Lecture 2.ppt

SCHOOL OF ENGINEERING

Introduction to Electrical Systems

Chapter 2 – Resistive Circuits

1

Resistive CircuitsSeries Resistances

• A series combination of resistance has an equivalent resistance equal to the sum of the original resistances

KVL: v=v1 +v2 +v3 = R1 i+ R2 i+ R3 i = (R1+R2+R3 )i= Req i

Req = R1+R2+R3

2

• A parallel combination of resistance has an equivalent resistance equal to the sum of the reciprocals of original resistances

KCL: i=i1 +i2 +i3 = v/R1+ v/R2 + v/R3 = (1/R1+1/R2+1/R3 ) v= 1/Req v

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

KCL: i=i1 +i2 +i3 = Geq v (G – conductance, [S])

1 Geq = G1+G2+G3

Resistive CircuitsParallel Resistances

3

Network analysis is the process of determining the current, voltage, and power for each element given the circuit diagram and the element values. The steps are:1.Begin by locating a combination of resistances that are in series or parallel. The best place to start is usually farthest from the source.2.Redraw the circuit using 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 (but not always) we end up with single source and a single resistance.4.Solve for the current and voltages in the final equivalent circuit. Then, transfer results back each step and solve for all unknown currents and voltages. Again, transfer the results back each step and solve. 5.Repeat until all of the currents and voltages are known in the original circuit.

Resistive CircuitsNetwork Analysis by using Series and

Parallel Equivalents

4

Example: Ladder Resistor Networks

5

Resistive CircuitsNetwork Analysis by Using Series and Parallel

Equivalents - 1

6

Resistive CircuitsNetwork Analysis by Using Series and

Parallel Equivalents - 2

7

.1;2;3

3

2

1

AiAiAi

Resistive CircuitsNetwork Analysis by Using Series and

Parallel Equivalents - 3

Transfer results back

v2=Req1i1= 20 3A = 60V

i2=v2/R2= 60V/30 = 2A

i3=v2/R3= 60V/60 = 1A

v1=R1i1=10 3A = 30V

ps=-vsi1 =-(90) 3A = -270 W(opposite to the passive configuration)

p1= R1i12 =10(3A)2 =90 W

p2= v22/R2=(60V)2/30 = 120 W

p3= v22/R3=(60V)2/60 = 60 W

ps+ p1 + p2 + p3 =0 – Energy conservation8

Resistive CircuitsVoltage Divider - 1

Principle of voltage division:

Req = R1+R2+R3

i=vtotal /Req

v1=R1 i = R1 (vtotal /Req )

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



9

total321

111 v

RRRRiRv

total321

222 v

RRRRiRv


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v

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v

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12.9;3

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4

3

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VvVv

VvVv

Resistive CircuitsCurrent Divider

Principle of current division:

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

Req = (R1 R2)/(R1 + R2)v=Req itotal

For two (only) resistances is parallel, the fraction of the total current flowing in a resistance is the ratio of the other resistance to the sum of two resistances. [If more than two in parallel, they must be combined only to two in the circuit.]

i2=v/R2 = [R1/(R1 + R2)] itotal

i1=v/R1 = [R2/(R1 + R2)] itotal

13total

total

iGG

Gi

iGG

Gi

21

22

21

11 ;

Application of the Current Division Principle

2060306030

32

32eq RR

RRR A10152010

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eq1

eq1

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total21

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11 i

RRR

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total21

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22 i

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total

total

total

iGGG

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Gi

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321

33

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11

;

;

Resistive CircuitsCurrent and Voltage Divider Example

15

Resistive Circuitsposition transducer potentiometer (Voltage Divider Example)

K is sensitivity of the device [ Volts / degree ]

16

A transducer produces a voltage proportional to a physical quantity of interest(such as distance, pressure, temperature, … )

Resistive Circuits Analysis• Series/parallel equivalents and the current/voltage division principles are not sufficient to solve all circuits - the general approach is to apply Kirchhoff equations• Example: 6 unknowns; 3 KVL and 3 KCL equations(below is the same circuit drawn in a different way)• Systematic methods simplify equations compilation andreduce the linear equations system order• Those systematic methods are:

Node Voltages analysis (Метод Узловых Потенциалов)Mesh Currents analysis (Метод Контурных Токов)

17

Resistive CircuitsNode Voltage Analysis - 1

Example: (note positive polarity at the head of the arrow)

Variables: node voltages -V2 and V3

For unknown Vx , using KVL for the loop with the unknown-V2 + Vx + V3=0 Therefore, the voltage across any floating element Vx = V2 - V3 is the difference between node voltages - Vy = V2 – V1 ; Vz = V3 – V1

