003 Chapter 2-1 Dc Circuit Analysis

7/29/2019 003 Chapter 2-1 Dc Circuit Analysis

1/67

1

CHAPTER 2

DC CIRCUITSAND ANALYSIS


2/67

CONTENTS

Ohms Law

Kirchhoffs Law

Series resistors & voltage division

Parallel resistors & current division

Y -

transformation

Method of Analysis Nodal and Mesh


3/67

INSPIRING CREATIVE AND INNOVATIVE MINDS


4/67

4

DC Circuits and method of analysis

Simplify dc circuits by combining resistances inseries and parallel or star-delta transformation(where applicable) circuit reduction.

- Apply the voltage-division and current-

division principles to determine thevoltages or currents.

- Usually, the circuit contains only one voltagesource or one current source.

Solve and simplify a more complex circuits usingthe node-voltage analysis and mesh-currentanalysis


5/67

5


6/67

Ohms Law

Property of a material to resist a flow of current known as resistance

A

lR - measured in ohms ()

- Resistivity of the material

l - length of the material

A - Cross section area of the material

Mathematically,

+ V

i


7/67

Ohms Law

Ohmss Law: A voltage across a resistor is directly proportional tothe current flowing through a resistor

+ V

i

v i

Constant of proportionality between v and i is the resistance, R ()

v = i R

Must comply with passive sign convention


8/67

Ohms Law

Fixed resistors

Wirewound type

carbontype type


9/67

Ohms Law

Variable resistors


10/67

Ohms Law

Two extreme values of resistance:

Short circuit

Open circuit

0i

0

i

vR

ov

ivR


11/67

Ohms Law

Conductance: reciprocal of resistance

v

i

R

1G - measured in siemens (S)

Conductance: ability of an element to conduct current


12/67

Ohms Law

Power in a Resistor

vip

+ V

i

Rii)iR(p 2R

v)

R

v(vp

2

Always absorbs power

Always positive


13/67

13

Kirchoffs Laws

Introduced in 1847 by Gustav RobertKirchoff German physicist

Formally known as

i) Kirchoffs current law (KCL)

ii) Kirchoffs voltage law (KVL)


14/67

Kirchhoffs Law

Network topology

A branch represents a single element such as a

voltage source or a resistor.


15/67

Kirchhoffs Law

Network topology

A node is the point of connection between two

or more branches.


16/67

Kirchhoffs Law

Network topology

A branch represents a single element such as a

voltage source or a resistor.

A node is the point of connection between two

or more branches.

A loop is any closed path in a circuit.


17/67

Kirchhoffs Law

Network topology

Two or more elements are inseries if they exclusively

share a single node and consequently share the same

current

Two or more elements are inparallel if they areconnected to the same two nodes and consequently

have the same voltage across them

1 & 2 - parallel

10V & 4 - parallel

5 in series with (1 and 2 in parallel)


18/67

Kirchhoffs Law

Kirchhoffs Current Law (KCL)

Kirchhoffs current law (KCL) states that the algebraic sum

of currents entering a node (or a closed boundary) is zero

01

N

n

niMathematically,


19/67

19

Kirchoffs current law (KCL) KCL states that the algebraic sum of currents entering a

node is zero.

01

N

nni

0)()()(54321

iiiii

i1 i2

i3

i4

i5

node

outin ii


20/67

20

a

b

IT

IT = I1 I2 + I3Equivalent circuit

IT

I1 I

2I3

a

b


21/67


Example

Determine the current I for the circuit shown in the figure below.

I + 4-(-3)-2 = 0

I = -5A

This indicates that the

actual current for I is

flowing in the opposite

direction.

We can consider the whole

enclosed area as one node.


22/67

Kirchhoffs Law

Kirchhoffs Voltage Law (KVL)

Kirchhoffs voltage law (KVL) states that the algebraic sum

of all voltages around a closed path (or loop) is zero.

Mathematically,0

1

M

m

nv


23/67

23

Kirchoffs voltage law (KVL)

KVL states that the algebraic sum of all voltagesaround a closed path (or loop) is zero.

01

M

mmv

0)()()(54321

vvvvv

V4V1

+ V2 - + V3 -

- V5 +

i


24/67



25/67

25

Example 1 Find Iand Vab

DC DC

DC

+

Vab

-

30V

3

10V

5

8V

I


26/67

SERIES RESISTOR

26

R1 R2R3 RN

V

+ V1 - + V2 - + V3 - + VN -

Current flow through each elementVoltage across each element (voltage

drop)

I


27/67

27

DC+ v1 - + v2 - + v3 - + vN -

R1 R2 R3 RNi

v

11 iRv 22iRv

33 iRv NN iRv KVL:

Nvvvvv ........

321

NiRiRiRiRv ......321

N

RRRRi ......321


28/67

28

NRRRRiv ......321

eqiR

Neq RRRRR ...321

N

nn

R1

eqR

vi

DC

+ v1 - + v2 - + v3 - + vN -

R1 R2 R3 RNi

v

DC

Req

v

i


29/67

29

DC

+ v1 - + v2 - + v3 - + vN -

R1 R2 R3 RN

i

v

1

321

11

... R

RRRRviRv

N

vRRRR

Rv

N

...

321

1

1

vRRRR

Rv

N

...

