ch2 resistive circuits

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Chapter 2: Resistive Circuits

BEE1133 : Circuit Analysis I

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LOGO Resistive Circuits: Syllabus

2.1 Resistive circuit: Series, parallel circuits and combination circuits

2.2 Principles of voltage division and current division

2.3 Delta-wye transformation 2.4 Equivalent resistance 2.5 Source transformation

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Identify series and parallel connections Apply voltage division and current division in circuit

problems

Resistive Circuits: Lesson Outcomes

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Series Circuits

Series Circuits

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• A series circuit is one that has only one current path

VS VS VS

R1

R1

R1

R2 R2

R2

R3

R3

R3

Series Circuits : Introduction

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Series circuit rule for current:

Because there is only one path, the current everywhere is the same.

For example, the reading on the first ammeter is 2.0 mA, What do the other meters read?

R 2

R 1

V S

+ _+ _

++ __

2.0 mA

ammeter

Series Circuits : Current

ammeter

ammeter

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When two or more voltage sources are in series, the total voltage is equal to the algebraic sum (including polarities of the sources) of the individual source voltages.For example, the voltage across node A and B is equal to 4.5 V which is equal to total voltages from all sources.

Series Circuits : Voltage

Series circuit rule for voltage:

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Voltage sources in seriesVoltage sources in series add algebraically. For example, the total voltage of the sources shown is

+

+

+

9 V

9 V

9 V

27 V

9 V

What is the total voltage if one battery is reversed?

Series Circuits : Voltage

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The total resistance of resistors in series is the sum of the individual resistors.

Series Circuits : Resistor

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Req =R1+R2+R3+..... +RN

Where n = the number of resistors

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For example, the resistors in a series circuit are 680 W, 1.5 kW, and 2.2 kW. What is the total resistance?

4.38 kW

R

R

2

3

R 1

V S

6 8 0

2 .2 k

1 .5 k 1 2 V

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Series Circuits : Resistor

Example

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Voltage Division

Series: Two or more elements are in series if they are cascaded or connected sequentially and consequently carry the same current.

The equivalent resistance of any number of resistors connected in a series is the sum of the individual resistances.

The voltage divider can be expressed as

N

nnNeq RRRRR

121

vRRR

Rv

N

nn

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10V and 5 are in series

Voltage Division

Example

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Series Circuit

Exercise 1 (prob 2.37)

R 10

+

_20 V

–

+30 V

+ 10 V -

Given voltage across resistor R, VR = 10 V, calculate the value of R.

Ans: R = 2.5

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Voltage Division

Exercise 2 (prob 2.35)

20

-

+

70 30

5

Vs = 50V +

V0

-

+

V1

-

Given voltage supply VS = 50 V, calculate the voltages across 70 resistor, V1 and 5 resistor, V0.

Ans: V0 = 8.0 V V1 = 42.0 V

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Parallel Circuits

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Parallel circuit rule for current:

The total current produced by all current sources is equal to the algebraic sum of the individual current sources.

For example, total current, A is 5 A because each parallel branch currents are added together.

Parallel Circuits : Current

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Parallel circuit rule for voltage:

The voltage across any given branch of a parallel circuit is equal to the voltage across each of the other branches in parallel.

For example, voltage across RN is equal to the voltages across R1, R2 and R3 because they are parallel to each other.

Parallel Circuits : Voltage

17Vab = VR1 = VR2= VR3 =..... VRN

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The total resistance of resistors in parallel is given by the equation below:

Parallel Circuits : Resistor

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LOGO Parallel Circuit : Resistor

Series Resistance, R = 15 + 20 + 13 = 48 Ω

Total resistance, Rab = R48||16

= 16(48)/(16+48) = 12 Ω

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Example

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Parallel: Two or more elements are in parallel if they are connected to the same two nodes and consequently have the same voltage across them.

The equivalent resistance of a circuit with N resistors in parallel is:

The total current i is shared by the resistors in inverse proportion to their resistances. The current divider can be expressed as:

Neq RRRR

1111

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n

eq

nn R

iR

R

vi

Current Division

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2, 3 and 2A are in parallel

Current Division

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Example

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Parallel Circuit

Exercise 3 (prob 2.26)

2 4 8

io

16

ix

Given io = 2 A, calculate ix and total power dissipated by the circuit.

Ans: ix = 30.0 A

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Exercise 4 (prob 2.31)

Current Division

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40 V

3

+

_

i1

4 1

i2

i42 i5

i3

For the circuit below, determine i1 to i5 by using current division technique.

