OpenStax College Physics
Instructor Solutions Manual
Chapter 20
CHAPTER 20: ELECTRIC CURRENT,
RESISTANCE, AND OHM’S LAW
20.1 CURRENT
1.
What is the current in milliamperes produced by the solar cells of a pocket calculator
through which 4.00 C of charge passes in 4.00 h?
Solution
Using the equation I 
Q
, we can calculate the current given in the charge and the
t
time, remembering that 1 A  1 C/s
I
Q 4.00 C  1 h 
4


  2.778  10 A  0.278 mA
t
4.00 h  3600 s 
2.
A total of 600 C of charge passes through a flashlight in 0.500 h. What is the average
current?
Solution
I
3.
What is the current when a typical static charge of 0.250 C moves from your finger
to a metal doorknob in 1.00 s ?
Solution
4.
Solution
5.
I
Q
600 C  1 h 


  0.3333 A  333 mA
t 0.500 h  3600 s 
Q 2.50 10 7 C

 0.250 A
t
1.00 10 6 s
Find the current when 2.00 nC jumps between your comb and hair over a 0.500 - s
time interval.
I
Q 2.00 10 9 C

 4.00 10 3 A  4.00 mA
7
t
5.00 10 s

A large lightning bolt had a 20,000-A current and moved 30.0 C of charge. What was
its duration?
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
Q
Q
30.0 C
 t 

 1.50 10 3 s  1.50 ms
t
I
2.00 10 4 A
Solution
I
6.
The 200-A current through a spark plug moves 0.300 mC of charge. How long does
the spark last?
Solution
I
Q
Q 3.00  10 4 C
 t 

 1.50  10 6 s  1.50 μs
t
I
200 A
7.
(a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a
10,000-V potential as in Figure 20.38. What is the resistance of the path? (b) The
defibrillator paddles make contact with the patient through a conducting gel that
greatly reduces the path resistance. Discuss the difficulties that would ensue if a
larger voltage were used to produce the same current through the patient, but with
the path having perhaps 50 times the resistance. (Hint: The current must be about the
same, so a higher voltage would imply greater power. Use this equation for power:
P  I 2 R .)
Solution
(a) Using the equation I 
V
, we can calculate the resistance of the path given the
R
V
current and the potential: I  , so that
R
V 10,000 V
R 
 1.667  103 Ω  1.67 kΩ
I
6.00 A
(b) If a 50 times larger resistance existed, keeping the current about the same, the
power would be increased by a factor of about 50, causing much more energy to
be transferred to the skin, which could cause serious burns. The gel used reduces
the resistance, and therefore reduces the power transferred to the skin.
8.
During open-heart surgery, a defibrillator can be used to bring a patient out of cardiac
arrest. The resistance of the path is 500  and a 10.0-mA current is needed. What
voltage should be applied?
Solution
V  IR  (10.0  10 3 A)(500 Ω)  5.00 V
9.
(a) A defibrillator passes 12.0 A of current through the torso of a person for 0.0100 s.
How much charge moves? (b) How many electrons pass through the wires connected
to the patient? (See Figure 20.38.)
OpenStax College Physics
Solution
(a) I 
Instructor Solutions Manual
Chapter 20
Q
 Q  It  (12.0 A)(0.0100 s)  0.120 C
t
(b) N e 
Q
 1 electron 
17
 (0.120 C)
  7.50 10 electrons
19
qe
 1.60 10 C 
10.
A clock battery wears out after moving 10,000 C of charge through the clock at a rate
of 0.500 mA. (a) How long did the clock run? (b) How many electrons per second
flowed?
Solution
(a) I 
Q
Q
10,000 C
 t 

 2.00 10 7 s
4
t
I
5.00 10 A
(b) N e 
Q
 1 electron 
15
 (5.00  10 4 C/s)
  3.13  10 electrons/ s
19
qe
1
.
60

