enz - resistive circuits · 2014-12-09 · ecet11 kirchhoff’s current law kirchhoff’s current...

ECET11ECET11

Resistive Circuits:

- Ohm’s Law

- Kirchhoff’s Laws

- Single-Loop Circuits

- Single-Node Pair Circuits

Resistive CircuitsResistive Circuits

1

- Single-Node Pair Circuits

- Series Circuits

- Parallel Circuits

- Series-Parallel Circuits

Enzo Paterno

ECET11ECET11

Material that opposes current flow is called resistance and is Material that opposes current flow is called resistance and is measured in Ohms [measured in Ohms [ΩΩ]. An electronic device that is purely resistive is called a ]. An electronic device that is purely resistive is called a resistor. A resistor is a passive element. Most popular and least expensive resistor. A resistor is a passive element. Most popular and least expensive resistors are normally carbon composition.resistors are normally carbon composition.Ohm’s law states that the voltage across a resistance is directly proportional to Ohm’s law states that the voltage across a resistance is directly proportional to the current flowing through it.the current flowing through it.

OHM’s LAWOHM’s LAW

0),()( ≥= RwithtRitv

2

A

V

ti

tvR

1

11

)(

)(=Ω⇒=

R

tvti

)()( =

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ECET11ECET11

The power supplied to the terminals is absorbed by the resistor and The power supplied to the terminals is absorbed by the resistor and the energy absorbed is dissipated in the form of heat.the energy absorbed is dissipated in the form of heat.

OHM’s LAWOHM’s LAW

Another important quantity is conductance, G, with units of Siemens Another important quantity is conductance, G, with units of Siemens [S]. Conductance is the reciprocal of resistance and used extensively in the [S]. Conductance is the reciprocal of resistance and used extensively in the analysis of parallel circuits.analysis of parallel circuits.

GR

RG

11=→=

V

AS

1

11 = G

titv

)()( =

)()( tGvti =

3

the energy absorbed is dissipated in the form of heat.the energy absorbed is dissipated in the form of heat.

)()()( tvtitp =

G

titptRitp

)()()()(

22 =⇒=

)()(

)( 22

tGvR

tvtp ⇒=

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ECET11ECET11OHM’s LAWOHM’s LAW

Let us look at a simple circuit with a variable resistor (i.e. Let us look at a simple circuit with a variable resistor (i.e. potentiometer).potentiometer).

0=∴↓ RRLet∞=∴↑ RRLet

4

0)()( == tRitv

Short circuitShort circuit

0)(

)( ==R

tvti

Open circuitOpen circuitEnzo Paterno

ECET11ECET11

Given current and resistanceFind the voltage

AI 2= +

Given Current and VoltageFind Resistance

−

+

][20 V

][4 AI =

I

VR =

Ω= 5R

RIV =

OHM’s LAW EXAMPLESOHM’s LAW EXAMPLES

AI 2=

Ω= 5R

−

=

+

][10 VV

IR =

Given Voltage and ResistanceCompute Current

−

+

][12 V Ω= 3R

R

VI =

][4 AI =

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ECET11ECET11

−

+

Determine I and PDetermine I and PRR

OHM’s LAW EXAMPLEOHM’s LAW EXAMPLE

mA6=

R

VRIVIP

22 ===

])[6])([12( mAVP = ][72 mW=

Enzo Paterno

ECET11ECET11

VP S

2

=

Ω==

k

V

R

VI

10

][6][6.0 mADetermine VDetermine VSS and Iand I


R

VP S=

)106.3)(1010( 332 WVS−×Ω×=

][6 VVS =

Enzo Paterno

ECET11ECET11

Determine VDetermine VSS and Pand PRR


G

IVIRV SS =⇒=

][1050

][105.06

3

S

AVS −

−

×

×= ][10 V=

G

IRIP

22 == ( )

=×

×=

−

−

][1050

][105.06

23

S

AP ][105.0 2

W−×

][5 mW

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ECET11ECET11

RIP 2=

Determine R and VDetermine R and VSS


RIP 2=

( )23

3

104

][1080

A

WR

−

−

×

×=

Ω= kR 5

IVP S=

][4

][80

mA

mWVS = ][5 V=

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ECET11ECET11

A A nodenode is a point of connection of two or more circuit elements.is a point of connection of two or more circuit elements. A A looploop is any closed path through the circuit in which a particularis any closed path through the circuit in which a particular

node is not encountered more than once.node is not encountered more than once. A A branchbranch is a portion of a circuit containing only at least oneis a portion of a circuit containing only at least one

element and the nodes at each end of that element.element and the nodes at each end of that element.

