chapter 3_ electrics & direct current_2016_reviewed

8/18/2019 Chapter 3_ Electrics & Direct Current_2016_Reviewed

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HAPTER 3

ELE TRI URRENT

AND DIRE T-

URRENT IR UITS


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CHAPTER 3 : ELECTRIC CURRENT AND

DIRECT-CURRENT CIRCUITS

3.1Electric Conduction3.2O!"# l$% $nd Re#i#ti&it'

3.3($ri$tion o) re#i#t$nce %it te!*er$ture

3.+Electro!oti&e )orce ,e!) intern$l re#i#t$nce $nd*otenti$l di))erence

3./Electric$l ener0' $nd *o%er

3.Re#i#tor# in #erie# $nd *$r$llel3.irco))"# L$%#

3.4Potenti$l di&ider

3.5Potentio!eter $nd 6e$t#tone 7rid0e


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3.1 Electric$l Conduction

3.1.1 Electric$l Current


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Area, A

Addin0 $ 8$tter' i!*o#e# $n electric *otenti$l

di))erence 8et%een te end# o) te loo* t$t $re

connected to te ter!in$l# o) te 8$tter'.

Te 8$tter' tu# *roduce# $n electric )ield %itin

te loo* )ro! ter!in$l to ter!in$l 9 te )ield c$u#e#

c$r0e# to !o&e $round te loo*.

Te !o&e!ent o) c$r0e# i# $ current I

e F

E I


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Te )lo% o) c$r0e , current *er#i#t# )or $# lon0 $#

tere i# $ *otenti$l di))erence.

Te direction o) te current i# o**o#ite te direction

o) )lo% o) electron#.

6itout $ *otenti$l di))erence no c$r0e )lo%# #ince

no electric )ield $nd electric )orce *roduced. Tere)oreno current *roduced.

Ti# i# $n$lo0ue to


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Te tot$l c$r0e Q )lo%in0 trou0 $n $re$ *er unit

ti!e t .

or

De)inition o) Electric current ,I

in#t$nt$neou# current $&er$0e current

electr

on

)lo%

Current I

Direction of electric current :Po#iti&e to ne0$ti&e ter!in$l

Direction of electron flows :Ne0$ti&e to *o#iti&e ter!in$l

t

Q I =

dt

dQ I =


7/89

SI Unit o) electric current i# A!*ere ,A Sc$l$r ;u$ntit' C$n 8e !e$#ured u#in0 $n $!!eter .

1 $!*ere of current is defined as one coulo!8 o)c$r0e *$##in0 trou0 te #ur)$ce $re$ in one

#econd.

OR

Note:

If the c$r0e !o&e $round $ circuit in te #$!e direction

$t $ll ti!e#, the current is called direct current ,dc, which is

*roduced 8' te 8$tter'.

1sC1

second1

coulomb1ampere1 −==


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• te#e

)ree electron# under0o r$ndo! !otion

trou0out te l$ttice #tructure.

• Ti# !otion i# $n$lo0ou# to te !otion o) 0$#

!olecule#.

3.1.2


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• Ho%e&er %en $ *otenti$l di))erence i#

$**lied $cro## te !et$l ,e? 8' !e$n# o) $

8$tter' $n electric )ield i# #et u* in te!et$l.

• Ti# )ield e?ert# $n electric )orce on te

)reel' !o&in0 electron#.


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• Te )reel' !o&in0 electron# tend to dri)t %it

con#t$nt $&er$0e &elocit' ,dri)t &elocit' vd

#lo%l' $lon0 te !et$l in $ direction o**o#ite

t$t o) te electric )ield.

• Te )lo% o) )ree electron# in one #*eci)ic

direction *roducin0 $ current.

> @


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E?$!*le

Tere i# $ current o) ./ A in $ )l$#li0t 8ul8 )or 2 !in.

Ho% !uc c$r0e *$##e# trou0 te 8ul8 durin0 ti#

ti!e BSolution

i&en : I ./ A t 2 !in 12 #

ro!:

A #il&er %ire c$rrie# $ current o) 3. A. Deter!ine

$ te nu!8er o) electron# *er #econd *$## trou0 te

%ire8 te $!ount o) c$r0e )lo%# trou0 $ cro##-#ection$l

$re$ o) te %ire in // #.

,i&en c$r0e o) electron e

1.×

1 15 C

Answer : (a) 1.88×10 19 electrons per second; (b) 165 C

ollo% U* E?erci#e

t I Q = )120(5.0= 60 C=


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3.2.1 O!"# L$%

#t$te# t$t te *otenti$l di))erence,&olt$0e $cro## $ conductor V i#

*ro*ortion$l to te current I )lo%in0trou0 it i) it# *'#ic$l condition# 9

te!*er$ture $re con#t$nt.