10
C


11.
The batteries of a submerged non-nuclear submarine supply 1000 A at full speed
23
ahead. How long does it take to move Avogadro’s number ( 6.02  10 ) of electrons at
this rate?
Solution
 1 electron 
21
I  1000 A  (1000 C/s )
  6.25  10 electrons/ s
19
 1.60  10 C 
1s


t  (6.02  10 23 electrons )
  96.3 s
19
 6.25  10 electrons 
12.
Electron guns are used in X-ray tubes. The electrons are accelerated through a
relatively large voltage and directed onto a metal target, producing X-rays. (a) How
many electrons per second strike the target if the current is 0.500 mA? (b) What
charge strikes the target in 0.750 s?
Solution
 1 electron 
15
(a) (0.500  10 3 C/s )
  3.125  10 electrons/ s
19
 1.60  10 C 
(b) I 
Q
 Q  It  (5.00 10 4 A)(0.750 s)  3.75 10 4 C
t
OpenStax College Physics
13.
Instructor Solutions Manual
Chapter 20
A large cyclotron directs a beam of He + + nuclei onto a target with a beam current of
++
0.250 mA. (a) How many He nuclei per second is this? (b) How long does it take for
1.00 C to strike the target? (c) How long before 1.00 mol of He ++ nuclei strike the
target?
Solution (a) Since we know that a He   ion has a charge oftwice the basic unit of charge:
(2.50 10 4 C/s)
(b) I 
1 He 
 7.811014 He  nuclei/s
19
2(1.60 10 C)
Q
Q
1.00 C
, so that t 

 4.00 10 3 s
I
2.50 10 4 A
t
 6.02 10 23 ions
(c) (1.00 mol He  )
mol

14.

1s

8

  7.7110 s
14

 7.8110 He ions 
Repeat Example 20.3, but for a wire made of silver and given there is one free
electron per silver atom.
Solution The number of atoms in a kg mol is 6.02  10 26 . Using the atomic mass of silver
107.9 kg/mol and the density 1.05  10 4 kg/m 3 , the electron density in the wire is:
6.02 10 26
(1.05 10 4 kg/m 3 )  5.86 10 28 electon/m 3
107.9 kg
I
20.0 A
vd 

  3.10 10 4 m/s
28
3
19
6
2
nqA (5.86 10 /m )( 1.60 10 C)(3.3110 m )
n
15.
Solution
Using the results of Example 20.3, find the drift velocity in a copper wire of twice the
diameter and carrying 20.0 A.
A   r 2   (2.053 10 3 m) 2  1.324 10 5 m 2
vd 
16.
I
20.0 A

  1.13 10 4 m/s
28
3
19
4
2
nqA (8.34 10 /m )( 1.60 10 C)(1.324 10 m )
A 14-gauge copper wire has a diameter of 1.628 mm. What magnitude current flows
when the drift velocity is 1.00 mm/s? (See Example 20.3 for useful information.)
OpenStax College Physics
Solution
vd 
Instructor Solutions Manual
Chapter 20
I

nqA
I  nq r 2 vd  (8.34 10 28 /m 3 )(1.60 10 19 C) (8.14 10 4 m) 2 (1.00 10 3 m/s )
vd  27.78 A  27.8 A
17.
Solution
SPEAR, a storage ring about 72.0 m in diameter at the Stanford Linear Accelerator
(closed in 2009), has a 20.0-A circulating beam of electrons that are moving at nearly
the speed of light. (See Figure 20.39.) How many electrons are in the beam?
First we calculate the circumference of the circular ring and use this value and the
speed of light to determine t , the time for the charge to complete one orbit around
the ring.
C
 (72.0 m)
C  2r  d ; d  72.0 m so that t  
 7.54  10 7 s
8
v 3.00  10 m/s
7
Thus, Q  It  (20.0 A)(7.54  10 s)  1.51  10 5 C
Let N e be the no. of electrons 
Q
1.5110 5 C

 9.42 1013 electrons
qe 1.60  10 19 C/electron
20.2 OHM’S LAW: RESISTANCE AND SIMPLE CIRCUITS
18.
What current flows through the bulb of a 3.00-V flashlight when its hot resistance is
3.60  ?
Solution
I
19.
Calculate the effective resistance of a pocket calculator that has a 1.35-V battery and
through which 0.200 mA flows.
Solution
I
20.
What is the effective resistance of a car’s starter motor when 150 A flows through it
as the car battery applies 11.0 V to the motor?
V 3.00 V

