Physics, 6th Edition
Chapter 27. Current and Resistance Chapter 27. Current and Resistance Electric Current and Ohm’s Law
27-1. 27-1.
How many many electrons electrons pass a point point every second second in a wire wire carryi carryin n a current current o! o! 2" #$ #$ How much much time is needed to transport %" C o! chare past this point$
C
1e -1* s 1.6 + 1" C )
Q = 2" Q = It = &2" C's(&1 s()
I
27-2. 27-2.
=
Q t
)
t =
I
%" C
= 2."" 2."" s
2" #
)
Q t
=
6"" C 0s
) I = 2" #
Q t
=
6*" C 12" s
)
Q 12"" C
3hat is the potential drop across a %- Ω resistor with a current o! 4 # passin throuh it$ V = IR = &4 #(&% Ω()
27-6.
I = .7 #
/! a current o! 2% # e+ists !or " s, how how many couloms couloms o! chare have have passed throuh throuh the wire$ Q = It = &2% #(&" s()
27-.
t = 2."" s
ind the current in amperes when 6*" C o! chare chare pass a iven point in in 2 min.
I =
27-%.
=
/! 6"" C o! chare chare pass a iven iven point point in 0 s, s, what is is the electric electric current current in amperes amperes$$
I =
27-0.
Q
1.2 + 1"2" electrons's
V = 02." 5
ind the resistance o! a rheostat i! the drop drop in potential is is %4 5 and the current current is % #. #.
11*
Physics, 6th Edition
Chapter 27. Current and Resistance V
R =
27-7.
I
%#
)
R = 12." Ω
etermine the current throuh a -Ω resistor that has a %"-5 drop in potential across it$
I =
27-4.
%4 5
=
V R
=
%" 5 Ω
) I = 4."" #
# 2-# !use is placed in a circuit with a attery havin a terminal voltae o! 12 5. 3hat is the minimum resistance !or a circuit containin this !use$
R =
7-*.
V I
=
12 5 2#
)
R = 6."" Ω
3hat em! is reuired to pass 6" m# throuh a resistance o! 2" 8 Ω$ /! this same em! is applied to a resistance o! 0""
Ω, what will e the new current$ E IR
6" + 1" -0 #(&2" + 1"0 Ω()
I =
E
R
=
E
12"" 5
12"" 5 0"" Ω )
I %."" #
Electric Power and Heat Loss
27-1". # solderin iron draws ".7 # at 12" 5. How much enery will it use in 1 min$ P = IV = &".7 #(&12" 5() P = *"." 3)
P =
Work ) t
t = 1 min *"" s
Energy = Pt = &*" 3(&*"" s( ;
12"
E = 41,""" 9
Physics, 6th Edition
Chapter 27. Current and Resistance
27-11. #n electric lamp has an 4"-Ω !ilament connected to a 11"-5 direct-current line. 3hat is the current throuh the !ilament$ 3hat is the power loss in watts$
I =
P =
V R
V 2 R
=
=
11" 5 4" Ω
) / 1.04 #
&11" 5(2 4" Ω
) P = 11 3
27-12. #ssume that the cost o! enery in a home is 4 cents per 8ilowatt-hour. # !amily oes on a 2-wee8 vacation leavin a sinle 4"-3 liht ul urnin. 3hat is the cost$ E = Pt = &4" 3(&2 w8(&7 day'w8(&2% h'day(&06"" s'h( 26.* 83 h E &26.* 83 h(&"."4 c'8w h(
:2.1
(Rates vary)
27-10. # 12"-5, direct-current enerator delivers 2.% 83 to an electric !urnace. 3hat current is supplied$ 3hat is the resistance$
I =
P V
=
2%""3 12" 5
R =
) I = 20 A;
V I
=
12" 5 2" #
) R = 6."" Ω
27-1%. # resistor develops heat at the rate o! 2" 3 when the potential di!!erence across its ends is 12" 5. 3hat is its resistance$
P =
V2 R
)
R=
V 2 P
=
&12" 5(2 2" 3)
R = 7.6 Ω
27-1. # 12"-5 motor draws a current o! %." #. How many ;oules o! electrical enery is used in one hour$ How many 8ilowatt-hours$
121
Physics, 6th Edition
Chapter 27. Current and Resistance P = VI = &12" 5(&%." #( %4" 3
P =
E ) t
E = Pt = &%4" 3(&06"" s() E = 1.70 <9
1 83 ⋅ h 0.6" + 1"6 9
E = 1.70 + 1"6 9
E ".%4" 83 h
27-16. # household hair dryer is rated at 2""" 3 and is desined to operate on a 12"-5 outlet. 3hat is the resistance o! the device$
V2
P =
R
)
R=
V 2 P
=
&12" 5( 2 2""" 3
)
R = 7.2" Ω
Resistivity
27-17. 3hat lenth o! copper wire 1'16 in. in diameter is reuired to construct a 2"- Ω resistor at 2""C$ 3hat lenth o! nichrome wire is needed$
ρ 1.74 + 1"-4 Ω m )
Copper = 1 16
nichroe =
!t "."62 in. 62. mil)
R =
ρ !
