chapter 2 resistive circuits - seoul national universityengineering.snu.ac.kr/lecture_pdf/ee/2005 -...

Chapter 2Resistive Circuits

1. Solve circuits by combining resistances in Series and Parallel.

2. Apply the Voltage-Division and Current-Division Principles.

3. Solve circuits by the Node-Voltage Technique.

4. Solve circuits by the Mesh-Current Technique.

5. Find Thévenin and Norton Equivalents.

6. Apply the Superposition Principle.

7. Understand the Wheatstone Bridge.

Goal

Resistance in Series

v2=R2iv1=R1i v3=R3iOhm’s Law

Using KVL, v = v1 + v2 + v3

v = (R1 + R2 + R3)i

Req = R1 + R2 + R3

Resistance in Parallel

v=R2iv=R1i v=R3iOhm’s Law

Using KCL, i = i1 + i2 + i3

i = (1/R1 + 1/R2 + 1/R3)v

Req = 1/R1 + 1/R2 + 1/R3

Example

Series Parallel

Series

Network Analysis using Series/Parallel Equivalents

1. Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source.

2. Redraw the circuit with the equivalent resistance for the combination found in step 1.

3. Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often end with a single source and a single resistance.

4. Solve for the currents and voltages in the final equivalent circuit. Transfer Results to Backward until Original Circuit

Network Analysis : The process of determining the Current, Voltage & Power for each element given in Circuit & Element Values

Step 1,2

Step 3

Step 1,2 & 3

Step 4

Voltage Division

total321

111 v

RRRRiRv

++== total

321

222 v

RRRRiRv

++==

Of the total voltage, the fraction that appears across a given resistance in a series circuit is the ratio of the given resistance to total series resistance Voltage Division Principle

Application of the Voltage-Division Principle

V5.1

156000200010001000

1000

total4321

11

=

×+++

=

+++= v

RRRRRv

Generalized Voltage Divider

total

in

ZVI =

21 ZZZ +=total

21

22 ZZ

ZVZIV+

== inout

Current Division

total21

2

11 i

RRR

Rvi

+== total

21

1

22 i

RRR

Rvi

+==

For two resistance in parallel, the fraction of the total current flowing in a resistance is the ratio of the other resistance to sum of the two resistance (Two Resistors Case) Current Division Principle

Application of the Current-Division Principle

Ω=+×

=+

= 2060306030

32

32eq RR

RRR

A10152010

20

eq1

eq1 =

+=

+= siRR

Ri

Position Transducer Based on Voltage-Division Principle

θKRR

Rvv so =+

=21

2

Typical Knob in Electrical System based on Analog Signalsuch as Amplifier, Adjustable Light, Volume

Series Sources

• Ideal independent voltage sources in series add algebraically

I+ - + - + - + -

I V1 V2 V3Vn

- VR +

R

Note : Parallel Voltage Sources are not Resolvable. WHY?

Parallel Sources

• Ideal independent current sources in parallel add algebraically

In

I3I2I1

R

+

V

_

IT

•Note : Series current sources are not resolvable. WHY?

Example

• R2 and R3 are effectively open circuited and therefore can be omitted• R7 and R8 are short circuited, and can be omitted• I1 and I2 are Algebraically added because they are in Series.• V1 and V2 are Algebraically added because they are in Parallel• R1 and R4 are in Series & R5 & R6 are in Parallel

Although they are veryimportant concepts,

series/parallel equivalents andthe current/voltage division

principles are not sufficient tosolve all circuits.

Wye - Delta Transformations

• Use Y to ∆ transformation

- Need to find equivalent resistance to determine current. - HOW?- (They are not in series, not in parallel)

Equating Resistance's• Resistance between X - Y• In ∆ ⇒ Ra // (Rb + Rc)

• In Y ⇒ R1 + R3

21)()( RR

RRRRRRR

cab

cabXZ +=

+++

=

RbRa

Rc

X

Y Z

R1 R2

R3

X

Y

Z

31)()( RR

RRRRRRR

cba

cbaXY +=

+++

=

32)()( RR

RRRRRRR

bac

bacYZ +=

+++

=

Similarly

Above 3 Equations are Linearly Solvable between R1 , R2 , R3 & Ra , Rb , Rc

X

Y Z

Y

X Z

∆ Y

With Voltage Source With Dependent Source

∆ Y Transformation : Not Enough for All Circuits

Node-Voltage Analysis

1. Select Reference Node2. Assign Node Voltages

If the reference node is chosen at one end of an independent voltage source,one node voltage is known at the start, and fewer need to be computed.