 0.833 A
R 3.60 Ω
V
V
1.35 V
, so that R  
 6.75  103 Ω  6.75 kΩ
4
R
I 2.00 10 A
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
Solution
R
21.
How many volts are supplied to operate an indicator light on a DVD player that has a
resistance of 140  , given that 25.0 mA passes through it?
Solution
V  IR  ( 2.50 102 A )(140 Ω)  350 V
22.
(a) Find the voltage drop in an extension cord having a 0.0600 -  resistance and
through which 5.00 A is flowing. (b) A cheaper cord utilizes thinner wire and has a
resistance of 0.300  . What is the voltage drop in it when 5.00 A flows? (c) Why is
the voltage to whatever appliance is being used reduced by this amount? What is the
effect on the appliance?
Solution
(a) V  IR  (5.00 A)(0.0600 Ω)  0.300 V
(b) V  IR  (5.00 A)(0.300 Ω)  1.50 V
(c) The voltage supplied to whatever appliance is being used is reduced because the
total voltage drop from the wall to the final output of the appliance is fixed. So, if
the voltage drop across the extension cord is large, the voltage drop across the
appliance is significantly decreased, so the power output by the appliance can be
significantly decreased, reducing the ability of the appliance to work properly.
23.
A power transmission line is hung from metal towers with glass insulators having a
9
resistance of 1.00  10  . What current flows through the insulator if the voltage is
200 kV? (Some high-voltage lines are DC.)
Solution
V 200 103 V
I 
 2.00 10-4  0.200 mA
9
R 1.00 10 Ω
V 11.0 V

 7.33 102 Ω
I 150 A
20.3 RESISTANCE AND RESISTIVITY
24.
Solution
What is the resistance of a 20.0-m-long piece of 12-gauge copper wire having a
2.053-mm diameter?
R
 L (1.72 108 Ω  m)(20.0 m)

 0.104 Ω
A
 (1.0265 103 m)2
OpenStax College Physics
25.
Solution
Instructor Solutions Manual
Chapter 20
The diameter of 0-gauge copper wire is 8.252 mm. Find the resistance of a 1.00-km
length of such wire used for power transmission.
We know we want to use the equation R 
L
, so we need to determine the radius
A
for the cross-sectional area of A  πr 2 . Since we know the diameter of the wire is
8.252 mm, we can determine the radius of the wire:
d 8.252  10 3 m
r 
 4.126  10 3 m . We also know from Table 20.1 that the
2
2
8
resistivity of copper is 1.72  10 Ωm . These values give a resistance of:
L (1.72  10 8 Ω  m)(1.00  10 3 m)
R

 0.322 Ω
A
 (4.126  10 3 m) 2
26.
Solution
27.
If the 0.100-mm diameter tungsten filament in a light bulb is to have a resistance of
0.200  at 20.0C , how long should it be?
R
L
A

L
Rr 2  (0.200 Ω)(5.00  10 4 m) 2

L


 2.81  10 2 m
2
8
r

5.6  10 Ω  m
Find the ratio of the diameter of aluminum to copper wire, if they have the same
resistance per unit length (as they might in household wiring).
Solution
 
A dA 2
d
R


 2 so that A   A 
dC  C 
AC
AA
C dC
28.
What current flows through a 2.54-cm-diameter rod of pure silicon that is 20.0 cm
3
long, when 1.00  10 V is applied to it? (Such a rod may be used to make nuclearparticle detectors, for example.)
Solution
C L
L
A L
1
2
 2.65  10 8 Ω  m 

 
8
 1.72 10 Ω  m 
(2300 Ω  m)( 2.00  10 1 m)
 9.078  10 5 Ω , so that
2
2
A
 (1.27  10 m)
V
1000 V
I 
 1.10  10 3 A
R 9.078  10 5 Ω
R

