chapter 18 direct currentstudent

7/27/2019 Chapter 18 Direct Currentstudent

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18.1 Electrical Conduction

18.2 Ohms law and Resistivity

18.3 Variation of resistance with temperature

18.4 Electromotive force (emf), internal

resistance and potential difference

18.5 Electrical energy and power

18.6 Resistors in series and parallel18.7 Kirchhoffs Laws

18.8 Potential divider

18.9 Potentiometer and Wheatstone Bridge

TOPIC 18 : ELECTRIC CURRENT AND

DIRECT-CURRENT CIRCUITS

1


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Conductors contain many free electrons and move randomly.

If a continuous wire is connected to the terminal of a battery, the

potential difference between the terminals of the battery sets up an

electric field inside the wire and parallel to it, directed from the positive

toward the negative terminal.

Thus free electrons are attracted into the positive terminal (are

forced to drift in one direction).

This direction is in the direction opposite to the field, E. The velocity of these free electrons is called drift velocity.

battery


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Thus free electrons at one end of the wire are attracted into

the positive terminal, and at the same time, electrons leave

the negative terminal of the battery and enter the wire at theother end.

There is a continuous flow of electrons through the wire that

begins as soon as the wire is connected to both terminals.

However, when the conventions of positive and negativecharge were advised two centuries ago, it was assumed that

positive charge flowed in a wire.

For nearly all purposes, positive charge flowing in one

direction is exactly equivalent to negative charge flowing in

the opposite direction.

Today we still use the historical convention of positive current

when discussing the direction of a current. So when we speak

of the current in a circuit, we mean the direction positive

charge would flow. 5


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SUBTOPIC :

LEARNING OUTCOMES :

a) State and use Ohms Law.

b) Define and use resistivity formulae,

At the end of this lesson, the students should

be able to :

RA

l

18.2 Ohms Law and Resistivity


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18.2 Resistivity and Ohms Law

Ohms Law

Ohms law states that the potential difference across a

conductor, Vis directly proportional to the current,I

through it, if its physical conditions and the temperature

are constant.

V I

constant

Ohm's Law

V

I

V R V IRI

18.2 Ohms Law and Resistivity


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V

I

constant

V

I

VRI

Ohmic conductorsare conductors which obey Ohmslaw. Examples: pure metals. (Figure A)

Non-ohmic conductorsdo not obey Ohms law.

Example: junction diode. (Figure B)

Figure A Figure B

V

I

10


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Resistivity

Resistivity is a measure of a materials ability to oppose

the flow of electric current through the material.

Resistivity is defined as the resistance of a sample of thematerial of cross-sectional area 1 m2 and of length 1m.

It is a constant value.

Its formulae is given by .

Its unit is m.

Its value depends on the material.

All conductors have smaller resistivity. Insulators have larger resistivity.

RA

l

where l= length of the conductor (m)A = area of cross-section of the conductor (m-2)



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Example 18.2

A wire (length=2.0 m, diameter=1.0 mm) has a resistance of

0.45 . What is the resistivity of the material used to make the

wire ?

12


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Example 18.3

What voltage will be measured across a 1000- resistor in a

circuit if we determine that there is a current of 2.50 mA flowing

through it ?

13


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14

SUBTOPIC :

LEARNING OUTCOMES :

a) Explain the effect of temperature on electrical

resistance in metals and superconductors.

b) Define and use temperature coefficient of resistivity,.

c) Apply resistanceR= Ro [1 + (T-To)].


be able to :

18.3 Variation of Resistance with

Temperature


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More collisons occur between free electrons and ions.

18.3 Variation of Resistance with Temperature

These electrons are slowed down thus increases theresistance.

The resistance of a metal can be represented by theequation below

R=Ro[1+(T)] ,R-Ro= RR=Ro+T

where R = the resistance at temperature T,

Ro= the resistance at temperature To = 20o C or 0oC,

= the temperature coefficient of resistance (o

C-1

)


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is a constant value and it is depends on the material.

o

RR

T

Temperature coefficient of resistance , is defined as the

fractional change in resistance per Celsius degree.


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Example 18.4

A platinum wire has a resistance of 0.50 at 0oC. It is

placed in a water bath, where its resistance rises to a

final value of 0.60 . What is the temperature of the

bath ? ( = 3.93 x 10-3oC -1)

18



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Example 18.5

A narrow rod of pure iron has a resistance of 0.10 at 20oC.

What is its resistance at 50 oC ? ( = 5.0 x 10-3oC -1)

19

18.3 Variation of resistance with temperature


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18.3 Variation of Resistance with TemperatureSuperconductor

As the temperature decreases, the resistance at first

decreases smoothly.