E?*re##ed !$te!$tic$ll' :

%ere V : *otenti$l di))erence ,&olt$0e

I : current )lo% trou0

R : Re#i#t$nce o) te conductor

3.2 O!"# l$% $nd Re#i#ti&it' ,F

I V ∝

IRV =


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• Not $ll !$teri$l# o8e' O!"# l$%.

•


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3.2.2 Re#i#t$nce , R

-- i# $ *ro*ert' %ic o**o#e# or li!it# current in $n

electric$l circuit.

An$lo0ue

to G

i# de)ined $# te r$tio o) *otenti$l di))erence ,V $cro##

conductor to te current ,I t$t )lo%# trou0 it.

De)inition o) Re#i#t$nce ,R

whereMathematically,

I

V R = (voltage)difference potential:V

current: I


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In 0ener$l re#i#t$nce o) $ conductor de*end# on :

1. T'*e o) !$teri$l it i# !$de2. It# len0t

3. It# cro## #ection$l $re$

+. It# te!*er$ture

It i# $ #c$l$r ;u$ntit' $nd it# unit i# o! ,Ω

or ( A 1

In $ circuit i) re#i#t$nce R i# con#t$nt $# V I.

RUN ANI


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o It i# $ #c$l$r ;u$ntit'

o Unit i# o! !eter ,Ω

mo It i# $ !e$#ure o) $ !$teri$l"# $8ilit' to

o**o#e te )lo% o) $n electric current.

(A)

(l )

3.2.3 Re#i#ti&it' , ρ

is defined as te re#i#t$nce o) $ unit cro##-#ection$l$re$ *er unit len0t o) te !$teri$l.

Mathematically,

where

RUN ANI


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Material Resistiity, ρ ( Ω m)

!iler ".#$ × "%−&

'oer ".& × "%−&

Aluminum *.&* × "%−&

Gold *.++ × "%−&

Glass "%"%−"%"+

o Resistiity deends on the t'*e o) te !$teri$l andon the te!*er$ture.

o A ood electric conductor# hae a ery lo%re#i#ti&itie# and ood in#ul$tor# hae ery i0re#i#ti&itie#.

o -ale elow shows the resistiity for ariousmaterials at *% °'.


20/89

E?$!*le

A con#t$nt$n %ire o) len0t 1.! $nd cro## #ection$l

$re$ o) ./ !!2

$# $ re#i#ti&it' o) +.5 ? 1 >

J!. indte re#i#t$nce o) te %ire.

Solution

U#in0:

i&en: 1. ! A ./ !!2 ./ ? 1 > !2

ρ +.5 ? 1 > J !

A

L R

ρ =

"

6

#.$ 10 (1.0)

0.5 10

−

−

×=×

Ω= $%.0 R


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T%o %ire# P $nd K %it circul$r cro## #ection $re

!$de o) te #$!e !et$l $nd $&e e;u$l len0t. I) te

re#i#t$nce o) %ire P i# tree ti!e# 0re$ter t$n t$t o)%ire K deter!ine te r$tio o) teir di$!eter#.

Solution :

i&en :

ro!:

E?$!*le

/nowin that :

& ' & 'same metal : same lengt: ρ ρ l l = =

'& R R =

2

'

2

&

d

d =

'

''

&

&&

A

l ρ

A

l ρ= #

2πd A =

2 2

& '

# #

P P Q Q ρ l ρ l

πd πd

=

'

&

d

d =


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3.3.1 E))ect o) te!*er$ture on re#i#t$nce

o Since te re#i#ti&it' o) $ !$teri$l de*end# on telen0t l $nd te cro##-#ection$l $re$ A %ic $re

$))ected $# te!*er$ture c$n0e# tu# re#i#ti&it'

$l#o c$n0e# $# te!*er$ture c$n0e#.


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Su*erconductor

-ale shows the critical temerature for arious

suerconductors.

In suerconductor, as the temerature decreases, the

resistance (or resistiity) at first decreases smoothly, li0e that

of any metal.

1ut then, $t $ cert$in critic$l te!*er$ture ! c te

re#i#t$nce ,or re#i#ti&it' #uddenl' dro*# to =ero.


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http://var/www/apps/conversion/tmp/scratch_5/The%20Awesome%20Levitating%20Train.flv


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o O&er $ #!$ll te!*er$ture r$n0e ,u* to 1°

C te

re#i#ti&it' o) $ !et$l c$n 8e re*re#ented

$**ro?i!$tel' 8' to te e;u$tion:

%ere

re#i#ti&it' $t #o!e te!*er$ture ! ,in de0ree# Cel#iu#

te re#i#ti&it' $t #o!e re)erence te!*er$ture ! o

re)erence te!*er$ture

te te!*er$ture coe))icient o) re#i#ti&it'.