• Ohm’s law is used to find currents when node voltages are known

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Reference node


• To find the current flowing out of node n through a resistance towards node k, we subtract the voltage at node k from voltage at node n and divide the difference by the resistance between the nodes

• We apply KCL, adding all the currents leaving (entering) node n and setting the sum to zero

• Repeat the same for all independent nodes in the circuit

• Straightforward for numerous current sources and “grounded” voltage sources (common reference node)

• More tricky if there are “floating” voltage sources (supernode)

19


siRvv

Rv

2

21

1

1

04

32

3

2

2

12

Rvv

Rv

Rvv

siRvv

Rv

4

23

5

3

Three equations, three unknowns

20

s

s

ivGGvGvGvGGGvG

ivGvGG

35424

34243212

22121

)(;0)(

;)(

s

s

i

i

vvv

GGGGGGGG

GGG0

0

0

3

2

1

544

44322

221

Node 1:

Node 2:

Node 3:

We use KCL to write an equation at each node.

IG V

IGV1

Matrix form:


21

s

s

i

i

vvv

GGGGGGGG

GGG0

0

0

3

2

1

544

44322

221

IG V

IGV1

3

2

1

3

2

1

333231

232221

131211

iii

vvv

ggggggggg

Circuit equation in Standard Form:

1. Circuit must contain only resistances and independent current sources 2. Diagonal terms of G sum of conductances connected to corresponding node3. Off diagonal terms of G negative conductance between connected nodes


svv 1

03

32

4

2

2

12

Rvv

Rv

Rvv

03

23

5

3

1

13

Rvv

Rv

Rvv

Two equations, two unknowns

22

12332342 )( vGvGvGGG

11335123 )( vGvGGGvG

Node 1:

Node 2:

Node 3:

11

12

3

2

3513

3342

vGvG

vv

GGGGGGGG

IG V

IGV1

Matrix form:

Resistive CircuitsSolving the Network Equations

• Once we have written the equations needed for the node voltages we put equations into standard form

• We group the node-voltage variables on the left-hand side of the equations and place terms that do not involve the node voltages on the right-hand sides

• Then we can solve for the node voltages by variety methods, such as substitution and determinants, MATLAB etc.

• I recommend on-line solver WIMS that is capable of making parametric (symbolic) calculations –

http://wims.unice.fr/wims/wims.cgi?session=Q753BAD623.1&+lang=en&+module=tool%2Flinear%2Flinsolver.en&+method=matrix&+cmd=resume

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Resistive CircuitsSolving the Network Equations – WIMS - 1

• WIMS numerical example – insert numbers and press “Solve”

• Check the system – is it indeed what you want to solve?

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• WIMS parametric (symbolic) – example 1

• Check the system – is it indeed what you want to solve

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• WIMS parametric (symbolic) – example 2

• A powerful tool

• There may be no solution at all

• If there is no unique solution – infinite set of solutions (matrix A rank is less than its size), WIMS will prompt about the structure of the solutions’ family

26

Resistive CircuitsSolving the Network Equations

• Example

Solve using WIMS or Matlab 27

Resistive CircuitsNode Voltage Analysis - Circuits with Voltage Sources - 1

• For this circuit it is impossible to write a current equation in terms of the node voltages for every node because of the floating voltage source• The circuit requires to form a supernode

• The supernode is formed by drawing a dashed line around several nodes, including the elements (voltage sources) connected between them• The modified KCL for supernodes: The net current flowing through any closed surface (enclosed by dash lines) must equal zero

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• The modified KCL for supernodes: The net current flowing through any closed surface (enclosed by dash lines) must equal zero.• Note v3 = -15 V because node 3 connected to the negative terminal of the source

• Then, for the supernode enclosing the 10-V source, we sum currents leaving the supernode surface through one of the resistors

01515

3

2

4

2

1

1

2

1

R

vRv

Rv

Rv

29


• Note, we obtain linearly dependent equations if we use all the nodes in writing current equations, i.e., if we would use current equations for both supernodes (they comprise all 4 nodes of the circuit) • To avoid dependency we can use KVL (clockwise sense) to the loop that include the voltage source

• These two equations form an independent set that can be used to solve for v1 and v2

1012 vv

01515

3

2

4

2

1

1

2

1

R

vRv

Rv

Rv

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Exercise 2.13 (Page 91 - 92)

• To solve for three voltages any two of the following three KCL equations can be used (one supernode and two normal nodes, including the reference node)

