Download Homework 4

Homework 4
Ch18: Q 7; P 7, 9, 19, 39, 43, 45, 49
Questions:
7.
If the resistance of a small immersion heater (to heat water for tea or soup, Fig. 18–
32) was increased, would it speed up or slow down the heating process? Explain.
Solution
We assume that the voltage is the same in both cases. Then if the resistance increases,
the power delivered to the heater will decrease according to P  V 2 R . If the power
decreases, the heating process will slow down.
Problems:
7.
(II) An electric clothes dryer has a heating element with a resistance of 9.6 .
(a) What is the current in the element when it is connected to 240 V?
(b) How much charge passes through the element in 50 min?
Solution
(a)
Use Eq. 18-2b to find the current.
V  IR  I 
(b)
R

240 V
9.6 
 25 A
Use the definition of current, Eq. 18-1.
I
9.
V
Q
t
 Q  I t   25 A  50 min  60 s min   7.5  10 4 C
(II) A bird stands on a dc electric transmission line carrying 2800 A (Fig. 18–34).
The line has 2.5 105  resistance per meter, and the bird’s feet are 4.0 cm apart.
What is the potential difference between the bird’s feet?
Solution
Find the potential difference from the resistance and the current.



R  2.5  105  m 4.0  102 m  1.0  10 6 


V  IR   2800 A  1.0  106   2.8  103 V
*19. (II) A 100-W lightbulb has a resistance of about 12  when cold (20°C) and 140 
when on (hot). Estimate the temperature of the filament when hot assuming an
average temperature coefficient of resistivity   0.0060 (Cº ) 1.
Solution
Use Eq. 18-4 multiplied by L A so that it expresses resistances instead of resistivity.
R  R0 1   T  T0  
T  T0 
1 R

1
 140  
o
 1   1798o C  1800o C
  1  20 C 
1 
o
  R0 

0.0060  C   12 
39. (II) A power station delivers 620 kW of power at 12,000 V to a factory through
wires with total resistance 3.0 . How much less power is wasted if the electricity is
delivered at 50,000 V rather than 12,000 V?
Solution
Find the current used to deliver the power in each case, and then find the power
dissipated in the resistance at the given current.
P  IV  I 
P
Pdissipated = I 2 R 
V
12,000 V
50,000 V
R
2
2
4
2
5
Pdissipated
V2
 6.20 10 W   3.0    8008 W

1.2 10 V 
 6.20 10 W   3.0    461W

 5 10 V 
5
Pdissipated
P2
4
2
difference  8008 W  461W  7.5  103 W
43. (I) An ac voltage, whose peak value is 180 V, is across a 330- resistor. What are
the rms and peak currents in the resistor?
Solution
Find the peak current from Ohm’s law, and then find the rms current from the
relationship between peak and rms values.
I peak 
Vpeak
R

180 V
330 
 0.54545 A
0.55 A
I rms  I peak
2   0.54545 A 
2  0.39 A
45. (II) The peak value of an alternating current in a 1500-W device is 5.4 A. What is
the rms voltage across it?
Solution
P  I rmsVrms 
I peak
2
Vrms  Vrms 
2P
I peak

2 1500 W 
5.4 A
 3.9  102 V
*49. (II) A 0.65-mm-diameter copper wire carries a tiny current of 2.3 A. What is the
electron drift speed in the wire?
Solution
We follow exactly the derivation in Example 18-14, which results in an expression for
the drift velocity.
vd 

I
neA

N 1 mole 
m 1 mole 
I
 D e  12 d 


2
4I m
N  D e d 2

m 1.60  10
4 2.3  10 6 A 63.5  10 3 kg
 6.02 10 8.9 10
23
3
kg
3
19

 
C  0.65  10 m
3

2
 5.1  1010 m s