Homework #4
8.
REASONING AND SOLUTION
a. The total charge that can be delivered is
 3600 s 
5
q  (220 A  h) 
  7.9 10 C
 1h 
b. The maximum current is
I
220 A  h
 350 A
 1h 
(38 min) 

 60 min 
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10.
REASONING
a. The resistance R of a piece of material is related to its length L and cross-sectional area A by
Equation 20.3, R   L / A , where  is the resistivity of the material. In order to rank the
resistances, we need to evaluate L and A for each configuration in terms of L0, the unit of length.
Resistance
a
b
c
 2 
   
L0  2 L0
 L0 
 1 
L0
R
  

2 L0  4 L0
 8L0 
 1 
2 L0
R
  

L0  4 L0
 2 L0 
R
4 L0
Rank
1
3
2
Therefore, we expect that a has the largest resistance, followed by c, and then by b.
b. Equation 20.2 states that the current I is equal to the voltage V divided by the resistance,
I  V / R . Since the current is inversely proportional to the resistance, the largest current
arises when the resistance is smallest, and vice versa. Thus, we expect that b has the largest
current, followed by c, and then by a.
SOLUTION
a. The resistances can be found by using the results from the REASONING:
a
b
c
 2 


2
R      1.50  102   m 
  0.600 
2
L
5.00

10
m


 0
 1 


1
2
R   
  1.50  10   m 
  0.0375 
2
8
L
8

5.00

10
m


 0
 1 


1
2
R   
  1.50  10   m 
  0.150 
2
2
L
2

5.00

10
m


 0






b. The current in each case is given by Equation 20.2, where the value of the resistance is
obtained from part (a):
a
b
c
3.00 V
V

 5.00 A
R 0.600 
3.00 V
V
I 
 80.0 A
R 0.0375 
3.00 V
V
I 
 20.0 A
R 0.150 
I
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20. REASONING The length L of the wire is related to its resistance R and cross-sectional area A by
L  AR /  (see Equation 20.3), where  is the resistivity of tungsten. The resistivity is known (see
Table 20.1), and the cross-sectional area can be determined since the radius of the wire is given.
The resistance can be obtained from Ohm’s law as the voltage divided by the current.
SOLUTION The length L of the wire is
L
AR

(1)
Since the cross-section of the wire is circular, its area is A   r 2 , where r is the radius of the wire.
According to Ohm’s law (Equation 20.2), the resistance R is related to the voltage V and current I by
R  V / I . Substituting the expressions for A and R into Equation (1) gives
L
AR


2   120 V 
 
3

 0.0030 10 m   
I 
 1.24 A   0.049 m

5.6 108   m
 r 2   V 
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24. REASONING AND SOLUTION The power delivered is P = VI, so that we have
a.
Pbd = VIbd = (120 V)(11 A) = 1300 W
b.
Pvc = VIvc = (120 V)(4.0 A) = 480 W
c. The energy is E = Pt, so that we have
Ebd
Evc

Pbdtbd
Pvctvc

(1300 W)(15 min)
 1.4
(480 W)(30.0 min)
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33. REASONING
a. The average power P delivered to the copy machine is equal to the square of the rms-current
2
Irms times the resistance R, or P  I rms
R (Equation 20.15b). Both Irms and R are known.
b. According to the discussion in Section 20.5, the peak power Ppeak is twice the average power, or
Ppeak  2 P .
SOLUTION
a. The average power is
2
2
P  I rms
R   6.50 A  18.6    786 W
(20.15b)
b. The peak power is twice the average power, so
Ppeak  2P  2  786 W   1572 W
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52.
REASONING
a. The three resistors are in series, so the same current goes through each resistor:
I1  I 2  I3 .
The voltage across each resistor is given by Equation 20.2 as V = IR. Because the current through
each resistor is the same, the voltage across each is proportional to the resistance. Since
R1  R2  R3 , we expect the ranking of the voltages to be V1  V2  V3 .
b. The three resistors are in parallel, so the same voltage exists across each: V1  V2  V3 . The
current through each resistor is given by Equation 20.2 as I = V/R. Because the voltage across each
resistor is the same, the current through each is inversely proportional to the resistance. Since
R1  R2  R3 , we expect the ranking of the currents to be I3  I 2  I1 .
SOLUTION
a. The current through the three resistors is given by I = V/Rs, where Rs is the equivalent
resistance of the series circuit. From Equation 20.16, the equivalent resistance is
Rs = 50.0  + 25.0  + 10.0   85.0  The current through each resistor is
I1  I 2  I 3 
24.0 V
V

 0.282 A
Rs 85.0 
The voltage across each resistor is
V1  I R1   0.282 A  50.0    14.1 V
V2  I R2   0.282 A  25.0    7.05 V
V3  I R3   0.282 A 10.0    2.82 V
b. The resistors are in parallel, so the voltage across each is the same as the voltage of the battery:
V1  V2  V3  24.0 V
The current through each resistor is equal to the voltage across each divided by the resistance:
I1 
24.0 V
V

 0.480 A
R1 50.0 
I2 
24.0 V
V

 0.960 A
R2 25.0 
I3 
24.0 V
V

 2.40 A
R3 10.0 
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70. REASONING The power P dissipated in each resistance R is given by Equation 20.6b as P  I 2 R ,
where I is the current. This means that we need to determine the current in each resistor in order
to calculate the power. The current in R1 is the same as the current in the equivalent resistance for
the circuit. Since R2 and R3 are in parallel and equal, the current in R1 splits into two equal parts at
the junction A in the circuit.
SOLUTION To determine the equivalent resistance of the circuit, we note that the parallel
combination of R2 and R3 is in series with R1. The equivalent resistance of the parallel combination
can be obtained from Equation 20.17 as follows:
1
1
1


RP 576  576 
or
RP  288 
This 288-Ω resistance is in series with R1, so that the equivalent resistance of the circuit is given by
Equation 20.16 as
Req  576   288   864 
To find the current from the battery we use Ohm’s law:
I
V
120.0 V

 0.139 A
Req
864 
Since this is the current in R1, Equation 20.6b gives the power dissipated in R1 as
P1  I12 R1   0.139 A   576    11.1 W
2
R2 and R3 are in parallel and equal, so that the current in R1 splits into two equal parts at the
junction A. As a result, there is a current of 12  0.139 A  in R2 and in R3. Again using Equation
20.6b, we find that the power dissipated in each of these two resistors is
P2  I 22 R2   12  0.139 A  
P3  I32 R3   12  0.139 A  
2
 576   
2.78 W
2
 576   
2.78 W