3.3.2 Te!*er$ture coe))icient o) re#i#ti&it' ,

)in$l te!*er$ture

( )1 OO T T ρ ρ α = + −

ρ

o ρ

oT

α

(0 C 20 C)o oor

α

T


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Te te!*er$ture coe))icient o) re#i#ti&it' ,α

i# te

r$tio o) te c$n0e o) re#i#ti&it' in $ !$teri$l due to $

c$n0e o) te!*er$ture o) 1°

C to it# re#i#ti&it' $t °

C.

2here:

: the chane in resistiity in the temerature interal, 3-

-he unit for Is OC-1 or >1

1ecause resistance is roortional to resistiity, thus we can write the

formulae of resistance as:

($riou# !$teri$l $&e &$riou# &$lue# o) α.

4or e5amle : !iler α = +."% × "%−6 / 7"

Mercury α = %.&$ × "%−6 / 7"

T ∆

∆= ρ

ρ α

0

1

0 ρ ρ ρ −=∆

α

( )1 OO R R T T α = + −


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E?$!*le

A *l$tinu! %ire $# $ re#i#t$nce o) ./ J $t MC. It i#

*l$ced in $ %$ter 8$t %ere it# re#i#t$nce ri#e# to $ )in$l&$lue o) . J. 6$t i# te te!*er$ture o) te 8$t B

Solution

ro!: *)(1+ oo T T R R −+= α

)( ooo T T R R R −+= α

ooo R RT T R −=− )(α

o

o

o T R

R RT +−=α

0.6 0.5 00.5(.$ 10 )−

−= +×

50.%$ CT = °

) ,"


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3.+.1 Electro!oti&e orce e.!.) ," $nd Potenti$l

di))erence ,V

e.m.f ," o) $ 8$tter' i# te !$?i!u! *.d $cro## it# ter!in$l#

%en it i# not connected to $ circuit#. , Refer Figure a

Ter!in$l &olt$0e V i# te *.d $cro## te ter!in$l# o) $ 8$tter'

%en tere i# $ current )lo%in0 trou0 it. , Refer Figure b

SI unit 9 V : &olt , V

Figure a

(

Figure b

3.+ Electro!oti&e orce e.!.) ," Intern$l

Re#i#t$nce ,r 9 Potenti$l Di))erence ,V


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3.+.2 Intern$l Re#i#t$nce o) $ 8$tter' ,r

o In re$lit' %en $ 8$tter' i# #u**l'in0 current it#ter!in$l &olt$0e i# le## t$n it# e.m.f ,

o Ti# reduce# o) &olt$0e i# due to ener0' di##i*$tion

in te 8$tter'. In e))ect te 8$tter' $# intern$l

re#i#t$nce ,r .

1attery (cell)

A ,

o Con#ider $ circuit con#i#tin0 o) $ 8$tter' ,cell t$t i#

connected 8' %ire# to $n e?tern$l re#i#tor R $#

#o%n in i0ure

I r ε

R

I


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o Te intern$l re#i#t$nce o) $ cell i# te re#i#t$nce due

to te ce!ic$l# in te cell.

o Te intern$l re#i#t$nce o) $ cell con#titute# *$rt o)

te tot$l re#i#t$nce in $ circuit.

o Te !$?i!u! current t$t c$n )lo% out )ro! $ cell i#deter!ined 8' te intern$l re#i#t$nce o) te cell.

o Te e!) o) $ 8$tter' i# con#t$nt 8ut te intern$l

re#i#t$nce o) te 8$tter' incre$#e# %it ti!e $# $re#ult o) ce!ic$l re$ction.


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E?$!*le

A 8$tter' $# $n e.m.f. o) 12 ( $nd $n intern$l

re#i#t$nce o) ./ J. It# ter!in$l# $re connected to $lo$d re#i#t$nce o) 3 J.

ind te current in te circuit 9 te ter!in$l &olt$0e o)

te 8$tter'.

Solution

U#in0 : r I R I +=ξ

)( r R I

+= ξ

05.012

+=

.$A I

=


33/89

Te ter!in$l &olt$0e

Anot#er alternat$%e :

U#in0 :

IRV =)($.=

11.% VV =

r I V +=ξ r I V −= ξ

)05.0($.12 −=11.% VV =


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4O88O2 9 ;


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3./ Electric$l Ener0' ,6 $nd Po%er ,P

Te electric ener0' 6 i# te $!ount o) ener0' 0i&en u* 8' $ c$r0e K in *$##in0 trou0 $n electric de&ice.