R =
A
)
ρ ! A
)
! =
! =
RA ρ
RA ρ
=
=
ρ 1"" + 1"-4 Ω m
A = &62. mil(2
= 0*"6 cmil
Ω(&0*"6 cmil( ) 1".% Ω ⋅ cmil'!t
! = 71" !t
Ω(&0*"6 cmil( ) 6"" Ω⋅ cmil'!t
! = 10" !t
&2"
&2"
122
Physics, 6th Edition
Chapter 27. Current and Resistance
27-14. # 0."-m lenth o! copper wire & ρ 1.74 + 1" -4 Ω m( at 2""C has a cross section o! mm2.3hat is the electrical resistance o! this wire$
R =
ρ ! A
%
> # % mm 2 % + 1" -6 m2 ?
&1.72 + 1"-4Ω ⋅ m(&0." m(
=
%."" + 1"-6 m2
)
R =12.* mΩ
27-1*. ind the resistance o! %" m o! tunsten & ρ . + 1"-4 Ω m( wire havin a diameter o! ".4 mm at 2""C$
A =
R =
π " 2
ρ ! A
=
%
=
π &"."""4 m(2
%
A = ."0 + 1"-7 m2
)
&. + 1"-4Ω ⋅ m(&%"." m( ."0 + 1"-7 m2
)
R = %.07 Ω
27-2". # certain wire has a diameter o! 0 mm and a lenth o! 1" m. /t has a resistance o! at 2""C. 3hat is the resistivity$
R =
ρ ! A
) ρ
RA !
=
0.""
> # π2/4 7."7 + 1"-7 m2. ?
&0 Ω(&7."7 + 1"-7 m2 ( 1" m
)
ρ 1.%1 + 1"-4 Ω m
27-21. 3hat is the resistance o! 2"" !t o! iron & ρ *. + 1" -4 Ω m( wire with a diameter o! ".""2 in. at 2""C$ &ρ *. + 1" -4 Ω m(. > 2"" !t 61." m) ".""2 in. ."4 + 1"- m ?
A =
π " 2
%
=
π &."4 + 1"- m(2
%
)
120
A = 2."0 + 1"-* m2
Ω
Physics, 6th Edition
Chapter 27. Current and Resistance
R =
ρ ! A
&*. + 1"-4Ω ⋅ m(&61." m(
=
."4 + 1"- m2
R = 246" Ω
)
*27-22. # nichrome wire &ρ 1"" + 1" -4 Ω m( has a lenth o! %" m at 2""C 3hat is the diameter i! the
total resistance is
R =
ρ ! A
A =
Ω$ A=
)
π " 2
%
)
ρ ! R
&1"" + 1"-4Ω ⋅ m(&%" m(
=
."" Ω
%A
"=
π
=
A 4 + 1" -6 m2
2&4 + 1"-6 m2 ( π
" 2.26 mm
*27-20. # 11-5 source o! em! is attached to a heatin element which is a coil o! nichrome wire & ρ 1"" +
1"-4 Ω m ( o! cross section 1.2" mm 2 3hat must e the lenth o! the wire i! the resistive power loss is to e 4"" 3$ ># 1.2" mm 2 1.2" + 1" -6 m2 ?
P =
R =
V2 R
ρ ! A
)
)
R=
! =
V 2 P
RA ρ
=
=
&11 5(2 4"" 3
=16. Ω
&16. Ω(&1.2" + 1"-6 m2 ( 1"" + 1"-4 Ω ⋅ m
R = 16. Ω
) ! = 1*.4 m
!emperature Coe""icient o" Resistance
27-2%. @he resistance o! a lenth o! wire & α ".""6'C"( is %."" Ω at 2""C. 3hat is the resistance at 4""C$ > ∆t 4""C A 2""C 6" C" ?
12%
Physics, 6th Edition
Chapter 27. Current and Resistance
∆ R = α Ro ∆t = &".""6' C" (&% Ω(&6" C" ( = 1.6 Ω )
R %."" Ω B 1.6 Ω .6 Ω
27-2. /! the resistance o! a conductor is 1"" Ω at 2""C, and 116 Ω at 6""C, what is its temperature coe!!icient o! resistivity$
α =
>
∆t 6""C A 2""C %" C" ?
∆ R 116 Ω - 1"" Ω = R" ∆t &1"" Ω(&%" C" ( )
α = ".""%"" 'C"
27-26. # lenth o! copper &α ".""%0'C"( wire has a resistance o! 4
Ω at 2""C.