Ans: i1 = 11.2 A i2 = 1.6 A i3 = 9.6 A i4 = 6.4 A i5 = 3.2 A

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Combination Circuit (Series & Parallel)

Exercise 5 (prob 2.38)

For the circuit below, determine equivalent resistance Req and current at io.

5 6

+

_

15 20 40 V

60

12

80

Req

io

Ans: Req = 12.5 io = 3.2 A

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LOGO Special Case for R: Short Circuit/Open Circuit

i = 0,

R = V/i = V/0 =

v = 0,

R = 0/i = 0

Neglect R!

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Exercise 6 (prob 2.44)

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Special Case for R: Short Circuit/Open Circuit

For the circuit below, obtain the equivalent resistance at terminal a-b.

Ans: Ra-b = 28.94

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Exercise 7 (prob 2.45)

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Special Case for R: Short Circuit/Open Circuit

Find the equivalent resistance at terminal a-b, of the circuit below.

a

b Ans: Ra-b = 32.5

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Simplify a circuit using delta-wye transformation Determine the equivalent resistance of a network Apply source transformation to simplify a circuit Construct DC circuit to understand the concept of

electrical quantities and validate circuit theorems to complete Laboratory 1

Resistive Circuits: Lesson Outcomes

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Wye-Delta Transformation

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LOGO Wye-Delta Transformations

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What is Wye circuit connection?

What is Delta circuit connection?

R1

R3

R2

Ra

Rc

Rb

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LOGO Wye-Delta Transformations

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Why do we need wye-delta tranformations? In some circuit analysis, the resistors are neither in

parallel nor in series This has to be simplified by using three terminal

equivalent networks

Our main objectives here:-How to identify them?How to apply wye-delta transformation in circuit

analysis?

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LOGO Wye-Delta Transformations

)(1cba

cb

RRR

RRR

)(2cba

ac

RRR

RRR

)(3cba

ba

RRR

RRR

1

133221

R

RRRRRRRa

2

133221

R

RRRRRRRb

3

133221

R

RRRRRRRc

Delta -> Star Star -> Delta

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Exercise 8 (prob 2.51)

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Wye-Delta Transformations

Obtain the equivalent resistance at the terminals a-b for each of the circuits below.

10

10

10

20

20

30

a

bAns: Ra-b = 9.23

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Exercise 9 (prob 2.54)

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Wye-Delta Transformations

Consider the circuit below. Find the equivalent resistance at terminals: (1) a-b (2) c-d

Ans: Ra-b = 130 Rc-d = 140

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LOGO Equivalent Resistance

• When two resistors R1 (= 1/G1) and R2 (= 1/G2) are in series, their equivalent resistance Req and equivalent conductance Geq are:

• When two resistors R1 (= 1/G1) and R2 (= 1/G2) are in parallel, their equivalent resistance Req and equivalent conductance Geq are:

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21

21

GG

GGGeq

21 RRReq

21

21

RR

RRReq

21 GGGeq

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Exercise 10 (prob 2.33)

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Equivalent Resistance

Obtain the values of Geq, v and i for the circuit below.

4 S 6 S

3 S2 S1 S9 A+v_

i

GeqAns: Geq = 3 S

i = 6 A

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Source Transformation

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LOGO Source Transformation

• Equivalent sources can be used to simplify the analysis of some circuits.

• An equivalent circuit is one whose v-i characteristics are identical with the original circuit.

• It is the process of replacing a voltage source vS in series with a resistor R by a current source iS in parallel with a resistor R, or vice versa.

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(a) Independent source transform

(b) Dependent source transform

•The arrow of the current source is directed toward the positive

terminal of the voltage source.

•The source transformation is

not possible when R = 0 for voltage source and R = ∞ for current source.

++

++

--

--

Source Transformation

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Find io in the circuit shown below using source transformation.

*Refer to in-class illustration, textbook, answer io = 1.78A

Source Transformation

Example

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Exercise 11 (prob 4.22)

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Source Transformation

4 52A

5

20V+

10

Use source transformation to find i.

i

Ans: i = 555.5 mA

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Exercise 11 (prob 4.23)

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Source Transformation

45 V9 A

8

10 6

3

By referring to figure below, use source transformation to find current that flows in 8 resistor.

Ans: i = 3 A

i

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LOGO Conclusion

• Series, parallel and combinations of resistors for both connections has been covered.

• Wye-delta transformation and source transformation help to simplify circuit analysis.

• Wye-delta transformation is applied to three terminal network (with three resistors) and then it is replaced with equivalent R values.

• A source transformation is the process of replacing a voltage source (in series with R) by a current source and a parallel R.

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The End

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