7ut K It #o :

no%in0 t$t ( IR #o:

3./.1 Electric$l Ener0' ,6

V QW =

)1(W VIt =

2W I Rt =2

V t W

R

=

(2)L

()L

3 / 2 P ,P


36/89

3./.2 Po%er ,Po Po%er P i# de)ined $# te ener0' li8er$ted *er unit

ti!e in te electric$l de&ice.

o Te electric$l *o%er P #u**lied to te electric$l de&icei# 0i&en 8'

o 6en te electric current )lo%# trou0 %ire orre#i#tor ence te *otenti$l di))erence $cro## it i#

ten te electric$l *o%er c$n 8e %ritten $#

o It i# $ #c$l$r ;u$ntit' $nd it# unit i# %$tt# ,!) or 2 s 1.

OR

t

VIt

t

W P == IV P =

IRV =

R I P 2= R

V P

2

=


37/89

". A wire of un0nown comosition has a resistance of 6#.% Ω

when immersed in the water at *%.%°'. 2hen the wire islaced in the oilin water, its resistance rises to +. Ω.

'alculate the temerature on a hot day when the wire has a

resistance of 6.& Ω.,P'#ic#t edition Cutnell 9 on#on K1/ *.35

ANS. : 3.4°

C

*. a. A attery of emf .% > is connected across a "% Ω

resistor. If the otential difference across the resistor is #.%

>, determine

i. the current in the circuit,ii. the internal resistance of the attery.

. 2hen a ".# > dry cell is short?circuited, a current of 6.% Aflows throuh the cell. 2hat is the internal resistance of

the cell@

ANS. : ./ A 2.Ω

./Ω

4O88O2 9 ;


38/89

6. A wire #.% m lon and 6.% mm in diameter has a resistance of

"%% Ω. A "# > of otential difference is alied across the

wire. Determine

a. the current in the wire,

. the resistiity of the wire,c. the rate at which heat is ein roduced in the wire.,Colle0e P'#ic#t edition 6il#on 7u))$ 9 Lou K/ *./45

ANS. : .1/ A 1.+1+ × 1 + Ω ! 2.2/ 6

+. A coer wire has a resistance of *# mΩ at *% °'. 2hen the

wire is carryin a current, heat roduced y the current

causes the temerature of the wire to increase y * °'.

a. 'alculate the chane in the wires resistance.

. If its oriinal current was "%.% mA and the otential differenceacross wire remains constant, what is its final current@

(Gien the temerature coefficient of resistiity for coer is

.&% × "%−6 °'−")

ANS. : ,$ +./5Q1 >3 J ,8 4.+/Q1 >3 A

3 R i t I S i A d P ll l


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3..1 Re#i#tor# in #erie#

3. Re#i#tor# In Serie# And P$r$llel

Te #$!e current I )lo%# trou0 e$c re#i#tor %ere

Con#ider tree re#i#tor# $re connected in #erie# to te

8$tter' $# #o%n in i0ure 8elo%.

21 I I I I

===


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Te *otenti$l di))erence ,&olt$0e $**lied $cro## te

#erie# co!8in$tion o) re#i#tor# %ill di&ide 8et%een te

re#i#tor#.

To re*l$ce te re#i#tor# 8' one re#i#tor %ic $# $ne;ui&$lent re#i#t$nce RE 9 !$int$in te #$!e current

%e $&e :

Su8#titute ,2 into ,1 :

21 V V V V ++=

- (2)V IR= L

)1(21 IR IR IRV ++=

- 1 2 IR IR IR IR= + +


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E?tendin0 ti# re#ult to te c$#e o) n re#i#tor# connectedin #erie# co!8in$tion:

C$ncelin0 te co!!on I"# %e 0et :

%ere R E i# te e))ecti&e ,e;ui&$lent re#i#t$nce

- 1 2 R R R R= + +

- 1 2 n... R R R R R= + + + +

3 2 R i t i ll l


42/89

3..2 Re#i#tor# in *$r$llel

S$!e *otenti$l di))erence ( $cro## te re#i#tor# %ere

Current di&ide# into di))erent *$t $t te unction.

21 I I I I ++=

)1(21

R

V

R

V

R

V I ++=

21 V V V V ===


43/89

Re#i#tor# connected in *$r$llel c$n 8e re*l$ced %it $n

e;ui&$lent re#i#tor RE t$t $# te #$!e *otenti$l

di))erence ( 9 te #$!e tot$l current I $# te $ctu$l

re#i#tor#.