3hat is the resistance at
*""C$ #t - 0""C$
∆ R = &".""%0' C" (&4 Ω(&7" C" ( = 2.%1 Ω )
R 4."" Ω B 2.%1 Ω 1".%1 Ω
∆ R = &".""%0 ' C" (&4 Ω(&-0""C − 2"" C( = −1.72 Ω )
R 4."" Ω - 1.72 Ω 6.24 Ω
27-27. @he copper windins &α ".""%0'C"( o! a motor e+perience a 2" percent increase in resistance over their value at 2" "C. 3hat is the operatin temperature$
∆ R R
= ".2) ∆t =
∆R R"α
=
".2 "
".""%0' C
= %6. C" ) t 2""C B %6. C" 66. "C
27-24. @he resistivity o! copper at 2" "C is 1.74 + 1" -4 Ω m. 3hat chane in temperature will produce a 2 percent increase in resistivity$
Challen#e Pro$lems
27-2*. # water turine delivers 2""" 83 to an electric enerator which is 4" percent e!!icient and has an output terminal voltae o! 12"" 5. 3hat current is delivered and what is the electrical resistance$ > Pout &".4"(&2""" 83( 16"" 83 ?
12
Physics, 6th Edition
Chapter 27. Current and Resistance
P = VI )
I =
R =
V I
P 16"" + 1"0 3 V
=
=
12"" 5
) I 100" #
12"" 5 10"" # )
R ".*"" Ω
27-0". # 11"-5 radiant heater draws a current o! 6." #. How much heat enery in ;oules is delivered in one hour$
P =
E t
= VI )
E = VIt = &11" 5(&6 #(&06"" s() E 2.04 <9
27-01. # power line has a total resistance o! % 8 Ω. 3hat is the power loss throuh the wire i! the current is reduced to 6." m#$
P = I 2 R = &".""6 #(2 &%""" Ω ()
P =1%% m3
27-02. # certain wire has a resistivity o! 2 + 1" -4 Ω m at 2""C. /! its lenth is 2"" m and its cross section is % mm2, what will e its electrical resistance at 1"" "C. #ssume that α ".""'C" !or this material. > ∆t 1"""C A 2""C 4" C" ?
R"
=
ρ ! A
=
&2 + 1"-4Ω ⋅ m(&2"" m( % + 1"-6 m2
) R0 1."" Ω
R = R" + α R" ∆t = 1."" Ω + &"."" ' C" (&1 Ω (&4" C" ()
126
at 2""C
R = 1.%" Ω
Physics, 6th Edition
Chapter 27. Current and Resistance
27-00. etermine the resistivity o! a wire made o! an un8nown alloy i! its diameter is ".""7 in. and 1"" !t o! the wire is !ound to have a resistance o! %."
Ω. R =
A = &7 mil( 2 %* cmil)
∆=
RA !
=
&%
> ".""7 in. 7 mil ?
Ω(&%* cmil( 1"" !t
)
ρ ! A
)
ρ =
RA !
ρ = 1.*6 Ω cmil'!t
27-0%. @he resistivity o! a certain wire is 1.72 + 1" -4 Ω m at 2""C. # 6-5 attery is connected to a 2"-m coil o! this wire havin a diameter o! ".4 mm. 3hat is the current in the wire$
A =
R = 27-0%. &Cont.(
ρ ! A
=
π " 2
%
=
π &"."""4 m(2
%
= ."0 + 1"-7 m2 )
&1.72 + 1"-4Ω ⋅ m(&2" m( ."0 + 1"-7 m2
I =
V R
=
R =
".64% Ω
A
R = ".64% Ω
)
6."" 5
ρ !
) I = 4.77 #
27-0. # certain resistor is used as a thermometer. /ts resistance at 2" "C is 26."" Ω, and its resistance at %""C is 26.2" Ω. 3hat is the temperature coe!!icient o! resistance !or this material$
α =
&26.2" Ω - 26."" Ω ( ∆ R ) = R" ∆t &26."" Ω(&%"" C - 2"" C(
127
α = 0.4 + 1"-%'C"
Physics, 6th Edition
Chapter 27. Current and Resistance
27-06. 3hat lenth o! copper wire at 2" "C has the same resistance as 2"" m o! iron wire at 2" "C$ #ssume the same cross section !or each wire. > Pro#$ct RA #oesn%t change . ?
