IT



38

Solution 3.7

A I

I

R I V

R

R

T

T

T T

T

T

9)10

3(30

)3

10

(30

current b)T otal

33.3

3

10

10

3

10

12

10

1

5

11

resistancenta)Equivale

W

++

30 V

+

V1

- 5 Ω

+

V2

- 10 Ω

I1 I2

IT



39

Solution 3.7(cont.)

30 V 10 Ω

+

V1

- 5 Ω

+

V2

-

I1 I2

IT

A A A

I I I CHECK

A R

V I

A

R

V I

R

V I IRV

T

9)36(9

:

310

30

6

5

30

21

2

2

1

1

+

+

resistor each through c)Currents



40

Solution 3.7(cont.)

W W W

P P T

P

CHECK

W V T

I T

P

W V I P

W V I P

IV P

270)90180(270

21

:

270309

source by thesuppliede)Power

9030322

18030611

power circuittotalandelementeachford)Power

+

+

30 V

+

V1

- 5 Ω

+

V2

- 10 Ω

I1 I2

IT



41

Current Divider Rules

The formula for total resistanceof 2 parallel branches,

21

21

R R

R R RT

+

I

VS

+ V1

-

R 1

+ V2

-

R 2

I1 I2

Current Divider formula for 2 branches

From Ohm’s Law,

T T T T

T T T T

T T

I R

R

R

R I

R

V I

I R R

R R I

RV I

R I R I R I V

222

2

111

1

2211

,

…..( Eq. 3.14)

…..( Eq. 3.15)



42


Substituting R1R2/(R1+R2) for RT and canceling result in

TI

T

R

yieldstermsgRearrangin

T

R T

I

inresultsX

Iforexpressionin theS

VforT

R T

IngSubstituti

T

R T

I

X R

x I

X R

x I

S V

X R

S V

x I

Current Divider formula for 2 branches

Current – divider formula for 2 branches

…..( Eq. 3.16)

…..( Eq. 3.17)



43


Substituting R1R2/(R1+R2) forRT and canceling result in

General Current Divider formula for any number of branches

Therefore, current divider formula for 2 branches are :

T T I R

R R

R R

I I R

R R

R R

I

+

+

2

21

21

2

1

21

21

1 ,

T T I

R R

R I I

R R

R I

+

+

21

1

2

21

2

1 ,

Current – divider formula

…..( Eq. 3.18)

I

VS +

V1

-

R 1

+ V2

-

R 2

I1 I2



44

Example 3.8

For the following circuit, find current foreach element using current divider rules.

12 V

+ V1

- 15 Ω

+ V2

- 30 Ω

I1 I2

I



45

Solution 3.8

12 V

1/Rall

= 1/R1

+1/R2= (R1 + R2)/(R1R2)

Rall = (R1R2)/(R1+R2)

= (15 x 30)/ (15 +30)= 450/45 = 10 Ω

Vall = V1 =V2 = 12V

I1 = V1/R1 = 12/15 = 0.8 A

I2 = V2/R2 = 12/30 = 0.4 A

+ V1

-

15 Ω

+ V2

- 30 Ω

I1 I2

I

Check!

Vall=IR allI=Vall/R all

I=12/10=1.2 A

KCL:

I = I1 + I2

1.2 A= (0.8 +0.4)A =1.2 A

leaving entering I I



46

3.5 Series and

Parallel Resistors



47

Combinational Series and Parallel Circuits

Define as a circuit that consist of a combinationseries and parallel circuit.

We can get the equivalent resistance by

looking the total resistance from the voltage orcurrent source.

Lets consider the following example for better

understanding.



S



49

Solution 3.9

Vs

R 1

R 2 R 3 R 6

R 4 R 5 R 7

(i)Rx(ii)Ry

(iii)Rz

(i) Series resistors, Rx =R1+R2 +R3

(ii)Series resistors Ry = R6 + R7

(i) Parallel resistors Rz = R4 //R5

1/Rz = 1/R4 +1/R5 + 1/RV

= ( R5RV +R4RV + R4R5 ) / ( R4R5RV )

So, Rz = ( R4R5RV ) / ( R5RV +R4RV + R4R5 )

Equivalent resistance = (i) + (iii) = RX +RZ

=(R1+R2 +R3) + ( R4R5RV ) / ( R5RV +R4RV + R4R5 )

=( R1+R2 +R3) + ( R4R5 (R6 + R7) ) /

( R5 (R6 + R7) +R4 (R6 + R7) + R4R5 )

E i 3 1



50

Exercise 3.1

Calculate the equivalent resistance Rab at terminals a-b.

a

b

W10

W80W60 W20 W30

A: R ab = 16 Ω



51

3.6 Wye-Delta

Transformations



52

DELTA – WYE transformation

R 1 R 2

R 3

R a

R bR c

a

b

c 321

31

c

321

32

b

321

21

a

RRR

RRR

RRR

RR

R

RRR

RRR

++

++

++

( Eq. 3.19)



53

WYE - DELTA transformation

a

cb

cb3

c

ba

ba2

b

ca

ca1

R

RRRRR

R

RRRRR

R

RRRRR

++

++

++R 1 R 2

R 3

R a

R bR c

a

b

c

( Eq. 3.20)



54

WYE - DELTA transformation

a

acc b ba3

c

acc b ba2

b

acc b ba1

R

R R R R R R R

R

R R R R R R R

R

R R R R R R R

++

++

++

R 1 R 2

R 3

R a

R bR c

a

b

c( Eq. 3.21)



55

20kΩ a

b c

d

12kΩ

18kΩ 6kΩ

12kΩ 12kΩ

Find the equivalent resistance.

Example 3.11



56

20kΩ a

b c

d

12kΩ

18kΩ 6kΩ

12kΩ 12kΩ

20kΩ a

b c

d

RcRb

12kΩ 12kΩ

Ra

Solution 3.11



57

W++W++

W++

W

++

W++-

k ) R Rak R

k k Rck Rb R

k

k

k

parallel all

parallel

875.3020(

875.7)12//()12(

612k 18k 6k

(12k)(18k) Rc

2

12k 18k 6k

(6k)(12k) Rb

312k 18k 6k

(6k)(18k) Ra :

Solution 3.11 (cont.)



Exercise 3.2Obtain the equivalent resistance at the terminals a-b

for the circuit below.

a

b

W30

W20

W20

W10

W10W10

Answer: R ab = 9.23 Ω

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Chapter 3a Soon