Su8#titute ,2 into ,1 :

C$ncelin0 te co!!on ("# %e 0et :

-

(2)V

I R

= L

- 1 2

V V V V

R R R R= + +

- 1 2

1 1 1 1

R R R R= + +


44/89

E?tendin0 ti# re#ult to te c$#e o) n re#i#tor# connected

in *$r$llel co!8in$tion:

E?$!*le

6$t i# te e;ui&$lent re#i#t$nce o) te re#i#tor# in

)i0ure 8elo% B

A

7R 1

R 2

R 3R +

R 1 R 2 R 3 R + 1

- 1 2 n

1 1 1 1 1... R R R R R

= + + +

S l ti


45/89

Anoter &ie% o) te circuit :

A

7

R1

R2

R3R+

Circuit $# 8een reduced to :

A

7

R1

R2

R3+

R3+ 9 R2 in #erie# tu#:

Solution

##

111

R R R+= 2

1

1

1

1=+=

#

1

2 R = Ω

#22# R R R +=

1 1

2 2

= + = Ω

Ci it 8 d d t


46/89

Circuit $# 8een reduced to :

A

7

R1R23+

E?$!*le

ind te current in 9 &olt$0e o) te 1 J re#i#tor #o%n

in i0ure.

- 1 2#

1 1 1

R R R= +

5

2

1

1=+=

5 E R = Ω

R

1 R

2 R

S l ti


47/89

Solution

1#t )ind te RE )or te circuit 9 0et te current I )lo% in

te circuit

p 1 2

1 1 1

R R R= +

2

1

10

1+=

10

6=

p 1.6" R = Ω

- p R R R= +0.56".1 +=

6.6" E R = Ω


48/89


49/89


50/89

In te #erie# c$#e te

#$!e current )lo%#

trou0 8ot 8ul8#.

I) one o) te 8ul8#

8urn# out tere %ill

8e no current $t $ll in

te

circuit $nd neiter

8ul8 %ill 0lo%

In te *$r$llel c$#e te *otenti$l

di))erence $cro## eiter 8ul8

re!$in# e;u$l i) one o) te 8ul8#

8urn# out. Tecurrent trou0 te )unction$l 8ul8

re!$in# e;u$l $nd te *o%er

deli&ered to t$t 8ul8 re!$in# te

#$!e . Ti# i# $noter o) te !erit# o)

$ *$r$llel $rr$n0e!ent o) li0t 8ul8#: I)

one )$il# te oter 8ul8# $re

un$))ected. Ti# *rinci*le i# u#ed in

ou#eold %irin0 #'#te!#

4O88O2 9 ;


51/89

%'467 8 sos te arrangement of five e/ualresistors in a circuit. Calculate

i. te e/uivalent resistance beteen point x and y.

ii. te voltage across point b and c.

iii. te voltage across point c and y.

ANS : ,i 4 ,ii 3 ( ,iii 5 (

(")

4O88O2 9 ;


52/89

4or the circuit aoe, calculate

a. the effectie resistance of the circuit,

. the current asses throuh the "* Ω resistor,

c. the otential difference across +.% Ω resistor,

d. the ower deliered y the attery.

-he internal resistance of the attery may e inored.

(*)

ANS : ,$ 1.24 ,8 ./A ,c 2 ( ,d 3 6

Ω0.#

Ω0.2

V0.%

Ω12

3 i ))" L

A #t$te!ent o)


53/89

3. irco))"# L$%

3..1 irco))"# 1#t L$% : unction Rule

Te #u! o) te current# enterin0 $n' unction in $ circuit!u#t e;u$l te #u! o) te current# le$&in0 t$t unction.

E?$!*le:

A #t$te!ent o)

con#er&$tion

o) electric

c$r0e

1 I

2 I I

21 III =+12 III =+

I 2 I 1 I

∑∑ = outin I I

3 2 i ))" 2 d L L R l


54/89

3..2 irco))"# 2nd L$% : Loo* Rule

Te $l0e8r$ic #u! o) te con#t$nt &olt$0e e.m.f # i#$l%$'# e;u$l to te &olt$0e dro*# $round $n' clo#ed

electric$l loo*.

Si0n con&ention )or 9 IR , &olt$0e dro*

direction o) loo*

@ > >@

direction o) loo*

ollo%# )ro!

te l$%# o)

con#er&$tion

o) ener0'

IRξ =∑ ∑

ε−ε+

ε ε

di ti ) l di ti ) l


55/89

direction o) loo* direction o) loo*

3..3 Pro8le! #ol&in0 #tr$te0' ,irco))"# L$%#

o Coo#e $nd l$8elin0 the current at each Bunction in the

circuit ien.o Coo#e $n' one unction in the circuit and $**l' te

irco))"# )ir#t l$%.o Coo#e $n' t%o clo#ed loo*# in the circuit and

desinate a direction (cloc%i#e OR $nticloc%i#e) totrael around the loo in $**l'in0 te irco))"##econd l$%.

o Sol&in0 te #i!ult$neou# e;u$tion to determine theun0nown currents and un0nown ariales.