NODES, LOOPS, BRANCHESNODES, LOOPS, BRANCHES

nodesnodes 5 5 1,2,3,4,51,2,3,4,5

10Enzo Paterno

loopsloops 9 9 (1(1--33--4), (14), (1--44--55--3)3)(1(1--22--3), (13), (1--22--55--3)3)(1(1--22--55--4), (24), (2--33--5), 5), (2(2--33--44--1), (21), (2--33--44--5),5),(3(3--44--5)5)

branchesbranches 8 8 (1(1--2), (12), (1--3), (13), (1--4)4)(2(2--3), (23), (2--5)5)(3(3--4), (34), (3--5)5)(4(4--5) 5)

ECET11ECET11KIRCHHOFF’s CURRENT LAW KIRCHHOFF’s CURRENT LAW -- KCLKCL

The algebraic sum of the currents entering and leaving any node is zeroThe algebraic sum of the currents entering and leaving any node is zeroAssume currents entering the node have a positive sign and the currentsAssume currents entering the node have a positive sign and the currentsleaving the node have negative signs.leaving the node have negative signs.

The sum of the currents entering a node is equal to the sum of the currentsThe sum of the currents entering a node is equal to the sum of the currentsleaving that nodeleaving that node

0)(1

=∑=

N

j

j ti

11Enzo Paterno

Current iCurrent ijj(t) enters the node through branch j with a total of N branches(t) enters the node through branch j with a total of N branchesconnected to that node.connected to that node.

For node 3:For node 3:

)()()()(

0)()()()(

7452

7542

titititi

titititi

+=+

=−+−


A5

A3

?=XI

a

b

c

d

Sum of currents into node is zeroSum of currents into node is zero

0)3(5 =−++ AIA X

AI X 2−=

12Enzo Paterno

d

b

a

c

d

e2A

-3A4A

Ibe = ? ?

4

3

,2

=

=

−=

=

be

bd

cb

ab

I

AI

AI

AI 0423 =−−+− beIAAA

AIbe 5−=


Devise a strategy to determineDevise a strategy to determineII1 1 ,I,I44 ,I,I55 and Iand I66

02060:1 1 =−− mAmAI

0:2 614 =+− III

04060:3 45 =−−+ mAIImA

04020:4 5 =−+ ImAmA

13Enzo Paterno

1)1) Calculate ICalculate I55 from 4from 4thth equationequation2)2) Calculate ICalculate I44 from the 3from the 3rdrd equationequation3)3) Calculate ICalculate I11 from the 1from the 1stst equationequation4)4) Calculate ICalculate I66 from the 2from the 2ndnd equationequation

04020:4 5


mAmAmAmAI

mAmAmAI

mAIImA

70405060

503020

04060

4

5

45

=−+=

=+=

=−−+

mAmAmAI 802060 =+=

Node 3:Node 3:

Node 1:Node 1:

Determine IDetermine I44 ,I,I11 and Iand I66

14Enzo Paterno

mAmAmAI 8020601 =+=

mAI

III

III

106

416

641

=

+=

+=Node 2:Node 2:

ECETECET1111

mAI 501 −= mAmAmAIT 204010 ++=

1I FindTI Find

KIRCHHOFF’s CURRENT LAW KIRCHHOFF’s CURRENT LAW -- KCLKCL

mAI 501 −= mAmAmAIT 204010 ++=

0410 1 =−− ImAmA

01241 =−+ mAmAI03 12 =−+ ImAI1I Find

21 I and I Find

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ECET11ECET11KIRCHHOFF’s VOLTAGE LAW KIRCHHOFF’s VOLTAGE LAW -- KVLKVL