3. Write Network Equations. - First, use KCL to write current equations for nodes and supernodes.

(Write as many current equations as you can without using all of the nodes.) - If not have enough equations because of voltage sources between nodes ,

(use KVL to write additional equations)- If the circuit contains dependent sources,

Find expressions for the controlling variables in terms of the node voltages.Substitute into the network equationsObtain equations having only the node voltages as unknowns.

4. Solve for the Node Voltages. (Equation in standard form) 5. Calculate currents or voltages of interest.

with the node voltages from Step 4

Select Reference Node & Assign Node Voltage

Node : Two More Circuit Elements are Joined TogetherReference Node : Node with Known & Referencing Voltage Ground Symbol : Zero Voltage

- Typically Reference Node is set to be Ground Symbol- Negative Polarity to Reference Node

Assign Element Unknown Voltages

12 vvvy −= 32 vvvx −=

- Assign Node Voltage across R1, R2 & R3- Use KVL In R2 In R3

13 vvvz −=In R1

Writing KCL Equations with Node Voltages

03

32

4

2

2

12 =−

++−

Rvv

Rv

Rvv

03

23

5

3

1

13 =−

++−

Rvv

Rv

Rvvsvv =1

3 Unknown Node Voltages 3 Equations

At Node 1 At Node 2

At Node 3

KCL

1133

22

Solution of Equations

Solving KCL equation as Standard form# of Unknown Node Voltage = # of Equation

2222121

1212111

ivgvgivgvg

=+=+

3333232131

2323222121

1313212111

ivgvgvgivgvgvgivgvgvg

=++=++=++

With 2 Unknown Node Voltages With 3 Unknown Node Voltage

Generally

ijij ivg =

Solution of Linear Simultaneous Equation gives Node Voltage

Calculation of Remaining Node Voltages & Currents

Tips on KCL Equation Writing

03

32

4

2

2

12 =−

++−

Rvv

Rv

Rvv

At Node 2

- Writing KCL Node not Connected to Voltage Source (note : Supernode) - Direction of Current from Node : Positive

(Node Voltage of Interest – Node Voltage Surrounding ) / Resistance = Current Output from Node of Interest to Node Surrounding- Sum of All Currents from Surrounding Nodes

Node 2 Node 1 Node 2 Ground Node 2 Node 3

02

21

1

1 =+−

+ siRvv

Rv 0

4

32

3

2

2

12 =−

++−

Rvv

Rv

Rvv

siRvv

Rv

=−

+4

23

5

3

KCL Equations with a Current Source

v2-v1 v2-v3

Node 1 Node 3Node 2

KCL Equations with 2 Current Sources

Node 1 02

21

1

31 =−−

+−

aiRvv

Rvv

Node 2 04

32

3

2

2

12 =−

++−

Rvv

Rv

Rvv

Node 3 01

13

4

23

5

3 =+−

+−

+ biRvv

Rvv

Rv

Different choice of Reference Node

Node 1 010520

21131 =−

++− vvvvv

Node 2 05

1010

3212 =−

++− vvvv

Node 3 051020

23313 =−

++− vvvvv

VvVvVv

45.4573.7227.27

3

2

1

−=−=−=

Aix 909.0=

Same Logic regardless of Reference Node

KCL Equations with Voltage & Current Sources

Node 1 01210

5121 =−−

+− vvv

Node 2 0510

105

1222 =−

+−

+vvvv

VvVv 129.632.10 21 ==

Same Logic for Current & Voltage Sources Containing Circuit

Node 3 : Not Writing KCL on , because of Voltage Source

Circuits with Voltage Sources

( ) ( ) 01515

3

2

4

2

1

1

2

1 =−−

++−−

+R

vRv

Rv

Rv

“The Net Current flowing through any closed surface must equal to Zero”Supernode : Combination of Node including Voltage Sources

If Write KCL equations with All Node & Supernodes

When a Voltage Source is connected between Nodes, KCL can not be written at Nodesbecause both Current & Resistance are not defined in Voltage Source Above Circuit can not Write KCL, because All node are connected to Voltage Sources