IR+

I

R IR−

I

R


56/89

or te circuit in i0ure /.24 Deter!ine te current$nd it# direction in te circuit.

E?$!*le

Ω1.15

Ω.226

Ω50.% Ω2V1.51

Ω#V5.01


57/89

Solution :

7' $**l'in0 te irco))"# 2nd l$% tu#

Loo* 1

,$nticloc%i#e

IRξ =∑ ∑

A"#.0= I I I I I I #50.%222.61.155.110.15 ++++=+

Ω1.15

Ω.226

Ω50.% Ω2V1.51

Ω#V5.01 I

I

I I

E l


58/89

E?$!*le

ind te &$lue o) current I re#i#t$nce R 9 te

e.m.f

D

A

E

7

C

I R

Ω1

Ω

1 1A I =

2 2A I =

12V

Solution


59/89

Con#ider unction 7 9 $**l' te irco))"# 1#t l$% :

7re$in0 te 0i&en co!*le? circuit into #e&er$l #i!*le clo#ed

circuit# 9 $**l' te irco))"# 2nd l$%:

or clo#ed loo* DCED : , tr$&er#e cloc

%i#e

out in I I Σ=Σ21 I I I += 21+=

A I =

I 1

I 2 I

IRΣ=Σξ )(12 2 I IR +=)(212 += R

2 R = Ω

l d l A7EA , t l i


60/89

or clo#ed loo* A7EA : , tr$&er#e cloc %i#e

* The –ve sign shows that the polarity of ξ isactually the reverse of that shown in

gure.

* If the value of current is –ve , this showsthat the actual direction of current ow is

reverse of that shown in gure.

IR

Σ=Σξ

)()1( 21 I I +−=−ξ 61+−=−ξ

5 V

5 V

ξ

ξ

− == −

E?$!*le


61/89

E?$!*le

Re)errin0 to te circuit in i0ure c$lcul$te te current

I1I2 $ndI3

Solution


62/89

Solution

Loo* 1 Loo* 2

a

b

c

d

e

At ti


63/89

At unction c :

Re)er Loo* 1 , $cd8$ > cloc%i#e

Re)er Loo* 2 , ce)dc> cloc%i#e

out in I I Σ=Σ

)1(21

I I I +=

IRΣ=Σξ

1121 102010100 I I I I +++= )2(10#00 21 I I +=

IRΣ=Σξ 2 101020101020 I I I I −++=−

)(10#010 2 I I −=

S 8 tit t ,2 i t ,1


64/89

Su8#titute ,2 into ,1:

Sol&e ,3 9 ,+ #i!ult$neou#l' :

-------------------------------------------------

Su8#tituteI2 .3333 A into ,3 %e 0et :

Su8#tituteI2 9I3 into ,1 %e 0et :

22 10)(#00 I I I ++=

)#(#0500 2 I I +=

)(10#010 2 I I −=

)#(#0500 2 I I +=

:)#()( −

26020 I −=−2 0.A I =

0.A I =

1 0.6666A I =

4O88O2 9 ;


65/89

4or the circuit aoe, determine

a. the currents I 1, I 2 and I ,

. the otential difference across the . Ω resistor,c. the ower dissiated from the ".* Ω resistor.

,1

ANS: ,$ I1 .2 A I21.3 A I 1./ A

,8 .5 (

,c 3.4 6

4O88O2 9 ;


66/89

,2

Te circuit #o%n in i0. ,$ cont$in# t%o 8$tterie# e$c

%it $n e!) $nd $n intern$l re#i#t$nce $nd t%o re#i#tor#.

ind

,$ te current in te circuit

,8 te *otenti$l di))erence ($8 %it re#*ect to 8 $nd

,c Te *o%er out*ut o) te e!) o) e$c 8$tter'

An# : ,$ ./ A ,8 5./ ( ,c 6 >2 6

ind.

1attery P with e.m.f ".6 > and internal resistance * C, attery K with e.m.f ".#,3


67/89

> and internal resistance %.& C and a + C resistor are connected in arallel.

(i) !0etch the circuit diaram.

(ii) 'alculate the current in attery P, attery K and the + C resistor.

(iii) 'alculate the otential difference across the + resistor.

Ans , (ii) 0.#-8A9 0.01 A9 0.31 A9 (iii) 1.# V

Referin to the circuit in 4iure elow, calculate

(a)-he current, I that flows in R resistor.

()-he resistance of R resistor.

(c)-he alue of emf

(d)-he current that flows in R resistance if the circuit is cut off at oint 5.

(Internal resistance of the emf source is neliile).

,+

Ans , (a) #A 9 (b) 5: 9 (c) # V (d) 3.5 A

(#)


68/89

e internal resistance of all te batteries in %'467

are negligible. Calculate te current I 1 I 2 and I en

sitc is

i. open. ii. closed.