• Node 3 equation

• Supernode (1,2) equations

13

32

2

31

1

1

R

vvR

vvRv

04

3

3

23

2

13

Rv

Rvv

Rvv

Referencenode

31

1021 vv

• If node 1 or 2 is selected as ground (reference), the solution does not require supernode

Resistive CircuitsNode Voltage Analysis - Circuits with Dependent Sources - 1

• Write KCL equations at each node, including the current of the controlled source the same as if it were an ordinary current source (independent node):

• Then use one additional equation for the dependent source current value ix in terms of node voltage

xs iiR

vv 21

21

03

32

2

2

1

12

Rvv

Rv

Rvv

024

3

3

23

xiRv

Rvv

3

23

Rvvix

32

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

Resistive CircuitsNode Voltage Analysis - Circuits with Dependent Sources - 2

Three equations, three unknowns

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Resistive CircuitsMesh Current Analysis - 1 (Метод Контурных Токов)

• Applying KVL to the loops with normal branch currents:

Applying KCL to the node:

Combined equations:

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123 iii

Resistive CircuitsMesh Current Analysis - 2

• If a network contains only resistances and independent voltage sources, we can write the required equations by following each current around its mesh and apply KVL• i1 and i2 are mesh currents, normally chosen to flow clockwise• When several mesh currents flow through one element, we consider the current in that element to be algebraic sum of the mesh currents • The current in R3 (referenced downwards) is i3 =i1 - i2 and v3 = R3 i3

• You can write down mesh current equations rightaway as

35

B

A

vv

ii

RRRRRR

2

1

323

331


• Mesh current analysis advantages seen for more complex networks

• Note actual current direction in the common resistor for two meshes

• For example, if current in R2 referenced to the right, then i2 is algebraic sum i1 - i3, if to the left, then i3 - i1

• After solving for mesh currents, actual currents may take negative actual values for selected reference directions

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Using this pattern for mesh 1

For mesh 2, we obtain: 024123 BviRiiR

For mesh 3, we have: 031132 BviRiiR

0213312 AviiRiiR 021441211 AviiRiiRiR

032612425 iiRiiRiR

043823637 iiRiiRiR

034814243 iiRiiRiR


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0 100 i 15 i- (i10

0 i- (i10 i20 150 -

212

211

)

)

100- i 25 i10 -

150 i10 - i30

21

21

What is the total current flowingthrough the 10 resistor ?

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A 2.31- iA 4.23 i

2

1

B

A

vv

ii

RRRRRR

2

1

323

331


0 100 i 15 i- (i10

0 i- (i10 i20 150 -

212

211

)

)

100- i 25 i10 -

150 i10 - i30

21

21

A 2.31- iA 4.23 i

2

1

39

A92.12.31- 4.23 i -ii 213

B

A

vv

ii

RRRRRR

2

1

323

331


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You can write down mesh current equations rightaway –

000

00

00

4

3

2

1

83282

88766

66544

24421 Av

iiii

RRRRRRRRRR

RRRRRRRRRR

Resistive CircuitsMesh Current Analysis with Current Sources - 1

• Classic mesh current analysis – voltage sources• Analysis for current sources is dual for node voltage analysis with voltage sources• Current source in an individual mesh explicitly defines respected mesh current (dual to a grounded voltage source that defines a node voltage)

• Example – mesh 1 current is equal to current source current

• KVL is applied to mesh 2

A21 i

0105)(10 212 iii41

Resistive CircuitsMesh Current Analysis with Current Sources - 2

• How do we write KVL for meshes 1 and 2 that have common current source?• Supermesh – combination of meshes 1 and 2 • KVL is applied to the supermesh first• Then KVL is applied to mesh 3 • Then for the current source :

• Finally, we can define all of the mesh currents from these equations

01042 32311 iiiii

0243 13233 iiiii

521 ii

42

026420 221 iii

421xvii

Resistive CircuitsMesh Current Analysis - Controlled Current Source Example

2/ 21 ii

A 2 i A 1 i

2

1

and

43

• Supermesh equation•

• Finally, we can define all of the mesh currents from these equations

22ivx

Resistive CircuitsThevenin Equivalent for Two-Terminal Circuits

• Two-terminal (single-port) circuit is one (that can be of any complex interconnections of resistances and sources) that has only two points or nodes that can be connected to other circuits

• The Thevenin equivalent of such circuits is one that consists of only an independent voltage source in series with a resistance

44

Resistive CircuitsThevenin Equivalent for Two-Terminal Circuits

• The Thevenin equivalent with open-circuited terminals has no current flowing through the circuit, therefore

• The Thevenin equivalent with short-circuited terminals has resistance value is the ratio of open-circuitvoltage Voc of the original circuit to its short-circuit current, isc

ocvVt

sc

oc

ivRt

45

Resistive CircuitsThevenin Equivalent for Two-Terminal

Circuits Example

A 0.1 50 100

15 R R

v i21

s1

volts 5 i R V V i2toc

A 15 0. 100v 15

Rv

1

s

sci

33.33 A 0.15

volts 5 iV R

sc

oct

46

• When zeroing a voltage source, it becomes a short circuit. When zeroing a current source, it becomes an open circuit.