( ) ( ) 01515

3

2

4

2

1

1

2

1 =−−

−−−−

−−R

vRv

Rv

Rv

Supernode 1 Supernode 2

Supernode 1

Supernode 2Dependant EqnsUse KVL at Supernode

( ) ( ) 01515

3

2

4

2

1

1

2

1 =−−

++−−

+R

vRv

Rv

Rv

010 21 =+−− vv

Solution by Supernode

KCL

KVL

Supernode

Independent Equation!! Solvable

at Supernode

Write KVL at Supernode Write KCL for Super node

010 21 =++− vv

13

32

2

31

1

1 =−

+−

+R

vvR

vvRv

04

3

3

23

2

13 =+−

+−

Rv

Rvv

Rvv

14

3

1

1 =+Rv

Rv

Supernode Example

KVL at Supernode

KCL around Supernode

KCL at Node 3

KCL at Ground

Same Equation!

Node-Voltage Analysis with a Dependent Source

First, write KCL equations at each node, including the current of the controlled source just as if it were an ordinary current source.

xs iiR

vv 21

21 +=−

03

32

2

2

1

12 =−

++−

Rvv

Rv

Rvv

024

3

3

23 =++−

xiRv

Rvv

1

3

2

Substitution yields

3

23

1

21 2R

vviR

vvs

−+=

−

03

32

2

2

1

12 =−

++−

Rvv

Rv

Rvv

023

23

4

3

3

23 =−

++−

Rvv

Rv

Rvv

Next, find an expression for the controlling variable ix in terms of the node voltages.

3

23

Rvvix

−=

Solvable!

Example with Dependant Voltage Source

050 21 =++− vv.v x 13 vvvx −=Supernode (Node 1 & 2) KVL

KCL

01

13

3

23

4

3 =−

+−

+R

vvR

vvRv

siRvv

Rvv

Rv

=−

+−

+3

32

1

31

2

1

Node 3

Ground siRv

Rv

=+4

3

2

1311 ,, vvvforSolution

Mesh-Current Analysis

1. Define the Mesh Currents (usually a clockwise direction)If necessary, redraw the network without crossing conductors or elements.

2. Write Network equations ( # of equations = # of mesh currents).

- Use KVL for meshes with no current sources. - If current sources, write expressions for their currents in terms of the mesh currents. - If common current source to two meshes, Use KVL for Supermesh.- If dependent sources,

Find expressions for the controlling variables Substitute into the network equations,Obtain equations having only the mesh currents as unknowns.

3. Solve for the mesh currents (Standard form equation)

4. Calculate any other currents or voltages of interest.

Basic Rule for Mesh Current

KVL for vA, R1 & R3 AviRiR =+ 3311

KVL for vB, R2 & R3 BviRiR −=+− 2233

KCL for Node 1321 iii +=

Node-Voltage Analysis Rearrange

AviiRiR =−+ )( 21311

BviRiiR =+−− 22213 )(

When Several Mesh Currents flow through one Elements,the Current in that Element to be Algebraic Sum of the Mesh Current

Choosing the Mesh Currents

Only for Planar Network : Network without Crossing Element

Mesh : Loop of Elements Mesh Current : Current flowing through MeshDefine Current Direction Clockwise for ConsistencyExample, Current flow in R2 to Left = i3 - i1

Current flow in R3 to Up = i2 - i1

Writing Equations to Solve for Mesh Currents

If Network with only resistances and independent voltage sources,write required equations by following each current around its mesh &applying KVL.

( ) 024123 =++− BviRiiR

( ) 031132 =−+− BviRiiR

( ) ( ) 021312 =−−+− As viiRiiRMesh 1

Mesh 2

Mesh 3

( ) ( ) 021441211 =−−+−+ AviiRiiRiR

( ) ( ) 032612425 =−+−+ iiRiiRiR

( ) ( ) 043823637 =−+−+ iiRiiRiR

( ) ( ) 034814243 =−+−+ iiRiiRiR

Example of Mesh Current with Voltage Source

Solution of Network Equation & Find Voltage at Node

Example of Mesh-Current Analysis

Mesh 1 20i1 + 10(i1 – i2) – 150 = 0

Mesh 2 10(i2 – i1)+ 15 i2 + 100 = 0

30i1 - 10(i1 – i2) = 150 -10 i1+ 25 i2 = -100

In standard form i1 = 4.231Ai2 = -2.308A

Current through R2 downward = i1 – i2 = 6.539A

Mesh Currents in Circuits with Current Sources

A common mistake made by beginning students isto assume that the voltages across current sources are zero.