(T, R)

At a certain critical temperature Tc (4.2 K for mercury)

the resistance suddenly drops to zero.

R

T

metal

R

T

superconductor

Graph of resistance Ragainst temperature T


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SUBTOPIC :

LEARNING OUTCOMES :

a) Define emf , .

b) Explain the difference between emf of a battery and

potential difference across the battery terminals.

c) Apply formulae V= Ir.


be able to :

18.4 Electromotive Force (emf), Internal

Resistance and Potential Difference


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18.4 Electromotive Force (emf), Internal

Resistance And Potential Difference

The e.m.f of a source is the p.d across the term inals o f

the source in open circui t(no current is flowing,I= 0).

The e.m.f of a source is the wo rk done per uni t charge.

The e.m.f of a source is defined as the electr ical energythat generated by a source so that the charges canflow

from one terminal to another terminalof the source

through any res is tor.

The e.m.f of a battery is the po tential di f ference across

i ts terminalwhen it is not connected to a circuit.

What is electromotive force,emf ( or)?

SI unit : Volt (V)


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18.4 Electromotive Force (emf), Internal Resistance and Potential Difference

What is in ternal resistance?

In a cell or battery, the negative ions are attracted by

anode and the positive ions are attracted by the cathode.

The flow of these ions produces current.

However the collisions between the ions and therecombination of opposite ions reduce the flow ofcurrent. This resistance in the cell is called internalresistance, r.

18 4 El i F ( f) I l R i d P i l Diff


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Suppose a battery of emf, and internal resistance r

is connected to an external resistor,R.

Total resistance in the circuit is (R + r).

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18 4 El i F ( f) I l R i d P i l Diff


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The e.m.fof this battery is given as

( )

ab

I R rIr IR

Ir V

ab dcV V

Vab = Vb

Va = terminal voltage (potential difference acrossthe battery terminals)

R = external resistance

r= internal resistance25


18 4 El t ti F ( f) I t l R i t d P t ti l Diff


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In a circuit diagram, this symbol

represents a res is torin a circuit that dissipateselectrical

energy.

A straight line represents a conducting wire

with negl igib le resistance.

26




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ab

ab

Ir V

V Ir

Notes:

a) Vab< when the battery of emf is connected to theexternal circuit with resistanceR.

b) Vab> when the battery of emf is being charged byother battery.

c) Vab= when the battery of emf has no internalresistance (r=0) and connected to the external circuit

with resistanceR..

27

terminal voltage (potential

difference across terminals)emf

potential difference

across internal resistance




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Example 18.6

28

A battery with a terminal voltage of 11.5 V when delivering0.50 A has an internal resistance of 0.10 . What is its emf?


18 4 Electromotive Force (Emf) Internal Resistance and Potential Difference


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Example 18.7

29

The battery in a circuit has an emf of 9.0 V. It is attached to aresistor and an ammeter that shows a current of 0.10 A. If a

voltmeter across the batterys terminals reads 8.9 V, what is

its internal resistance ?

18.4 Electromotive Force (Emf), Internal Resistance and Potential Difference

SUBTOPIC


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SUBTOPIC :

LEARNING OUTCOMES :

a) Apply formulaP=IVand electrical energy, W=VIt.


be able to :

18.5 Electrical Energy and Power


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18 5 Electrical Energy and Power


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The electrical (potential) energy, Wis the energy gained by

the charge Q from a voltage source (battery) having a

terminal voltage V.

W= QV (the work done by the source on the charge)

But Q=It, then W= VI t

Unit : Joule (J)

The rate of energy delivered to the external circuit by the

battery is called the electric powergiven by,

Unit : watt ( 1 W = 1J/s) 32

@

,W QV

P Q I t t t

P I V P I




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The energy dissipated per second in an electric device

(rate of energy dissipated) is given as

for any deviceW VIt

P VIt t

A passive resistor is a resistor which converts all the

electrical energy into heat. For example, a metal wire.

22

but

or

P VI V IR

V

P I R P R

only for resistor


18 5 El t i l E d P


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Calculate the resistance of a 40 W automobile headlight

designed for 12 V?

