# Ch2 Resistive Circuits

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BEE1133 : Circuit Analysis I

Chapter 2: Resistive Circuits

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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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Resistive Circuits: Lesson Outcomes

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

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Series Circuits

Series Circuits

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Series Circuits : Introduction

A series circuit is one that has only one current path

R1

R1 R2 R3

VS R3

R2

VS

R1

R2

R3

VS

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Series Circuits : Current

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?+ 2.0 mA _ VS _ammeter +

R1

+ ammeter _ R2 _ammeter +

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Series Circuits : Voltage

Series circuit rule for voltage: 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.

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Series Circuits : Voltage

Voltage sources inin series add algebraically. For example, Voltage sources series the total voltage of the sources shown is+

27 V

9V + 9V

What is the total voltage if one battery is reversed?

+ 9V

9V

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Series Circuits : Resistor

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

Req =R1+R2+R3+..... +RNWhere n = the number of resistors9

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Series Circuits : Resistor

Example For example, the resistors in a series circuit are 680 W,1.5 kW, and 2.2 kW. What is the total resistance?R1

680 ;VS 12 V R3

R2 1.5 k;

2.2 k;

4.38 kW10

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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.N

Req ! R1 R2 R N ! Rnn !1

The voltage divider can be expressed asRn v vn ! R1 R2 R N11

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Voltage Division

Example

10V and 5; are in series12

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Series Circuit

Exercise 1 (prob 2.37) Given voltage across resistor R, VR = 10 V, calculate the value of R.R 10 ;

+ 10 V +

20 V

30 V_ +

Ans: R = 2.5 ;

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Voltage Division

Exercise 2 (prob 2.35) Given voltage supply VS = 50 V, calculate the voltages across 70 ; resistor, V1 and 5 ; resistor, V0.

70 ;+

+

V1-

30 ;

Vs = 50V

-

+

20 ;

V0-

5;

Ans: V0 = 8.0 V V1 = 42.0 V14

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Parallel Circuits

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Parallel Circuits : Current

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.

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Parallel Circuits : Voltage

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.

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

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Parallel Circuits : Resistor

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

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Parallel Circuit : Resistor

Example

Series Resistance, R = 15 + 20 + 13 = 48 Total resistance, Rab = R48||16 = 16(48)/(16+48) = 1219

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Current Division

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:1 1 1 1 ! Req R1 R2 RN

The total current i is shared by the resistors in inverse proportion to their resistances. The current divider can be expressed as: v iReq in ! ! Rn Rn 20

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Current Division

Example

2;, 3; and 2A are in parallel21

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Parallel Circuit

Exercise 3 (prob 2.26) Given io = 2 A, calculate ix and total power dissipated by the circuit.ix io2; 4; 8; 16 ;

Ans: ix = 30.0 A22

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Current Division

Exercise 4 (prob 2.31) For the circuit below, determine i1 to i5 by using current division technique.3;

i1 i3

i2+

40 V

_

4;

1;

i4

2;

i5

Ans: i1 = 11.2 A i2 = 1.6 A i3 = 9.6 A i4 = 6.4 23 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.60 ; 12 ;

io

5;

6; 80 ;

+

40 V

15 ;

20 ;

_

Ans: Req = 12.5 ; io = 3.2 A

Req

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Special Case for R: Short Circuit/Open Circuiti = 0, R = V/i = V/0 =

g

v = 0, R = 0/i = 0 Neglect R!

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Special Case for R: Short Circuit/Open Circuit Exercise 6 (prob 2.44)

For the circuit below, obtain the equivalent resistance at terminal a-b.20 a 10 b (a)Ans: Ra-b = 28.94 ;

20

5

15

20

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Special Case for R: Short Circuit/Open Circuit Exercise 7 (prob 2.45)

Find the equivalent resistance at terminal a-b, of the circuit below.30

12

a

5

20

25

60 10

b

15 (b)

Ans: Ra-b = 32.5 ;27

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Resistive Circuits: Lesson Outcomes

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

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Wye-Delta Transformation

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Wye-Delta Transformations

What is Wye circuit connection?R1 R2 R3

What is Delta circuit connection?Rc Rb Ra

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Wye-Delta Transformations

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?31

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Wye-Delta Transformations

Delta -> Star

Star -> Delta

Rb Rc R1 ! ( Ra Rb Rc ) Rc R a R2 ! ( R a Rb Rc ) R3 ! Ra Rb ( Ra Rb Rc )

Ra !

R1 R2 R2 R3 R3 R1 R1

R1 R2 R2 R3 R3 R1 Rb ! R2 R1 R2 R2 R3 R3 R1 Rc ! R332

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Wye-Delta Transformations

Exercise 8 (prob 2.51) Obtain the equivalent resistance at the terminals a-b for each of the circuits below.a10 ; 30 ; 10 ; 20 ;

10 ;

20 ; Ans: Ra-b = 9.23 ;33

b

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Wye-Delta Transformations

Exercise 9 (prob 2.54) Consider the circuit below. Find the equivalent resistance at terminals: (1) a-b (2) c-d50 a 100 b 150 Ans: Ra-b = 130 ; Rc-d = 140 ;34

15

60 c 100 d

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Equivalent Resistance

When two resistors R1 (= 1/G1) and R2 (= 1/G2) are in series, their equivalent resistance Req and equivalent conductance Geq are: G1G2 Geq ! Req ! R1 R2 G1 G2 When two resistors R1 (= 1/G1) and R2 (= 1/G2) are in parallel, their equivalent resistance Req and equivalent conductance Geq are: R1 R2 Geq ! G1 G2 Req ! R1 R235

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Equivalent Resistance

Exercise 10 (prob 2.33) Obtain the values of Geq, v and i for the circuit below.i 4S 6S

9A

+ v _

1S

2S

3S

Geq

Ans: Geq = 3 i = 6A

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Source Transformation

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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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Source Transformation+ + The ar

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