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CHAPTER 2

Resistive Circuits

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2.1 Ohms Law

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Ohms Law

The voltage across a resistance is

directly proportional to the current

flowing through it.

Resistance

The constant of proportionality between the voltageand current

Resistor

A circuit element whose electrical characteristic isprimarily resistive

2.1 Ohms Law

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Resistor

2.1 Ohms Law

- C composition, wire wound, thick oxide,

thin metal films, diffused in semiconductor IC

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The mathematical relationship of Ohms law

We have tacitly assumed that the resistor has a constant value.

There are some practicalelements that exhibit a nonlinear

resistance characteristic. A typical characteristic for a light bulbs

To represent ohms,

2.1 Ohms Law

0 RwhereRi(t),v(t)

1V/A1

linear nonlinear

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Energy dissipation

The charge moves from the higher to the lower

potential as it passed through the resistor.

Energy is absorbed by the resistor.

The energy absorbed is dissipated by the resistorin the form of heat.

Instantaneous powerof the resistor

The rate of energy dissipation

Always a positive quantity (Resistor is passive element)

2.1 Ohms Law

R

tvtRititvtp

)()()()()(

22

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Conductance

The reciprocal of resistance

The unit of conductance The siemens

Additional expressions

2.1 Ohms Law

RG

1

11A/V11 S

)()( tGvti

)()(

)(2

2

tGvG

titp

7

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Variable resistance

Short circuit

As the resistance is decreased,

we finally reach a point where

the resistance is zero.

Open circuit

If the resistance is increased,

we finally reach a point where

it is essentially infinite.

2.1 Ohms Law

GR ,0

0, GR

0)()( tRitv

0/)()( Rtvti8

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Example 2.1

Determine the current and the power absorbed by the resister.

2.1 Ohms Law

R

VRIVIP

22 mW72)mA6)(V12( P

mA6)kV)/(2(12/ RVI

12V

9

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Example 2.2

Determine the voltage and the current in the circuit.

The other solution

2.1 Ohms Law

R

VP S

2

W)10)(3.610(10 332 PRVS

V6 SV mA0.6k10V6

RVI s

V6SV mA0.6I

Vs

10

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Example 2.3

Find the value of the voltage source and the power absorbed bythe resistance.

2.1 Ohms Law

IGRIVS

1

V10S1050

A105.06

3

SV

G

IRIP

22 mW5

S1050

A105.06

232

G

IP

11

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Example 2.4

Find R and Vs.

2.1 Ohms Law

2I

PR

IRVS V20)105)(A104(33 SV

Vs

k5A104

W1080

23

3

RmA4, I

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2.2 Kirchhoffs Law

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Lumped-parameter circuit

The interconnection is performed by perfect

conductors (wires).

Because the wires have zero resistance, the energy

is in essence lumped in each element.

2.2 Kirchhoffs Law

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2.2 Kirchhoffs Law

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Node

A point of connectionof two or more circuit elements

2.2 Kirchhoffs Law

Same node

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Loop

Any closed path through the circuit in which no node

is encountered more than once

2.2 Kirchhoffs Law

Loop NOT a loop

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Branch

A portion of a circuitcontaining only a single element

and the nodes at each end of the element

2.2 Kirchhoffs Law

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Lumped-parameter circuit

The interconnection is performed by perfect conductors(wires).

Because the wires have zero resistance, the energy is inessence lumped in each element.

Node A point of connectionof two or more circuit elements

Loop

Any closed path through the circuit in which no node is

encountered more than once

Branch

A portion of a circuitcontaining only a single element and

the nodes at each end of the element

2.2 Kirchhoffs Law

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Kirchhoffs current law (KCL)

The algebraic sum of the current entering any node iszero.

In mathematical form the law appears as

Where is the th current entering the node through

branch and is the number of branches connected to the

node

2.2 Kirchhoffs Law

N

j

j ti1

0)(

)(tij j

j N

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Kirchhoffs current law (KCL)

2.2 Kirchhoffs Law

The algebraic sum of the

currents enteringa node

is zero.

The algebraic sum of the

currents leaving a node is

zero.

The sum of the currents

entering a node is equal

to the sum of the currents

leaving the node.

0)()()()( 7542 titititi

)()()()(7452

titititi

0)()()()( 7542 titititi

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Kirchhoffs current law (KCL)

2.2 Kirchhoffs Law

Recall that current is charge in motion.

Charges cannot be stored at a node

If we have a number of charges

enteringa node, then an equalnumber must be leavingthat same

node

The sum of the currents entering a

node is equal to the sum of the

currents leaving the node.

KCL is based on the principle of

conservation of charge.

NODE

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Example 2.5

Write the KCL for every node.

2.2 Kirchhoffs Law

The fifth equation is the sum of the first four.

The first four equation is linearly independent,

but the set of five equations is not linearly independent

So, The fifth equation is redundant.24

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Example 2.6

Find the unknown current in the network

2.2 Kirchhoffs Law

mA801I

mA704I

mA505I

mA106 I

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Example 2.7

Write the KCL equation for the circuit

2.2 Kirchhoffs Law

The first three equation is linearly

independent,

but the set of four equations is

not linearly independent

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Example 2.8

Find and in the network

KCL to surface 2

KCL to surface 1

2.2 Kirchhoffs Law

4I 1I

004.003.002.006.04 I

mA704 I

002.006.01 I

mA801 I

Surface: Set of elements completely contained

within the surface that are interconnected

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Kirchhoffs voltage law (KVL)

The algebraic sum of the voltages around any loop is

zero.