A21 =i

0105)(10 212 =++− iiiKVL to Mesh 2

Combine meshes 1 and 2 into a Supermesh. Write KVL equation around the periphery of meshes 1 & 2

( ) ( ) 01042 32311 =+−+−+ iiiii( ) ( ) 0243 13233 =−+−+ iiiii

512 =− ii

Mesh Currents with Voltage & Current Sources

KVL for Mesh 1 & 2 is impossible,because Voltage across 5A is unknown

Mesh 3

Current Source 5A

( ) 0100510 212 =−+− iii

Ai 51 −=

01020105 21 =−++ ii

Aii 112 =−

Supermesh

Example of Mesh-Current Analysis

026420 221 =+++− iii

124iivx −=

22ivx =

Mesh-Current with Dependant Source

Supemesh

Dependant Source

- Write Equations exactly the same as Independent Source- Express Dependant Variable with Mesh Current- Substitute into Network Equation

Thévenin Equivalent Circuits

=

Real Complex CircuitEasy UnderstandableEquivalent Circuit

a

a

b

a

b

Thévenin Equivalent Circuits

ocvVt =

sc

oc

ivRt =

Example of Thévenin Equivalent Circuits

ARR

vi s 1.021

1 =+

=

ViRvoc 51.05012 =×==

ARvi s

sc 15.010015

1

===

Ω=== 3.3315.05

sc

oct i

vR

To find Open Circuit Voltage To find Short Circuit Current

Ω=+

=+

= 3.3350100

5000

501

1001

1eqR Rt equals Req with Zero Voltage!

Example of Thévenin Equivalent Circuits

Aioc 5=

ViRvoc 5051011 =×==

ARR

RAisc 1521

1 =+

=

Ω=== 50150

sc

oct i

vR

To find Open Circuit Voltage To find Short Circuit Current

ioc

voc i1

isc

Req = 10 + 40 = 50Ω Rt equals Req with Zero Current!

Finding the Thévenin Resistance Directly

If No Dependant Source,When zeroing a Voltage source, it becomes a Short circuitsWhen Zeroing a Current Source, it becomes a Open Circuit

In Zeroing Independent Source, Replace a Voltage Source with Short Circuit Replace a Current Source with Open Circuit.

When the Source is Zeroed, the resistance seen from the Circuit terminals is equal to the Thévenin Resistance

Short Circuit

Direct Finding the Thévenin Resistance

b) Zeroing SourceΩ=

+=

+= 4

201

51

111

1

21 RR

Req

c) Short Circuit

02 =isciii +=+ 21 2 A

Rvi s 4

520

11 === AiSC 6=

ViRV sctt 2464 =×==

Exercise 2.24

Find Thévenin Resistance

Zero

Rt=30Ω

Zero

Ω=+

+= 14

201

51

110eqR

Ω=+

+= 10

201

51

161eqR Ω=102

eqR Ω=+

= 5

101

101

1eqR

Zero Zero

Thévenin Resistance with Dependant Source

102 oc

xxvii =+

Can not use Zeroing Source Technique due to dependant source

Open Circuit : Node-Voltage Tech

KCL at Node 1KVL Loop ocx vi −=105

Vvoc 57.8=

Short Circuit

AAix 25

10== Aii xsc 63 ==

Ω=== 43.1657.8AV

ivR

sc

oct

Norton-Equivalent Circuit Analysis

Norton Equivalent Circuit : Independent Current Source +Thévenin Resistance

In is isc in Thévenin Equivalent Analysis

Step-by-step Thévenin/Norton-Equivalent-Circuit Analysis

1. Perform two of these:a. Determine the open-circuit voltage Vt = voc.b. Determine the short-circuit current In = isc.c. Zero the sources and find the Thévenin resistance Rt

looking back into the terminals.2. Use the equation Vt = Rt In to compute the remaining value.3. Thévenin equivalent : a voltage source Vt in series with Rt .4. Norton equivalent : a current source In in parallel with Rt .