1
2
 1.24 .
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
29.
(a) To what temperature must you raise a copper wire, originally at 20.0C , to
double its resistance, neglecting any changes in dimensions? (b) Does this happen in
household wiring under ordinary circumstances?
Solution
(a) R  R0 (1  T )  2 R0  T  1  T  1 

1
 256 C
3.9  10 3 /C
T  20.0 C  256 C  276 C
(b) No, under ordinary circumstances this temperature is unreachable in household
wiring. The resistance is more likely to double because of a decrease in radius of
the wire.
30.
Solution
A resistor made of Nichrome wire is used in an application where its resistance cannot
change more than 1.00% from its value at 20.0C . Over what temperature range can
it be used?
R  R0 (1  T )  1.01 R0  T  0.0100 
0.100
 25 C.

0.4  10 3 / 0 C
Range  5 C to 45 C.
T 
0.0100

(Note: T expressed to two digits to avoid significant round-off errors.)
31.
Of what material is a resistor made if its resistance is 40.0% greater at 100C than at
20.0C ?
Solution Since R  R0 (1  T )  1.400 R0 ,
T  1.400  1   
0.400
0.400

 5.00  10 3 / C
T
80.0 C
So, based on the values of in Table 20.2, the resistor is made of iron.
32.
An electronic device designed to operate at any temperature in the range from
10.0C to 55.0C contains pure carbon resistors. By what factor does their
resistance increase over this range?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 20
R  R0 (1  T )  R  R0 [1   (30.0 C)] and
R -  R 0 [1   (25.0 C)], so that
R  1  (5 10 4 / C)( 30.0 C)

 1.028  1.03
R - 1  (5 10 4 / C)( 25.0 C)
33.
(a) Of what material is a wire made, if it is 25.0 m long with a 0.100 mm diameter and
has a resistance of 77.7  at 20.0C ? (b) What is its resistance at 150C ?
Solution
L L
π r 2 R  (5.00 10 5 ) 2 (77.7 Ω)

 2.44 10 8   m
(a) R    2   
A πr
L
25.0 m
The wire is gold.
(b) R  R0 (1  T )  (77.7 Ω)[1  (3.4  10 3 / C)(130 C)]  112 Ω  1.1 10 2 Ω
34.
Assuming a constant temperature coefficient of resistivity, what is the maximum
percent decrease in the resistance of a constantan wire starting at 20.0C ?
Solution At absolute zero, T  293.16 C , so that
R
 1  T  1  (2  10 6 / C)( 2.93.16 C)
R0
 99.94%  0.06% decrease
35.
Solution
A wire is drawn through a die, stretching it to four times its original length. By what
factor does its resistance increase?
V  πri Li  πrf Lf  πrf 4 Li 
2
Ri 
36.
Li
Ai
2

Li
πri
2
and Rf 
2
Lf
Af

rf
2
ri
2
2

r
1
2
 rf  i , so that
4
4
 (4 Li )
2
π(ri / 4)

16 Li
πr1
2
 16 R i
A copper wire has a resistance of 0.500  at 20.0C , and an iron wire has a
resistance of 0.525  at the same temperature. At what temperature are their
resistances equal?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 20
At 20 C, T  0  , RC  0.500 Ω and RI  0.525 Ω
R  RC (1   C T )  RI (1   I T ) so that
RC  RC C T  RI  RI I T and T ( RI I - RC C )  RC  RI
T 
RC  RI
0.500 Ω  0.525 Ω

 37 C
RI I  RC C (0.525 Ω)(5.0  10 3 / C)  (0.500 Ω)(3.9  10 3 / C)
T  20 C  37 C   17 C
37.
(a) Digital medical thermometers determine temperature by measuring the resistance
of a semiconductor device called a thermistor (which has   0.0600 / C ) when it is
at the same temperature as the patient. What is a patient’s temperature if the
thermistor’s resistance at that temperature is 82.0% of its value at 37.0C (normal
body temperature)? (b) The negative value for  may not be maintained for very low
temperatures. Discuss why and whether this is the case here. (Hint: Resistance can’t
become negative.)