The algebraic sum of the voltages around any loop is zero.The algebraic sum of the voltages around any loop is zero. Kirchhoff’s voltage law is one of the fundamental conservation of energyKirchhoff’s voltage law is one of the fundamental conservation of energy

laws in electrical engineering laws in electrical engineering -- “energy cannot be created nor destroyed”“energy cannot be created nor destroyed”

Let VLet VR1R1 and Vand VR2R2 for the circuit below known to be 18 V and 12 V respectively.for the circuit below known to be 18 V and 12 V respectively.Find VFind VR3R3..

Enzo Paterno 16

Find VFind VR3R3..

015530 321 =−+−+− RRR VvVvVv

vvvvvVR 1218155303 −−+++=

vVR 203 =

ECETECET1111KIRCHHOFF’s VOLTAGE LAW KIRCHHOFF’s VOLTAGE LAW -- KVLKVL

The circuit below has three closed loops:The circuit below has three closed loops:

Left loop:Left loop:

Right loop:Right loop:

01624 41 =+−− vVVv RR

0816 324 =−−−+− vVVVv RRR

Enzo Paterno 17

Outer loop:Outer loop:

0816 324 =−−−+− vVVVv RRR

0824 321 =−−−−+ vVVVv RRR

Note that the outer loop equation is the sum of the left and right loop equations.Note that the outer loop equation is the sum of the left and right loop equations.These three equations are not These three equations are not linearly independentlinearly independent. Thus, only the first two . Thus, only the first two equations are needed to solve the voltages in the circuit.equations are needed to solve the voltages in the circuit.

ECET11ECET11EQUIVALENT FORMS FOR VOLTAGE LABELINGEQUIVALENT FORMS FOR VOLTAGE LABELING

A

R

R

groundtorespectwithAatVoltagev

RacrossVoltagev

RacrossVoltagev

⇒

⇒

⇒

2

1

2

1

AA

BB

vvoutout

Labeling Conventions:Labeling Conventions:

Enzo Paterno 18

A

ABRout

AB

B

A

vtv

vvv

BandAbetweenVoltagev

groundtorespectwithBatVoltagev

=

==

⇒

⇒

)(

1

ECETECET1111SINGLE LOOPS CIRCUITSSINGLE LOOPS CIRCUITS

The circuit below is a single loop of elements. The The circuit below is a single loop of elements. The same currentsame currentflows through all the elements in that loop and as a result we say that theseflows through all the elements in that loop and as a result we say that theseelements are connected in elements are connected in seriesseries..

Applying KVL: Applying KVL:

21

21

)(

0)(

RR

RR

vvtv

vvtv

+=∴

=−−

Applying Ohm’s law:Applying Ohm’s law:i(t)

Equivalent circuitEquivalent circuit

++

i(t)

v(t) R

Enzo Paterno 19

[ ]

T

R

R

R

tv

RR

tvti

RRtitiRtiRtv

tiRv

tiRv

)()()(

)()()()(

)(

)(

21

2121

2

1

2

1

=+

=∴

+=+=∴

=

=

Applying Ohm’s law:Applying Ohm’s law:

and substituting:and substituting:

In a purely resistive series circuit, the In a purely resistive series circuit, the totaltotalresistanceresistance, R, RTT, of N resistors is the sum of the, of N resistors is the sum of theindividual resistanceindividual resistance

∑=

=N

i

iT RR1

i(t)--

v(t) RT

ECET11ECET11VOLTAGE DIVIDER RULE VOLTAGE DIVIDER RULE -- VDRVDR

21

2

1

)()(:

)(

)(

2

1

RR

tvtithatrecalling

tiRv

tiRv

R

R

+=

=

=

Recalling Ohm’s law:Recalling Ohm’s law:

i(t)

Enzo Paterno 20

21

2

21

2

21

1

21

1

21

)()(

)()(

:VDR - ruledivider voltageget theWe

2

1

RR

Rtv

RR

tvRv

RR

Rtv

RR

tvRv

RR

R

R

+=

+=

+=

+=

+

VDRVDRFor a purely resistive series circuitFor a purely resistive series circuitcomprising of N resistors:comprising of N resistors:

