chapter 3_ electrics & direct current_2016_reviewed
TRANSCRIPT
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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$l
t
Q I =
dt
dQ I =
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SI Unit o) electric current i# A!*ere ,A Sc$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 *% °'.
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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
+=
.$A I
=
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Te ter!in$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
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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
=
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". 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 ;
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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
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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 $t te unction.
21 I I I I ++=
)1(21
R
V
R
V
R
V I ++=
21 V V V V ===
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Re#i#tor# connected in *$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$ncelin0 te co!!on ("# %e 0et :
-
(2)V
I R
= L
- 1 2
V V V V
R R R R= + +
- 1 2
1 1 1 1
R R R R= + +
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E?tendin0 ti# re#ult to te c$#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
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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
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Circuit $# 8een reduced to :
A
7
R1R23+
E?$!*le
ind te current in 9 &olt$0e o) te 1 J re#i#tor #o%n
in i0ure.
- 1 2#
1 1 1
R R R= +
5
2
1
1=+=
5 E R = Ω
R
1 R
2 R
S l ti
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Solution
1#t )ind te RE )or te circuit 9 0et te current I )lo% in
te circuit
p 1 2
1 1 1
R R R= +
2
1
10
1+=
10
6=
p 1.6" R = Ω
- p R R R= +0.56".1 +=
6.6" E R = Ω
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In te #erie# c$#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 ;
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%'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 ;
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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$te!ent o)
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3. irco))"# L$%
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$r0e
1 I
2 I I
21 III =+12 III =+
I 2 I 1 I
∑∑ = outin I I
3 2 i ))" 2 d L L R l
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3..2 irco))"# 2nd L$% : 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&$tion
o) ener0'
IRξ =∑ ∑
ε−ε+
ε ε
di ti ) l di ti ) l
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direction o) loo* direction o) loo*
3..3 Pro8le! #ol&in0 #tr$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
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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
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Solution :
7' $**l'in0 te irco))"# 2nd l$% tu#
Loo* 1
,$nticloc%i#e
IRξ =∑ ∑
A"#.0= I I I I I I #50.%222.61.155.110.15 ++++=+
Ω1.15
Ω.226
Ω50.% Ω2V1.51
Ω#V5.01 I
I
I I
E l
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E?$!*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
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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$&er#e cloc
%i#e
out in I I Σ=Σ21 I I I += 21+=
A I =
I 1
I 2 I
IRΣ=Σξ )(12 2 I IR +=)(212 += R
2 R = Ω
l d l A7EA , t l i
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or clo#ed loo* A7EA : , tr$&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?$!*le
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E?$!*le
Re)errin0 to te circuit in i0ure c$lcul$te te current
I1I2 $ndI3
Solution
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Solution
Loo* 1 Loo* 2
a
b
c
d
e
At ti
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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
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Su8#titute ,2 into ,1:
Sol&e ,3 9 ,+ #i!ult$neou#l' :
-------------------------------------------------
Su8#tituteI2 .3333 A into ,3 %e 0et :
Su8#tituteI2 9I3 into ,1 %e 0et :
22 10)(#00 I I I ++=
)#(#0500 2 I I +=
)(10#010 2 I I −=
)#(#0500 2 I I +=
:)#()( −
26020 I −=−2 0.A I =
0.A I =
1 0.6666A I =
4O88O2 9 ;
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4or the circuit aoe, determine
a. the currents I 1, I 2 and I ,
. the otential difference across the . Ω resistor,c. the ower dissiated from the ".* Ω resistor.
,1
ANS: ,$ I1 .2 A I21.3 A I 1./ A
,8 .5 (
,c 3.4 6
4O88O2 9 ;
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,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
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> 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
(#)
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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$l Di&ider
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3.4 Potenti$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∴ = = +
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-- 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$rl':
o
1 2
(1)( )
V I
R R⇒ =
+ K
1
1 o
1 2( ) RV V
R R=
+
11: IRV From =
2
2 o
1 2( )
R
V V R R= +
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4or the circuit in 4iure elow aoe,
a. calculate the outut oltae.. If a oltmeter of resistance +%%% Ω is connected across
the outut, determine the readin of the oltmeter.
E?$!*le
Ω000#
V21
Ω000%
outV
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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
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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
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• 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!*$re te e!)# o) t%o cell# or )indunno%n e!)
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unno%n e!) o In this case, a otentiometer is set u as illustrated in 4iure
elow, in which A1 is a wire of uniform resistance and H is aslidin contact (Boc0ey) onto the wire.
o An accumulator 4 maintains a steady current I throuh thewire A1.
,2
,1S
4
,A I
I
2ξ
I I
1ξ
C 1l
2l
o Initially, switch ! is connected to the terminal (") and the Boc0ey
moed until the emf ! e5actly alances the otential difference ( d )
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o After that, the switch ! is connected to the terminal (*) and the
Boc0ey moed until the emf ! 2 alances the .d. from the
accumulator at oint D. Fence
where and
,1
then
where and
,2then
moed until the emf ! 1 e5actly alances the otential difference (.d.)
from the accumulator (alanometer readin is ero) at oint '.
Fence
,2
,1S
1 ACV ξ =
ACAC IRV = A
ρl R 1AC =
11
ρl I
A
ξ =
2 A
V ξ =
AA IRV = A
ρl R 2A =
22
ρl I
A
ξ =
4
,A I
I
2ξ
I I
1ξ
C 5
• 1 di idi (") d (*) th
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• 1y diidin eE. (") and eE. (*) then
!ince
;Euation aoe can e written as
1 1
2 2
l
l ξ ξ
=
1
1
22
ρl I
A ρl
I A
ξ ξ
=
1 ρl R R l A
= ⇒ ∝
1 1
2 2
R
R
ξ
ξ =
E?$!*le
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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 &$lue o) te unno%n
e.m.f.B
2
1.01%6 6
#%ξ =
2 1.5% Vξ =
1 1
2 2
R
R
ξ
ξ =
4O88O2 9 ;
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". In 4iure ", J is a uniform wire of lenth ".% m and
resistance "%.% Ω.
T
i0ure 1
ξ 1 is an accumulator of emf *.% >
and neliile internal resistance.
R1 is a "# Ω resistor and R2 is a #.%
Ω resistor when !" and !* oen,
alanometer G is alanced whenJ- is *.# cm. 2hen oth !" and
!* are closed, the alance lenth is
"%.% cm. 'alculate
a. the emf of cell ξ 2
.
. the internal resistance of cell ξ 2.
ANS. : ,$ ./ ( ,8 ./Ω
4O88O2 9 ;
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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
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0 W X
-- u#ed to o8t$in $n $ccur$te !e$#ure!ent o) $n
unno%n re#i#t$nce R?
ε
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-- 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
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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?$!*le
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A %e$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 ;
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". -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 ;
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-he alication of the 2heatstone ride is
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• -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
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– 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
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Next Chapter…
CHAPTER 4 :Magnetic field