R =
! 2
=
ρ ! ) RA = ρ ! ) A
ρ 1!1 ρ 2
=
! 2
ρ & R& = ρ 2! 2;
&*. '1"−4 Ω ⋅ m(&2"" m( 1.72 + 1"-4Ω ⋅ m
=
ρ 1! 1 ρ 2
) ! 2 = 11"" m
27-07. @he power loss in a certain wire at 2" "C is %"" 3. /! α ".""06'C ", y what percentae will the power loss increase when the operatin temperature is 64 "C$
∆ R = α R" ∆t )
∆ R R
= &".""06 ' C" (&64" C - 2"" C( ".170
Dince P /2R, the power loss increases y same percentae= 17.0
Critical !hin%in# Pro$lems
27-04.
# 1"- Ω resistor at 2" "C is rated at 2." 3 ma+imum power. 3hat is the ma+imum voltae that can e applied across the resistor with e+ceedin the ma+imum allowale power$ 3hat is the current at this voltae$
P =
V 2 R
)
V
=
PR
=
&2."" 3(&1" Ω() V = 17.0 5
124
Physics, 6th Edition
Chapter 27. Current and Resistance
27-0*. @he current in a home is alternatin current, ut the same !ormulas apply. Duppose a !an motor operatin a home coolin system is rated at 1" # !or a 12"-5 line. How much enery is reuired to operate the !an !or a 2%-h period$ #t a cost o! * cents per 8ilowatt-hour, what is the cost o! operatin this !an continuously !or 0" days$ P = VI = &11" 5(&&1" #( 11"" 3)
E = Pt = &11"" 3(&2% h( 26.% 83 h
E = &26.% 83 h(&06"" s'h(&1""" 3'83() E *." <9
Cost = 26.%
*27-%".
83 ⋅ h :"."4 day
&0" days() 83 ⋅ h
Cost = :0.06
@he power consumed in an electrical wire & α ".""%'C"( is %" 3 at 2" "C. /! all other !actors are held constant, what is the power consumption when &a( the lenth is douled, &( the diameter is douled, &c( the resistivity is douled, and &d( the asolute temperature is douled$ & Poer !oss is proportiona! to resistance R (
R =
ρ ! ) A
P ∝ !) P ∝
&a( "o$*!e !ength an# #o$*!e poer !oss;
1 A
)
P ∝ ∆R ∝ ∆
+oss = 2&%" 3( 4" 3
(*) #o$*!ing #iaeter gives ,A 0 an# one-.o$rth poer !oss/
+oss = &%" 3( 1" 3
&c( "o$*!ing resistivity #o$*!es resistance1 an# a!so #o$*!es poer !oss/ Foss 4" 3 27-%". &Cont.(
(#) &2"" B 270"( 2*0 G) ∆ = 2 = ; ∆ 2*0 G 2*0 C "
I. a*so!$te teperat$re #o$*!es1 the ne resistance is given *y/
R = R" &1 + α ∆ ()
R R"
= 1 + &".""% ' C" (&2*0 C" ( = 2.172)
12*
Physics, 6th Edition
Chapter 27. Current and Resistance P P"
=
R R"
= 2.172)
Foss 2.172&%" 3() +oss = 46.* 3
his o. co$rse pres$es that resistivity reains !inear1 hich is not !ike!y3
27-%1. 3hat must e the diameter o! an aluminum wire i! it is to have the same resistance as an eual lenth o! copper wire o! diameter 2." mm$ 3hat lenth o! nichrome wire is needed to have the same resistance as 2 m o! iron wire o! the same cross section$
R =
2 a
"
ρ ! A
=
R
)
!
ρ a "c2 ρ c
=
ρ A
= const.)
) "a
R =
ρ c Ac
ρ c
= "c
! n
=
ρ n
=
A=
)
Aa
& "c (
2
=
ρ a
& "c ( 2
= const.)
ρ n !n
= ρ i !i )
! n
)
=
"a 1.7 mm
ρ i ! i ρ n
&1.72 + 1"-4Ω ⋅ m( &1"" + 1"-4Ω ⋅ m( )
".""%0'C"( has a resistance o! .%"
∆t =
%
ρc
)
2.4" + 1"-4Ω ⋅ m
27-%2. #n iron wire &α ".""6'C "( has a resistance o! 6.""
same resistance$ > Conditions=
π " 2
1.72 + 1"-4Ω ⋅ m
= &2 mm(
ρ a
ρ ! ) RA = ρ ! A
ρ i ! i
=
ρa
Ω at 2""C.
! n = 1.72 cm
Ω at 2""C and a copper wire & α
#t what temperature will the two wires have the
αiR oi∆ti - αcR "c∆tc
6 Ω −5.4 Ω -".6".Ω. ?
−".6"" Ω −".6"" Ω = α i Roi − α c Roc &".""6 ' C" (&6." Ω ( - &".""%0' C" (&.% Ω ( ) ∆t -04." C"
10"
Physics, 6th Edition
Chapter 27. Current and Resistance
∆t t! A 2"" -04." C")
101
t! -14.""C