Nx

R

Rtvv

N

i

i

xRx

L1)(

1

==

∑=

ECETECET1111VOLTAGE DIVIDER RULE VOLTAGE DIVIDER RULE -- VDRVDR

RR11 is a variable resistor (i.e. potentiometer) such as the volumeis a variable resistor (i.e. potentiometer) such as the volumecontrol for an electronic device. Let Vcontrol for an electronic device. Let VSS = 9 V, R= 9 V, R22 = 30 k= 30 kΩΩ..Find I, VFind I, VR2R2 and Pand PR2R2 for both values of Rfor both values of R11 is set to 1) 60 kis set to 1) 60 kΩΩ and 2) 15 kand 2) 15 kΩΩ

vkk

kV

uAkk

vI

33060

309

1003060

9

2 =

+=

=+

=RR11::

60 k60 kΩΩ

Enzo Paterno 21

( ) mWkxP

kk

R 3.03010100

3060

26

2==

+−

( ) mWkxP

vkk

kV

uAkk

vI

R 2.13010200

63015

309

2003015

9

26

2

2==

=

+=

=+

=

−

RR11::15 k15 kΩΩ

Explain why VExplain why V22 doubles when Rdoubles when R11

decreases by a factor of 4decreases by a factor of 4

Give another reason why:Give another reason why: VV22 = 3v when R= 3v when R11 = 60k= 60k VV22 = 6v when R= 6v when R11 = 15k= 15k

ECET11ECET11VOLTAGE DIVIDER RULE VOLTAGE DIVIDER RULE -- VDRVDR

The load of a highThe load of a high--voltage dc transmission facility is183.5voltage dc transmission facility is183.5ΩΩ,,find the power loss in the line.find the power loss in the line.

IITT MODELMODEL

RRLINELINE

Enzo Paterno 22

kVkVVLOAD 24.3665.165.183

5.183400 =

+=

( )MW

x

R

VP

LOAD

LOADLOAD 734

5.183

10366262

===

MWPPP LOADSloss 66734800 =−=−=

Note: PNote: PLOSSLOSS = P= PRLINERLINE

To minimize PTo minimize PLOSSLOSS for a certain Pfor a certain PLOADLOAD

it is desirable to get a higher supplyit is desirable to get a higher supplyvoltage & smaller line current rathervoltage & smaller line current ratherthan a larger line current & smallerthan a larger line current & smallersupply voltage. WHY?supply voltage. WHY?

S

L

PP=ε

LINETloss RIP2=

ECETECET1111MULTIPLEMULTIPLE--SOURCE RESISTOR NETWORKSSOURCE RESISTOR NETWORKS

The analysis of multipleThe analysis of multiple--source resistor networks can be simplified source resistor networks can be simplified using an equivalent circuit.using an equivalent circuit.