( )

Ans , (a) '1;'# ; 1.3 A '3 ; 0A 9 (ii) #.5# A 1.15A 1.38 A

3 4 Potenti$l Di&ider


69/89

3.4 Potenti$l Di&ider

-- u#ed to o8t$in $n' de#ired #!$ller *ortion o) &olt$0e

)ro! $ #in0le &olt$0e #ource V o

-- 2 re#i#tor# R 1 9 R 2 $re connected to $ &olt$0e #ource

, 8$tter' %it &olt$0e V o

oV sourceVoltage

1 R 2 Ra c

b

1V

o - - 1 2 ereV IR R R R∴ = = +


70/89

-- te current trou0 R1 9 R2 $re #$!e.

Su8#titute ,1 into te e;u$tion:

-- 8' u#in0 di))erent &$lue# o) R 1 9 R 2 di))erent

&olt$0e c$n 8e o8t$ined )ro! $ &olt$0e #ource V o

,8$tter'.

Si!il$rl':

o

1 2

(1)( )

V I

R R⇒ =

+ K

1

1 o

1 2( ) RV V

R R=

+

11: IRV From =

2

2 o

1 2( )

R

V V R R= +


71/89

4or the circuit in 4iure elow aoe,

a. calculate the outut oltae.. If a oltmeter of resistance +%%% Ω is connected across

the outut, determine the readin of the oltmeter.

E?$!*le

Ω000#

V21

Ω000%

outV


72/89

Solution :

a. -he outut oltae is ien y

. -he connection etween the oltmeter and +%%% Ω resistor is

*$r$llel, thus the eEuialent resistance is

Fence the new outut oltae is ien y

-herefore the re$din0 o) te &olt!eter i# 2.+ (.

V12#000%000 21 =Ω=Ω= V R R

V R R

RV += 212out

V0.#out =V

#000

1

#000

11

e/

+= R

12#000%000

#000out +=V

Ω= 2000e/ R

V#.2out =V 122000%000

2000out

+=V

3.5 Potentio!eter $nd 6e$t#tone 7rid0e


73/89

0

3.5.1 Potentio!eter

• 'onsider a otentiometer circuit is shown in 4iure elow

• -he otentiometer is 8$l$nced when the Boc0ey (slidin contact)

is at such a osition on wire A1 that there is no current

trou0 te 0$l&$no!eter . -hus

,Dri&er cell -$ccu!ul$tor

oce'

?

,Unno%n &olt$0e

V

,A C

3V

I I

I I

l t di


74/89

• 2hen the otentiometer in alanced, the unno%n

&olt$0e ,*otenti$l di))erence 8ein0 !e$#ured i#e;u$l to te &olt$0e $cro## AC.

• otentiometer can e used to – co!*$re te e!)# of two cells.

– !e$#ure $n unno%n e!) of a cell.

– !e$#ure te intern$l re#i#t$nce of a cell.

$l&$no!eter re$din0

?

AC3 V V =

V

,AC

3

V

I I

I I

A**lic$tion : Co!*$re te e!)# o) t%o cell# or )indunno%n e!)


75/89

unno%n e!) o In this case, a otentiometer is set u as illustrated in 4iure

elow, in which A1 is a wire of uniform resistance and H is aslidin contact (Boc0ey) onto the wire.

o An accumulator 4 maintains a steady current I throuh thewire A1.

,2

,1S

4

,A I

I

2ξ

I I

1ξ

C 1l

2l

o Initially, switch ! is connected to the terminal (") and the Boc0ey

moed until the emf ! e5actly alances the otential difference ( d )


76/89

o After that, the switch ! is connected to the terminal (*) and the

Boc0ey moed until the emf ! 2 alances the .d. from the

accumulator at oint D. Fence

where and

,1

then

where and

,2then

moed until the emf ! 1 e5actly alances the otential difference (.d.)

from the accumulator (alanometer readin is ero) at oint '.

Fence

,2

,1S

1 ACV ξ =

ACAC IRV = A

ρl R 1AC =

11

ρl I

A

ξ =

2 A

V ξ =

AA IRV = A

ρl R 2A =

22

ρl I

A

ξ =

4

,A I

I

2ξ

I I

1ξ

C 5

• 1 di idi (") d (*) th


77/89

• 1y diidin eE. (") and eE. (*) then

!ince

;Euation aoe can e written as

1 1

2 2

l

l ξ ξ

=

1

1

22

ρl I

A ρl

I A

ξ ξ

=

1 ρl R R l A

= ⇒ ∝

1 1

2 2

R

R

ξ

ξ =

E?$!*le


78/89

Solution

U#in0 :

