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Homework 4
Ch18: Q 7; P 7, 9, 19, 39; Ch19: Q1; P17, 37, 45
Ch18
Question:
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
R  R0 1   T  T0  
T  T0 
1 R

1
 140  
o
o
o
  1  20 C 
 12   1  1798 C  1800 C

1
o
  R0 

0.0060  C  
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
V
Pdissipated = I R 
2
;
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
Ch19
Question:
1. Explain why birds can sit on power lines safely, whereas leaning a metal ladder up
against a power line to fetch a stuck kite is extremely dangerous.
Solution
The birds are safe because they are not grounded. Both of their legs are essentially at the
same voltage (the only difference being due to the small resistance of the wire between
their feet), and so there is no current flow through their bodies since the potential
difference across their legs is very small. If you lean a metal ladder against the power
line, you are making essentially a short circuit from the high potential wire to the low
potential ground. A large current will flow at least momentarily, and that large current
will be very dangerous to anybody touching the ladder.
Problems:
17.
(II) Determine (a) the equivalent resistance of the circuit shown in Fig. 19–39,
and (b) the voltage across each resistor.
Solution
(a)
The equivalent resistance is found by combining the 820  and 680  resistors in
parallel, and then adding the 470  resistor in series with that parallel combination.
1
 1
1 
Req  

  470   372   470   842   840 
 820  680  
(b)
The current delivered by the battery is
I
V
Req

12.0 V
842 
 1.425 102 A .
This is the current in the 470  resistor. The voltage across that resistor can be found by
Ohm’s law
V470  IR  1.425  102 A   470    6.7 V .
Thus the voltage across the parallel combination must be
12.0 V  6.7 V  5.3 V .
This is the voltage across both the 820  and 680  resistors, since parallel resistors have
the same voltage across them. Note that this voltage value could also be found as
follows.


Vparallel  IRparallel  1.425  10 2 A  372    5.3 V
37. (I) A 3.00-F and a 4.00-F capacitor are connected in series, and this combination is
connected in parallel with a 2.00-F capacitor (see Fig. 19–52). What is the net
capacitance?
Solution
The series capacitors add reciprocally, and then the parallel combination is found by
adding linearly.
1
1 1 
Ceq      C3
 C1 C2 
45.
1
1
1


6
6


  2.00  10 F  3.71 10 F  3.71  F
6
6
3.00

10
F
4.00

10
F


(II) A 0.40-F and a 0.60-F capacitor are connected in series to a 9.0-V battery.
Calculate (a) the potential difference across each capacitor, and (b) the charge on
each. (c) Repeat parts (a) and (b) assuming the two capacitors are in parallel.
Solution
When the capacitors are connected in series, they each have the same charge as the net
capacitance.
1
(a)
1 1 
1
1


Q1  Q2  Qeq  CeqV     V  


6
6
 0.40  10  F 0.60 10  F 
 C1 C2 
1
 9.0 V 
 2.16  106 C
V1 
Q1
C1

2.16  106 C
0.40  106 F
 5.4 V
V2 
Q2
C2

2.16 10 6 C
0.60 10 6 F
 3.6 V
(b)
Q1  Q2  Qeq  2.16 106 C  2.2 106 C
When the capacitors are connected in parallel, they each have the full potential
difference.
(c)
V1  9.0 V
V2  9.0 V



Q1  C1V1  0.40  106 F  9.0 V   3.6 106 C

Q2  C2V2  0.60  106 F  9.0 V   5.4 106 C