• We can find the Thévenin resistance by zeroing the sources in the original network and then computing the resulting resistance between the terminals.

Resistive CircuitsFinding Thevenin Resistance for Two-Terminal Circuits

Directly [1]

47

Resistive CircuitsExample: Thevenin Equivalent with a

Dependent Source• We use node - voltage analysis for the open-circuit voltage (not direct)

• and

• Short-circuit

Voc = 8.57 v

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Resistive CircuitsNorton Equivalent for Two-Terminal Circuits

• The Norton equivalent consists of an independent current source In

in parallel with the Thevenin resistance

• If we zero the Norton current source (disconnecting it), the Norton equivalent becomes a resistance because if we zero the voltage source in the Thevenin equivalent (by short circuiting) it also becomes a resistance, that is equivalent to the internal resistance of the original circuit.

49

Resistive CircuitsNorton Equivalent for Two-Terminal Circuits

• If we place a short circuit across the Norton equivalent, the Norton current becomes equal to the short-circuit current In = iSC

Perform TWO of the following steps:1. First determine the open-circuit voltage Vt =vOC

2. Next determine the short-circuit current In = iSC

3. Zero the current source and find the Thevenin resistance, looking back into the terminalsUse Vt = Rt In to compute the remaining value

50

Voc = 4.62 volts

isc = 15 v / 20 = 0.75 A

Vx = 0 volts

Resistive CircuitsNorton Equivalent for Two-Terminal Circuits Example

6.15 A 0.75

volts4.62 IV R

SC

OCTH

0 20V

2015 - V

4V ocOCx KCL

15

V - V 5

V xOCx

0 20

V 20

15 - V 16V OCOCOC

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Source Transformations

i1 = 10 V / 15 = 0.67 A

i2 = 5 A * 5 / 15 = 1.67 A

i3 = 10 A * 5 / 15 = 3.33 A

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Resistive CircuitsMaximum Power Transfer

• Question: What load resistance RL should be connected to a two-terminal circuit in order to maximize the power delivered to the load?

• The current through the resistor • The power delivered• Derivative of the power with respect to resistance

• Solution is• Actual maximum power is• The load resistance that absorbs the maximum power from a two-terminal

circuit is equal to the Thévenin resistance.

53

Resistive CircuitsSuperposition Principle

• Suppose we have a circuit composed resistors, linear dependent (which follow the linear function) sources and n independent sources• The current flowing in each element is a response to the independent sources• Consider zeroing all the independent sources (current sources become open circuits and voltage sources become short-circuited) except the first source• The response ( while sources are zeroed) for that source is then r1 (could be either current or voltage response)• If we keep only the second source the response becomes r2

• We can repeat the process for each source in the circuit until response to the nth source rn is obtained • The superposition principle states that he total response is the sum of the responses due to each of the independent sources acting individually:

nT rrrr 21

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Resistive CircuitsSuperposition Principle Example

1.Only voltage source active (apply voltage division principle)

2. Only current source active (resistors in parallel)

3. Voltage across due to the current source

4. Adding the individual responses

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Resistive CircuitsWheatstone Bridge

• Wheatstone bridge is a circuit to measure unknown resistances

•The bridge is capable of responding to a very small currents (less than 1 µA) through a detector resistance Rx. Resistors R2 and R3 are adjustable and can be tuned until the detector indicates zero current and no voltage between terminals a and b.

• In this conditions we say the bridge is balanced

56

Resistive CircuitsWheatstone Bridge

• In balanced condition (ig =0, vab = 0) by applying KCL at nodes a and b respectively KVL applied around the loop R1, R2 and detector and since vab = 0

• Similarly KVL around R3, R4 and detector

•Dividing one equation by the other we obtain the resistors ratio for the balanced bridge:

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Summary: Step-by-step Thévenin/Norton-Equivalent-Circuit

Analysis1. 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. Cannot zero dependent sources.2. Use the equation Vt = Rt In to compute the remaining value.3. The Thévenin equivalent consists of a voltage source Vt in series with Rt.4. The Norton equivalent consists of a current source In in parallel with Rt.

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Lecture 2.ppt

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