This preview shows page 1. Sign up to view the full content.
Unformatted text preview: × 10 A = 1.7 × 10 A Np 13 ( ) (22.13) c. The average power delivered by the wall socket is the product of the primary current and
voltage: ( ) Pp = I p Vp = 17 × 10−3 A (120 V ) = 2.0 W (20.15a) The average power delivered to the batteries is the same as that coming from the wall
socket, so Ps = 2.0 W .
______________________________________________________________________________
63. SSM REASONING The ratio ( I s / I p ) of the current in the secondary coil to that in the primary coil is equal to the ratio ( N p / Ns ) of the number of turns in the primary coil to that in the secondary coil. This relation can be used directly to find the current in the primary
coil.
SOLUTION Solving the relation ( I s / I p ) = ( N p / Ns ) (Equation 22.13) for Ip gives
N 1
I p = I s s = (1.6 A ) = 0.20 A Np 8 ______________________________________________________________________________ 64. REASONING The generator drives a fluctuating current in the primary coil, and the
changing magnetic flux that results from this current induces a fluctuating voltage in the
secondary coil, attached to the resistor. The peak emf of the generator is equal to the peak
voltage Vp of the primary coil. Given a peak voltage in the secondary coil of Vs = 67 V, the
peak voltage Vp can be found from
Vs
Vp = Ns
Np (22.12) In Equation 22.12, Np and Ns are, respectively, the number of turns in the primary coil and
the number of turns in the secondary coil. 1230 ELECTROMAGNETIC INDUCTION SOLUTION Solving Equation 22.12 for Vp, we obtain Np Vp = Vs N s (1) Since there are Np = 11 turns in the primary coil and Ns = 18 turns in the secondary coil,
Equation (1) gives 11 Vp = ( 67 V ) = 41 V 18 65. REASONING AND SOLUTION The resistance of the primary is (see Equation 20.3)
Rp = ρ Lp
A (1) The resistance of the secondary is Rs = ρ Ls
A (2) In writing Equations (1) and (2) we have used the fact that both coils are made of the same
wire, so that the resistivity ρ and the crosssectional area A is the same for each. Division of
the equations gives
Rs Ls 14 Ω 1
=
=
=
Rp Lp 56 Ω 4 Since the diameters of the coils are the same, the lengths of the wires are proportional to the
number of turns. Therefore,
Ns Ls
1
=
=
4
N p Lp
______________________________________________________________________________
66. REASONING The power that your house is using can be determined from P = I rmsξ rms
(Equation 20.15a), where Irms is the current in your house. We know that ξ rms is 240 V. To IS NP
(Equation 22.13) to convert the current given for the
I P NS
primary of the substation transformer into the current in the secondary of the substation
transformer. This secondary current then becomes the current in the primary of the find Irms, we must apply = Chapter 22 Problems 1231 transformer on the telephone pole. We will use Equation 22.13 again to find the current in
the secondary of the transformer on the telephone pole. This secondary current is Irms.
SOLUTION According to Equation 20.15a, the power that your house is using is P = I rmsξ rms (1) Applying Equation 22.13 to the substation transformer, we find IS N
=P
I P NS
1 24
43 or N 29 IS = I P P = 48 ×10−3 A = 1.4 A
N 1 S ( ) Substation Thus, the current in the primary of the transformer on the telephone pole is 1.4 A. Applying
Equation 22.3 to the telephonepole transformer, we obtain
IS N
=P
I P NS
1 24
43 or N
IS = I P P
N
S 32 = (1.4 A ) = 45 A 1 Telephone pole Using this value of 45 A for the current in Equation (1) gives P = I rmsξ rms = ( 45 A )( 240 V ) = 1.1× 104 W 67. SSM REASONING The power used to heat the wires is given by Equation 20.6b: 2 P = I R . Before we can use this equation, however, we must determine the total resistance
R of the wire and the current that flows through the wire.
SOLUTION
a. The total resistance of one of the wires is equal to the resistance per unit length multiplied
by the length of the wire. Thus, we have
(5.0 × 10 –2 Ω/km)(7.0 km) = 0.35 Ω
and the total resistance of the transmission line is twice this value or 0.70 Ω. According to
Equation 20.6a ( P = IV ), the current flowing into the town is I= P 1.2 ×106 W
=
= 1.0 ×103 A
V
1200 V 1232 ELECTROMAGNETIC INDUCTION Thus, the power used to heat the wire is ( P = I 2 R = 1.0 × 103 A ) 2 ( 0.70 Ω ) = 7.0 × 105 W b. According to the transformer equation (Equation 22.12), the steppedup voltage is N 100 5
Vs = Vp s = (1200 V ) = 1.2 × 10 V Np 1 According to Equation 20.6a (P = IV), the current in the wires is I= P 1.2 × 106 W
=
= 1.0 × 101 A
V 1.2 × 105 V The power used to heat the wires is now ( P = I 2 R = 1.0 × 101 A ) 2 ( 0.70 Ω ) = 7.0 × 101 W ______________________________________________________________________________
68. REASONING The turns ratio is related to the current IP in the primary and the current IS in
N
I
the secondary according to P = S (Equation 22.13).
IS N P
The turns ratio is related to the voltage VP in the primary and the voltage VS in the secondary
V
N
according to the transformer equation, which is S = S (Equation 22.12).
VP N P
Given a value for the power P us...
View Full
Document
 Spring '13
 CHASTAIN
 Physics, The Lottery

Click to edit the document details