Equivalent circuitEquivalent circuit

Enzo Paterno 23

04232115 =−−+−−+− vvvvvvv RR

( ) 2154321 RR vvvvvvv +=−−+−

( ) 21)( RReq vvtvv +==

21

)()(

RR

tvti

+=

ECET11ECET11MULTIPLEMULTIPLE--SOURCE RESISTOR NETWORKSSOURCE RESISTOR NETWORKS

Find Find II, V, Vbdbd, P, P30k30kΩΩ, and V, and Vbcbc

Enzo Paterno 24

mAI

kI

kIkIkI

1.0

660

0301220106

−=

−=

=−−−−mA

kI 1.0

60

6−=

−=

ECETECET1111MULTIPLEMULTIPLE--SOURCE RESISTOR NETWORKSSOURCE RESISTOR NETWORKS

Find I, Find I, VVbdbd, P, P30k30kΩΩ, and V, and Vbcbc

mAI

VkIeq

kIVkIeq

bd

bd

1.0

02012:2

030106:1

−=

=−+

=−−−

Enzo Paterno 25

VVbd 10=

mAk

I 1.060

6−=

−=

ECET11ECET11MULTIPLEMULTIPLE--SOURCE RESISTOR NETWORKSSOURCE RESISTOR NETWORKS

Find I, VFind I, Vbdbd, , PP30k30kΩΩ, and , and VVbcbc

mAk

I 1.060

6−=

−=

Enzo Paterno 26

mWARIP 30)10*30()10( 3242 =Ω−==

Ω−

RESISTOR30k ON POWER ( ) Vkk

kVbc 2

4020

206 −=

+−=

ECETECET1111VOLTAGE DIVIDER RULEVOLTAGE DIVIDER RULE

+

-

1R

2RS

V

−

+

OV

SOV

RR

RV

21

2

+=

VOLTAGE DIVIDERVOLTAGE DIVIDERVOLTAGE DIVIDERVOLTAGE DIVIDER

OSV

R

RRV

2

21+

=

"INVERSE" DIVIDER"INVERSE" DIVIDER"INVERSE" DIVIDER"INVERSE" DIVIDER

Enzo Paterno 27

Ω=+

= kVS

5003.458220

20220

DIVIDER INVERSE""

Find VFind VSS

ECET11ECET11CIRCUIT SIMULATION WITH MULTISIM CIRCUIT SIMULATION WITH MULTISIM

Enzo Paterno 28

ECETECET1111CIRCUIT SIMULATION WITH PSPICECIRCUIT SIMULATION WITH PSPICE

Enzo Paterno 29

ECET11ECET11SINGLESINGLE--NODENODE--PAIR CIRCUITSPAIR CIRCUITS

A singleA single--node pair circuit is shown below. The voltage across each node pair circuit is shown below. The voltage across each branch is the same, and therefore , are said to be in parallel.branch is the same, and therefore , are said to be in parallel.

−

+

V

For example:For example:

Enzo Paterno 30

)()(

)(1

)(

21

21 tiRR

RRtv

tvR

tip

+=

=

RRpp is the equivalent resistance of the two is the equivalent resistance of the two resistors in parallelresistors in parallel

ECETECET1111CURRENT DIVIDER RULE CURRENT DIVIDER RULE -- CDRCDR

21

2

1

1 )()()(RR

Rti

R

Rtiti

p

+==

2

2

)()(

R

tvti =

1

1

)()(

R

tvti =

21

21)()()(RR

RRtiRtitv p +

==

Enzo Paterno 31

mAI 1)5(41

11 =

+=

)5(51

412 +

=−= III

21

1

2

2 )()()(RR

Rti

R

Rtiti

p

+==

ECET11ECET11CURRENT DIVIDER RULE CURRENT DIVIDER RULE -- CDRCDR

O21 VI I FIND ,,

Equivalent circuit:Equivalent circuit:

Enzo Paterno 32

vIkVO 24*80 2 =Ω=

ECETECET1111CURRENT DIVIDER RULE CURRENT DIVIDER RULE -- CDRCDR

O21 VI I FIND ,,


mAtiti 9.0)()( =+

Enzo Paterno 33

mAtiti 9.0)()( 21 =+

branch1 has ½ the resistance of branch2branch1 has ½ the resistance of branch2

)(2)( 21 titi =∴

mAtimAti

mAti

mAtiti

6.0)(3.0)(

9.0)(3

9.0)()(2

12

2

22

=⇒=

=

=+

Bra

nch

1B

ran

ch

1 Bra

nch

2B

ran

ch

2


CAR CAR CAR CAR CAR CAR CAR CAR STEREO STEREO STEREO STEREO STEREO STEREO STEREO STEREO AND CIRCUIT MODELAND CIRCUIT MODELAND CIRCUIT MODELAND CIRCUIT MODELAND CIRCUIT MODELAND CIRCUIT MODELAND CIRCUIT MODELAND CIRCUIT MODEL

Enzo Paterno 34

POWER PER SPEAKERPOWER PER SPEAKERPOWER PER SPEAKERPOWER PER SPEAKERPOWER PER SPEAKERPOWER PER SPEAKERPOWER PER SPEAKERPOWER PER SPEAKER