0154 321

=+

+−

+RR

vR

vv ococxKCL at Node

KVL at Loopococx vv

RRRv 25.0

32

3 =+

=

Vvoc 62.4=

Current along R2 & R3 is Zero : vx = 0

ocs

sc vRvi 75.0

2015

1

===

Ω=== 15.675.062.4

sc

oct i

vR

Example of Norton-Equivalent-Circuit Analysis

Maximum Power TransferThe load resistance that absorbs the maximum power from a two-terminal circuit is equal to the Thévenin resistance.

Lt

tL RR

Vi+

= LL RipL

2=( )2

2

Lt

LtL RR

RVp+

=

( ) ( )( )

024

222

=+

+−+=

Lt

LtLtLtt

L

L

RRRRRVRRV

dRdp

At Power Maximum

Lt RR =t

tmaxL R

Vp4

2

=

SUPERPOSITION PRINCIPLE

The total response is the sum of the responses to each of the Independent Sources acting individually with Other

Independent Source Zeroed. When Zeroed, Current Source become Open and Voltage Source become short.

nT rrrr +++= .......21

In equation form, this is

In Circuit with Resistance, Linear Dependant Source, Independent Source

Superposition Principle

Response : Voltage or Current

221

1sx

TsT iKiRv

Rvv

=++−

2Rvi T

x =

2121

211

121

2ssT i

KRRRRRv

KRRRRv

+++

++=

1121

21 sv

KRRRRv++

=

2121

212 siKRRR

RRv++

=

Example of Superposition PrinciplevT is Interesting

KCL at Top

If is2 is set to Zero, Response due to vs1 :

If vs1 is set to Zero, Response to is2 :

21 vvvT +=In General, dependant sources do not contribute a separate term to the total

response , and we must zero dependant source in applying superposition

WHEATSTONE BRIDGE

The Wheatstone bridge is used by mechanical and civil engineers to measure the resistances of strain gauges in experimental stress studies of machines and buildings.

31

2 RRRRx =

Wheatstone Bridge

42

4

31

3

RRVRV

RRVRV ba +

=+

=

( ) ( )4231

4123

42

4

31

3

RRRRRRRRV

RRVR

RRVRV

+⋅+−

=

+−

+=∆

4123@ RRRRVoltageZero =

ba VVV −=∆ Va = Voltage @ aVb = Voltage @ b

DC Wheatstone bridgeV : Bridge Supply VoltageD : Voltage Detector

GTh

ThG RR

VI+

=

42

42

31

31

RRRR

RRRRRTh +

++

=

( )( )4231

4123

RRRRRRRRVVTH ++

−=

Bridge Circuits(Galvanometer)

Galvanometer BridgeV : Bridge Supply VoltageG : Voltage Detector

G

Lead Compensation for Remote Sensor Contact Resistance of each Resistor

Lead Compensation

( ) ( )L

L

RRRRRRRRRRVV2

)2(

4231

4123

++⋅++−

=∆

LRRR 244 +→Lead Problem

- RL makes ∆V change- Environmental Effect

Temperature, Stress,Vapor- Transient effect

( ) ( )4231

4123

RRRRRRRRVV+⋅+

−=∆

RL

RL

( ) ( )4231

4123

RRRRRRRRVV+⋅+

−=∆

g

Lead Compensated Circuit

Lead Compensation by Power Lead ExtensionConnection (3) : g→c

No Change on ∆VConnection (1) & (2) : R3 → R3 + RL , R4 → R4 + RL

Same Condition : R3 & R4 are identically changedFinally No Change in Bridge

RL

RL

RL

( ) 54254 & RRRRR >>+>>

( )5

542

54 IRRRR

RRVVb +++

+=

( )5

542

54

31

3 IRRRR

RRVRR

VRV −++

+−

+=∆

Current Balance Bridge

1. Splitting R4→ R4 +R52. Current on R5 <<R4 to make ∆V null

by adjusting magnitude and polarity of current I manually or electronically

To make null bridge circuit

axC VVV +=

baxbC VVVVVV −+=−=∆

042

4

31

3 =+

−+

+RR

VRRR

VRVx

( ) 05542

54

31

3 =−++

+−

++ IR

RRRRRV

RRVRVx

Potential Measurement

Make ∆V zero (null condition)

Or current balance combination

05 =− IRVx

Bridge off-null voltage ∆V(a) nonlinear for large-scale changes in resistance(b) nearly linear at small ranges of resistance change

Linearity of Bridge Circuit

( ) ( )4231

4123

RRRRRRRRVV+⋅+

−=∆