34

Example 18.818.5 Electrical Energy and Power



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The current through a refrigerator of resistance 12 is 13

A. What is the power consumed by the refrigerator?

35




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An electric iron with a 15-ohm heating element operates at

120 V. How many joules of energy does the iron convert toheat in 1.0 h ?

36


SUBTOPIC


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SUBTOPIC :

LEARNING OUTCOMES :

a) Deduce and calculate effective resistance of resistors

in series and parallel.


be able to :

18.6 Resistors in series and parallel


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1R 2R 3R

V

1V 2V 3V

I I

Resistors in Series

The properties of resistors in series are given below.o The same currentI flows through each resistor

where

321 IIII

battery , r=0

18 6 Resistors in series and parallel


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o The sum of the voltages around a circuit loop (thatis, the gains and losses with + and - ,respectively) is

zero.

(Assuming that the connecting wires have no resistance)

1 2 3

0ii

ii

V V

V V

V V V V

total potential difference

;22

IRV ;33

IRV 1 1

bur ;V I R

321eq IRIRIRIR eq

IRV

321eq RRRR

where resistance)(effectiveequivalent:eqR



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Resistors in Parallel

1R

3R

V1V

2V

3V

I I

2R2I

1I

3I The properties of resistors inparallel are given below.o There is the same potential

difference, Vacross eachresistor where

321 VVVV

o Charge is conserved, thereforethe total currentIin the circuitis given by

321 IIII




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;22 R

V

I ;33 RV

I 1 1but

;

V

I R

321eq R

V

R

V

R

V

R

V

eqR

V

I

321eq R

1

R

1

R

1

R

1




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42

Example 18.10

.04.02

V06 .

12

Calculate :

a. the total resistance of the circuit.

b. the total current in the circuit.c. the potential difference across 4.0 resistor.




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Solution 18.10

.04.02

V06 .

12

p



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Solution 18.10

b. Total current,

c. The potential difference acrossR1=2.0 is

Therefore the potential difference acrossR3=4.0

is given by

p



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Example 18.11

For the circuits shown below, calculate the equivalent

resistance between points x and y.

.03

.01.01

x

y

.02

.02

.018

.016

.08

yx .09

.016

.06

.020

(0.79 ) (8.0 )

p

SUBTOPIC :


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SUBTOPIC :

LEARNING OUTCOMES :

a) State and use Kirchhoffs Laws.


be able to :

18.7 Kirchhoffs Laws


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Kirchhoffs first law (junction/current law)

It states that the algebraic sum of the currents at any

junction of a circuit is zero,

0

or

i n out

I

I I

18.7 Kirchhoffs Laws.


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Kirchhoffs second law (loop/voltage law)

I R

It states that the algebraic sum of the voltages across

all of the elements of any closed loop is zero.or

It states thatin any closed loop, the algebraic sum of

e.m.fs is equal to the algebraic sum of the products

of current and resistance.

Sign convention

IR

+IR

across resistor

-IR

+

-

across battery



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Example 18.12

Using Kirchhoffs rules, find the current in each resistor.

R1 = 10 R2 = 20

210 V

1 20 V

18.7 Kirchhoff s Laws



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Solution 18.12

R1 = 10 R2 = 20

210 V

1 20 V

Step

1. Draw current. (arbitrary)

2. Draw loop. (arbitrary)

3. Apply Kirchhoffs laws.

0 ,I I R

I



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Example 18.13

Apply Kirchhoffs rules to the circuit in figure below and find

the current in each resistor.

R1 = 3.0

R2 = 3.0

R3 = 3.0

R4 = 3.0

2 2 0 V.

1 2 0 V.



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Solution 18.13

R1 = 3.0

R2 = 4.0

R3 = 5.0

R4 = 2.0

2 3 0 V.

1 6 0 V.

I

2nd KL,

( )I R



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Example 18.14

Find the current in each resistor in the circuit shown below.

R2 = 4.0

R3 = 4.0

R1 = 4.0 3 5 0 V.

1 10 V

2 5 0 V.

I1=3.75 A up, I2= 1.25 A left, I3= 1.25 A right


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Example 18.16

Calculate the currentsI1,I2 andI3. Neglect the internalresistance in each battery.

1R1

.10R3

V151

.50R2

V102

V033 .

1I

2I

3I

;. A6917I1

;. A6214I2

A073I3

.

E l 18 1718.7 Kirchhoffs Laws


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Example 18.17

Given 1=8 V, R2=2 , R3=3 ,

R1 =1 andI=3 A. Ignore the internal resistance in

each battery.

Calculate

a. the currentsI1and I2.b. the e.m.f. 2.

Ans. : 1 A, 4 A , 17 V

3R

1

2R2

1I

2I

I

1R

SUBTOPIC :


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57

SUBTOPIC :

LEARNING OUTCOMES :

a) Explain the principle of a potential divider.

b) Apply equation of potential divider


be able to :


11

1 2

.R

V V

R R


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A potential divider is used to tap a fraction of the voltage

supplied by a source of e.m.f.