ECETECET1111


CURRENT DIVIDER RULE CURRENT DIVIDER RULE -- CDRCDR

Enzo Paterno 35

)()()()(

)()(

tiR

Rti

R

tvti

tiRtv

O

k

p

K

k

k

OP

=⇒

=

=

Np GGGG +++= ...21


Equivalent circuit:Equivalent circuit:++=

1111

Find IFind ILL through the load Rthrough the load RLL

Enzo Paterno 36


Ω=

++=

kR

kkkR

p

p

4

12

1

9

1

18

11

mAIT 1214 =−−=

mAkk

kxI L 25.0

124

4)101( 3 =

+= −

ECETECET1111SERIESSERIES--PARALLEL CIRCUITSPARALLEL CIRCUITS

Find RFind RABAB

Enzo Paterno 37

ECET11ECET11SERIESSERIES--PARALLEL CIRCUITSPARALLEL CIRCUITS

Enzo Paterno 38


Enzo Paterno 39


Enzo Paterno 40


Enzo Paterno 41


Find IFind I11,I,I22, I, I33, I, I44, I, I55, V, Vaa, V, Vbb, and V, and Vcc

Enzo Paterno 42


Enzo Paterno 43


1 mA1 mA

Enzo Paterno 44

mAk

vI 1

12

121 ==

Ohm’s LawOhm’s Law


Find IFind I11,I,I22, I, I33, I, I44, and I, and I55

1/2 mA1/2 mA

1/2 mA1/2 mA

1 mA1 mA

3 V3 V

Enzo Paterno 45

mAmAI 5.012

612 == mAII

21

22 ==

CDRCDR CDR or ProportionsCDR or Proportions

3 V3 V


Find IFind I11,I,I22, I, I33, I, I44, and I, and I55

1 mA1 mA 1/2 mA1/2 mA

1/2 mA1/2 mA

3 V3 V 1.5 V1.5 V 3/8 V3/8 V

Enzo Paterno 46

mAmAI8

316

12

21

4 == mAmAmAI 818

32

15 =−=

CDRCDR KCLKCL


Find VFind Voo

Enzo Paterno 47

= ½ mA= ½ mA


Find VFind Voo

3 v3 v

1.5 mA1.5 mA3 mA3 mA

18 v18 v

Enzo Paterno 48

= ½ mA= ½ mA1 mA1 mA

1.5 mA1.5 mA

3 v3 v

3 mA3 mA

12 12 vv

36 v36 v


Enzo Paterno 49

kkk 1066 ++

ECET11ECET11RESISTOR TOLERANCERESISTOR TOLERANCE

1/8 Watt1/8 Watt

1/4 Watt1/4 Watt

1/2 Watt1/2 Watt

Resistors have a power rating that specifies the maximum power Resistors have a power rating that specifies the maximum power dissipation it can tolerate.dissipation it can tolerate.

Because of manufacturing process, a resistorBecause of manufacturing process, a resistorvalue has a deviation (error) called value has a deviation (error) called tolerancetolerance

example: example: 2700 ohms with a 10% tolerance2700 ohms with a 10% tolerance

)7.2(9.024302702700 Ω=Ω=Ω−Ω kExample:Example:

Enzo Paterno 50

Ω≤≤Ω 29702430 R

)7.2(9.024302702700 Ω=Ω=Ω−Ω k

)7.2(1.129702702700 Ω=Ω=Ω+Ω k

RANGERANGE

Example:Example:Let R (Nominal) = 2.7 kLet R (Nominal) = 2.7 kΩΩ

@ @ ±± 10%10%

mAI 704.37.2

10nominal == mWP 04.37

7.2

10 2

nominal ==

mAI

mAI

115.47.29.0

10

367.37.21.1

10

max

min

=×

=

=×

= 15.41max mWP =

R R ½ Watt½ Watt

ECET11ECET11RESISTORS COLOR CODERESISTORS COLOR CODE

Enzo Paterno 51

enz - resistive circuits · 2014-12-09 · ecet11 kirchhoff’s current law kirchhoff’s current...

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