In mathematical form the law appears as

Where is the voltage across the th branch (with a

proper reference direction) in a loop containing voltages. The lumped-parameter circuit is conservative,

meaning that the work required to move a unit charge

around any loop is zero.

2.2 Kirchhoffs Law

N

j

j tv1

0)(

)(tvj j

N

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Kirchhoffs voltage law (KVL)

In applying KVL, we must traverse any loop and sum

to zero the increases and decreases in energy level.

We will consider a decrease in energy level as

positiveand an increase in energy level as negative.

2.2 Kirchhoffs Law

S

V1

V2

V 03V

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Example 2.9

Find

Suppose,

2.2 Kirchhoffs Law

3RV

V181RV V12

2RV

LOOP abcdefa

V203 RV

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Convention for voltage

The voltage of point a with respect to point b

The variable for the voltage between point a and point b, with

point a considered positive relative to point b.

Equivalent forms for labeling voltage.

double-subscript notation

single arrow notation

+ andnotation

2.2 Kirchhoffs Law

abV

abV

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Example 2.11

Find

2.2 Kirchhoffs Law

ecae VV ,

Given the choice, use the

simplest loop.

V14 aeV

V10 ecV

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Example 2.12

Write the KVL equation for two loop abda, bcdb

2.2 Kirchhoffs Law

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Polarity and direction

For applying Ohms law

Be careful with the polarity of the voltage and the

direction of the current.

For

BUT If the direction of the current orthe polarity of the

voltage, but not both, is reversed,

2.2 Kirchhoffs Law

IRV

IRV

IRV

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2.3 Single-Loop Circuits

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Voltage division

2.3 Single-Loop Circuits

Voltage division The source voltage is dividedbetween R1and R2in direct

proportional to their resistances.

Ohms law

KVL

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Example 2.13

Find the voltage across R2and the power absorbed by R2.Suppose that

and R1 is changed from to

2.3 Single-Loop Circuits

k30V9 2RVS ,

k90 k51

k90when 1R

mW169.0

k)30(k120

92

2

2

RIP

mW.21

k)30(5k4

92

2

2

RIP

k15when 1R

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Example 2.14

Determine both the power delivered to the load and the powerlosses in the line.

2.3 Single-Loop Circuits

||

516400041250resistanceLine ..

load516200kA2

kV400R .

5183load .R

MW800kA2kV400 inP

For decreasing power loss low current

increasing transfer power high voltage39

S C

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Multiple-source network

2.3 Single-Loop Circuits

)()()()()()()( tvtvtvtvtvtvtv RR21

54321

015432 21 )()()()()()()( tvtvtvtvtvtvtv RR

)()()( tvtvtv RR21

Voltage sources in series can

be algebraically addedto form

an equivalent source. .

)()()()()()( 54321 tvtvtvtvtvtv

40

2 3 Si l L Ci it

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Multiple-resistor network

2.3 Single-Loop Circuits

The equivalent resistance of

N resistors in seriesis

the sum of the N resistances.

In each resistor i,

(voltage division)

||

)()( tiRtvS

)(tvR

Rv

S

i

Ri

NS RRRR

21

)()( tiRtviR

i

41

2 3 Si l L Ci it

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Example 2.15

Find and the power absorbed by the resistor: and

2.3 Single-Loop Circuits

bdVI, k30

0k2012 IVbd

V10bdV

mW301030A10 3242k30 .)()(

RIP

Using voltage division, find

V26

k40k20

k20

)(bcV

bcV

bcV

42

2 3 Si l L Ci it

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Example 2.16

If , find the voltage at the sending end of theline and the power loss in the line.

Inverse DIVIDER

2.3 Single-Loop Circuits

kV3458.loadV

kV5003.458220

20220

SV

MW97.86line2

line RIP L

kA083.2220/k3.458 LI

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2.4 Single-Node-Pair Circuits

44

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Current division

Apply KCL

Use Ohms law

The equivalent resistance of two resistors connected in parallel

is the product of their resistances divided by their sum

2.4 Single-Node-Pair Circuits

21

21

RR

RRRP

Current division The current i(t) from the source

divides between the two

branches.

45

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Example 2.17

Find and

Applying current division

Ohms law

2.4 Single-Node-Pair Circuits

21 II , OV

V24k80 2 IVo

46

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Example 2.18

The output circuit of the audio amplifier is in essence a 430mAcurrent source, and each speaker has a resistance of 4.

Determine the power absorbed by the speakers

By each speaker,

2.4 Single-Node-Pair Circuits

mA215)mA430(44

4

I

47

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Multiple-source network

KCL

Current sources in parallel can be algebraically addedto form a single equivalent source.

2.4 Single-Node-Pair Circuits

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Multiple-source network

KCL

Ohms law

In each resistor i,

2.4 Single-Node-Pair Circuits

N

i ip RR1

11 The equivalent conductance of N resistors in

parallelis the sum of the N conductances. Note that Rp is smaller than the smallest Ri

(current division))()( tiR

Rti

o

i

p

i

)()( tiRtv opi

iR

tvti

)()(

49

2 4 Single Node Pair Circuits

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Example 2.17 (sol2)

Find

Equivalent resistor

Ohms law

Voltage divider

2.4 Single-Node-Pair Circuits

OV

k80k40

1

k60

11

pR k40pR

parallel

V36 k4010903

1

).(V

V4240k80k

80k1

VVo

50

2 4 Single Node Pair Circuits

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Example 2.19

Find the current in the 12 kload resistor.

Combine the sources

Combine the resistors

Current division

Notice the minus sign!!

2.4 Single-Node-Pair Circuits

mA1mA2mA4mA1)( ti

k4PR