Con#ider $ *otentio!eter. I) $ #t$nd$rd 8$tter' %it $n

e.m.f.o) 1.14 ( i# u#ed in te circuit. 6en te

re#i#t$nce i# 3 J te 0$l&$no!eter re$d# =ero. I) te#t$nd$rd 8$tter' i# re*l$ced 8' $n unno%n e.m.f.te

0$l&$no!eter re$d# =ero %en te re#i#t$nce i#

$du#ted to +4 J. 6$t i# te &$lue o) te unno%n

e.m.f.B

2

1.01%6 6

#%ξ =

2 1.5% Vξ =

1 1

2 2

R

R

ξ

ξ =

4O88O2 9 ;


79/89

". In 4iure ", J is a uniform wire of lenth ".% m and

resistance "%.% Ω.

T

i0ure 1

ξ 1 is an accumulator of emf *.% >

and neliile internal resistance.

R1 is a "# Ω resistor and R2 is a #.%

Ω resistor when !" and !* oen,

alanometer G is alanced whenJ- is *.# cm. 2hen oth !" and

!* are closed, the alance lenth is

"%.% cm. 'alculate

a. the emf of cell ξ 2

.

. the internal resistance of cell ξ 2.

ANS. : ,$ ./ ( ,8 ./Ω

4O88O2 9 ;


80/89

a uniform wire O! usin Boc0eys < and K as shown in 4iure *.

a. the otential difference across OK when OK = #.% cm,

. the otential difference across OK when K touches ! and the

alanometer is alanced,

c. the internal resistance of the cell A,

d. the emf of cell A.

i0ure 2

V

-he lenth of the uniform wire O! is".%% m and its resistance is "* Ω.

2hen OK is #.% cm, the

alanometer does not show any

deflection when O


81/89

0 W X

-- u#ed to o8t$in $n $ccur$te !e$#ure!ent o) $n

unno%n re#i#t$nce R?

ε


82/89

-- Te 8rid0e i# !$de u* o) + re#i#tor $r!# R1 R2 R3 R?

$nd *oint# CD $re oined 8' $ center lin or Y8rid0e"

%ic cont$in# $ 0$l&$no!eter.

-- one o) te no%n re#i#tor , )or e?$!*le R1 i#

&$ried until te 0$l&$no!eter re$din0 i# =ero.

,A

C

1 R2

R

R 3 R

0=

I I

2 I

1 I

2 I

1 I

-- No current )lo%# trou0 C to D tu# tere %ill 8e


83/89

0

no &olt$0e di))erence 8et%een *oint# C 9 D (CD

-- under ti# condition te 8rid0e i# #$id to 8e

8$l$nced.-- Tere)ore

-- Re#i#tor# R1R2 c$rr' te #$!e current I1

-- Re#i#tor# R3R c$rr' te #$!e current I2

"#$# A" A$ V V V V ==

)1(211 R I R I =

)2(221 % R I R I =(2)

:(1)

3 2

) 1

R R

R R=

E?$!*le


84/89

A %e$t#tone 8rid0e i# u#ed to !$e $ *reci#e

!e$#ure!ent o) te re#i#t$nce o) $ %ire connector. I)

R 1 J 9 te 8rid0e i# 8$l$nced 8' $du#tin0 P #uct$t P 2./ K %$t i# te &$lue o) B

Solution

U#in0 :

1 10

2.5

%

Q Q×=

Ω= #00 %

1 R

2 R

) R

% R

2

1

% R R

R R=

4O88O2 9 ;


85/89

". -he circuit shown in fiure is 0nown as a 2heatstone ride.

Determine the alue of the resistor R such that the currentthrouh the .% Ω resistor is ero.

,P'#ic#3t edition $!e# S. 6$ler K53 *.31

ANS. : ./ Ω

4O88O2 9 ;


86/89

-he alication of the 2heatstone ride is


87/89

• -he metre ride is 8$l$nced when the

Boc0ey H is at such a osition on wire A1

that there is no current trou0 te0$l&$no!eter . -hus the current I 1 flows

throuh the resistance R3 and R ut

current I 2 flows in the wire A1.

• 8et V 3 : .d. across R3 and V : .d. across R,

At l diti


88/89

– At alance condition,

1y alyin Ohms law, thus

Diidin ies

and

and

where and

A73 V V = 7,V V =

A7231 R I R I = 7,21 R I R I =

A

ρl R 1A7 =

7,2

A72

1

31

R I

R I

R I

R I =

A

ρl R 27, =

=

A

ρl

A

ρl

R

R

2

1

3

Rl

l R

=

2

13


89/89

Next Chapter…

CHAPTER 4 :Magnetic field

chapter 3_ electrics & direct current_2016_reviewed

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