Potential divider circuit

11

1 2

RV V

R R

Potential difference across l1 orR1 is

V

I

2V1V

1RI

2ReqRVI

Two resistors are connected in series.

The current flows in each

resistor is the same ;

1 2,

eqR R R

21

RR

VI

11 IRV

18.8 Potential divider.


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V

I

2l1lI

ba c

2V1V

Vll

lV

21

11

lR

A

Potential difference across l1 is

11

1 2

RV V

R R

ResistanceR1 andR2 are replaced by a uniformhomogeneous wire as shown in figure below.

E l 18 1818.8 Potential divider


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60

Example 18.18

Resistors of 3.0 and 6.0 are connected in series to a

12.0 V battery of negligible internal resistance. What are the

potential difference across the (a) 3.0 and (b) 6.0

resistors ?

4.0 V, 8.0 V


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18 9 Potentiometer and Wheatstone Bridge


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A potentiometer is mainly used to measure potentiometer.

It consists of a uniform wire. Basically a potentiometer circuit consists of a uniform wire

AB of length 100.0cm, connected in series to a driver cell

with emfVof negligible internal resistance.

V

I I

BA C

xV

I

G

+ -

I

(Unknown Voltage)

Jockey

(Driver cell -accumulator)

Potentiometer



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The potentiometer is balanced when the jockey (sliding

contact) is at such a position on wire AB that there is no

current through the galvanometer. Thus

When the potentiometer in balanced, the unknown

voltage (potential difference being measured) is equal tothe voltage across AC.

Galvanometer reading = 0

ACx VV

Potentiometer can be used to :i) Measure an unknown e.m.f. of a cell.ii) Compare the e.m.f.s of two cells.

iii) Measure the internal resistance of a cell.


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i) Measure an unknown e.m.f. of a cell.

V

I I

BAC

I

G

+ -

I

(Unknown cell)

Jockey

(Driver cell -accumulator) When the potentiometer is

balanced, IG = 0

VAC=

VAC= IRAC1

AC

AC

lR

A

2 , ABAB

lR

A

3

2 , ...43

AC

A C A B

AB

lR R

l

...5AB

VI

R

4 and 5 into 1,AC

AC AB

AB AB

AC

AB

lVV R

R l

lV

l

Balance length = lAC


ii) C th f f t ll


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ii) Compare the e.m.f.s of two cells.

V

I I

BAI

G

I

(2)

(1)

2

1

S

CJ

D

When the potentiometer is

balanced, IG = 0

Balance length,

lAC= l1 for1 and lCD= l2 for2

AC

AB

l Vl

From , thus

1 2

2 1

22 1

1

...1 and ...2

2 1 ,

AC CD

AB AB

CD

AC

l lV V

l l

l

l

l

l

1l2l

Wheatstone Bridge18.9 Potentiometer and Wheatstone Bridge


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Wheatstone Bridge

BA G

C

D

1R 2R

3R XR

I

2I

I

1I

2I

1I

0

It is used to measure the unknown resistance of the resistor.

Figure below shows the Wheatstone bridge circuit consists of a cell

of e.m.f. (accumulator), a galvanometer , known resistances (R1,R2

andR3) and unknown resistanceRx.

The Wheatstone bridge is said to be balanced when no current

flows through the galvanometer. Hence

Then ,

Therefore

Since

1CBAC III

2DBAD III

and

Potential at C = Potential at D

ADAC VV and BDBC VV

IRV thus

3211 RIRI X221 RIRI and

X2

32

21

11

RI

RI

RI

RI 3

1

2X R

R

RR



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BA G

C

D

1R 2R

3R XR

I

2I

I

1I

2I

1I

03

1

2

X RR

R

R

XR

A

GB

R

J

Thick copper

strip

(Unknown

resistance)

Jockey

(resistance box)

Accumulator

Wire of uniform

resistance

0

1I 1I

2I I

I

1

2

XR l

R l

2

3 1

XR R

R R


Example 18 19


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Example 18.19

An unknown length of platinum wire 0.920

mm in diameter is placed as the unknown

resistance in a Wheatstone bridge as

shown in figure below.

Resistors R1 and R2 have resistance of

38.0 and 46.0 respectively.

Balance is achieved when the switch

closed and R3 is 3.48 . Find the

length of the platinum wire if its

resistivity is 10.6 x 10-8 m.

chapter 18 direct currentstudent

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