321

2

2

v3 = ? ; vN= ?


30/67

30

Series resistor

vRRRR

R

vN

n

n ...

321

Voltage division


31/67

31

Example 2 Find

a)Voltage across 2 and 3

b)Power absorbedc)Power supplied DC 20V

2

3


32/67

32

DC 20V

2

3

I

+ V2 - +

V3

-

KVL :

A4

2032020 32

I

IIVV

V1243

V842

3

2

V

V

Voltage division

V82032

22

V V1220

32

33

V


33/67

33

Parallel resistors

DC

Current flow through each elementVoltageacross each element (voltage

drop)

21 vvv

21iii KCL

V R1R2

i1

i i2

+

v1

-

+

v2

-


34/67

34

2211RiRiv

;22

2

2

11

1

1R

v

R

vi

R

v

R

vi

eqR

v

RRv

R

v

R

v

iii

2121

21

11

V R1R2

i1

i i2

+

v1

-

+

v2

-


35/67

35

21

111

RRReq

21

21

RR

RRR

eq

21 GGGeq

DC v

+

vReq

-

Req

i

V R1R2

i1

i i2

+

v1

-

+

v2

-


36/67

36

21

21

RR

RiRiRv

eq

21

2

1

1i

RR

R

R

vi

21

1

2

2i

RRR

Rvi

iGG

Gi

21

1

1

iGG

G

i21

2

2

Current divider

V R1R2

i1

i i2

+

v1

-

+

v2

-


37/67

37

NRRRR

1111

R

1

321eq

NeqGGGGG ....

321

R1 RNR3R2V+

V1

-

+

V2

-

+

V3

-

+

VN

-

i

i3 i

Ni

2i

1

NVVVVV 321


38/67

38

iGGGG

Gi

N

....

321

1

1

iGGGG

Gi

N

N

N

....

321

R1 RNR3R2V+

V1

-

+

V2

-

+

V3

-

+

VN

-

i

i3 i

Ni

2i

1


39/67

39

Example 3 For each figure,

calculate:

- Currents- Power on each

elements.

10 212A

i1 i2

10 212A

i1 i2 i3

5

10 212A

i1 i2 i3

5

3


40/67

40

10 212A

i1 i2 A212210

21

i

A1012210

10

2

i

Current Division

A212)21()101(

1011

i

A1012)101()21(

212

i

POWER


41/67

41

Example 3 Find v1 and v2, P3k and P20k, and

Power supplied by the source


42/67

42

i1

i2i3

mA510)20/1()5/1()41(

411

i

V15

)105)(103( 331

v

mA110)20/1()5/1()4/1(

20/13

i

V20

)101)(1020( 332

v


43/67

43

Series-parallel circuit

a

b

R1

R2 R3Rab= Req

R1

R2//R3

32

32

1

RRRRRRR

eqab


44/67

44

Example 4


45/67

45

Example 5


46/67

46

Find v1 and v2, i1 and i2 ,P12 and P40

Example 6


47/67

47

Example 7

1

3

1.6

2

4

a

b

Find the resistance Rab


48/67

48

Example 8

Calculate Vo andIo in the circuit below:


49/67

49

I0I1

I2

I1 = I0 + I2

0 V

70 30

20 5 +V0

-


50/67

50

Example 9

For the circuit in Figure below, obtain the

Equivalent resistance at terminal a-b.


51/67

51

Example 10

Find Reqand io in the following circuit


52/67

52

Example 11 Find i and V0 in the circuit below:


53/67

53

Example 12 FindRab


54/67

54

Example 13

FindRab

3k

1k

400

600

a b


55/67

55

Example 14 Find the equivalent resistance at terminals a-b .

Y t f ti


56/67

Y transformationStar delta transformation

How can we combine R1 to R7 ?

Y transformation


57/67

)(1

cba

cb

RRR

RRR

)(2

cba

ac

RRR

RRR

)(3

cba

ba

RRR

RRR

1

133221

R

RRRRRRRa

2

133221

R

RRRRRRRb

3

133221

R

RRRRRRRc

Delta -> Star Star -> Delta


Y transformation


58/67


example

Y transformation


59/67


example


60/67


Obtain the equivalent resistance Rab for the circuit below

Example 15


61/67



62/67



63/67

Source transformation

Another circuit simplifying technique

Itis the process of replacing a voltage source vS in serieswith a resistor R by a current source iS in parallel with a

resistor R, or vice versa

+

R

vs

a

b

Terminal a-b sees:

Open circuit voltage: vs

Short circuit current: vs/R

For this circuit to be equivalent, it

must have the same terminal

charateristics

Ris

a

b


64/67

Source transformation

Another circuit simplifying technique

Itis the process of replacing a voltage source vS in serieswith a resistor R by a current source iS in parallel with a

resistor R, or vice versa

+

R

vs

a

b

Ris

a

b

ix

iy

Note: current through R (hence power) for both circuits is not the same

i.e. ix iy

l


65/67

Find vo in the circuit shown below using source transformation

Example 16

l 1


66/67

Example 17

E l 18


67/67

Find io in the circuit shown below using source transformation

Example 18

003 Chapter 2-1 Dc Circuit Analysis

Documents

Transcript of 003 Chapter 2-1 Dc Circuit Analysis