Solution (a) R  R0 1   (T  37.0 C)  0.820 R0 , where   0.600 / C.
Dividing by R0 ,1   (T  37.0 C)  0.820, so that 0.180   (T  37.0 C)
 0.180
 3.00 C.

 0.0600 / C
Finally, T  37.0 C  3.00 C  40.0 C
giving (T  37.0 C) 
 0.180

(b) If  is negative at low temperatures, then the term 1   (T  37.0 C) can
become negative, which implies that the resistance has the opposite sign of the
initial resistance , or it has become negative. Since it is not possible to have a
negative resistance, the temperature coefficient of resistivity cannot remain
1
negative to low temperatures. In this example the magnitude is  
37.0 C  T
38.
Integrated Concepts (a) Redo Example 20.6 taking into account the thermal
expansion of the tungsten filament. You may assume a thermal expansion coefficient
6
of 12  10 / C . (b) By what percentage does your answer differ from that in the
example?
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
Solution (a)
L'
R '0 
; L  L(1  T ); A'  A(1  2T ) so that
A'
 1  (1.2  10 5 / C)( 2830 C) 
(1  T )
R'0  R0
 (0.350 Ω) 
  0.339 Ω
5
1  2T
1  2(1.2  10 / C)( 2830 C) 
(from just thermal expansion effects), so that


R'  (0.339 Ω) 1  (4.5  10 3 /C)( 2830 C)  4.65 Ω  4.7 Ω (total)
(b)
39.
4.65 
 0.970  3.0% decrease.
4.8 
Unreasonable Results (a) To what temperature must you raise a resistor made of
constantan to double its resistance, assuming a constant temperature coefficient of
resistivity? (b) To cut it in half? (c) What is unreasonable about these results? (d)
Which assumptions are unreasonable, or which premises are inconsistent?
Solution (a) Using the equation R  R0 (1  T ) and setting the resistance equal to twice the
initial resistance, we can solve for the final temperature:
R  R0 (1  T )  2R0  T   (T  T0 )  1. T0  20C
So the final temperature will be:
1
1
T  T0  
 5  105 C  T  5  105 C
α 2  10 6 / C
(b) Again, using the equation R  R0 (1  T ) , we can solve for the final
temperature when the resistance is half the initial resistance:
R
1
R  R0 (1  T )  0  (T  T0 )   , so the final temperature will be:
2
2
 0.5
T  T0 
 2.5 10 5 C or T   2.5 10 5 C.
2 10 6 / C
(c) In part (a), the temperature is above the melting point of any metal. In part (b) the
temperature is far below 0 K , which is impossible.
(d) The assumption that the resistivity for constantan will remain constant over the
derived temperature ranges in part (a) and (b) above is wrong. For large
temperature changes,  may vary, or a non-linear equation may be needed to
find  .
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
20.4 ELECTRIC POWER AND ENERGY
40.
What is the power of a 1.00  10 2 MV lightning bolt having a current of 2.00  10 4 A
?
Solution
P  IV  (2.00 104 A)(100 106 V)  2.00 1012 W
41.
What power is supplied to the starter motor of a large truck that draws 250 A of
current from a 24.0-V battery hookup?
Solution
P  IV  (250 A)(24.0 V)  6.00 kW
42.
A charge of 4.00 C passes through a pocket calculator’s solar cells in 4.00 h. What is
the power output, given the calculator’s voltage output is 3.00 V? (See Figure 20.40.)
Solution From Exercise 20.1, I  0.2778 mA. Thus,
P  IV  (2.778  10 4 A)(3.00 V)  8.33  10 4 W
43.
2
How many watts does a flashlight that has 6.00  10 C pass through it in 0.500 h use
if its voltage is 3.00 V?
Solution From Exercise 20.2, I  0.3333 A . Thus, P  IV  (0.33 A)(3.00 V)  1.00 W
44.
Find the power dissipated in each of these extension cords: (a) an extension cord
having a 0.0600 -  resistance and through which 5.00 A is flowing; (b) a cheaper
cord utilizing thinner wire and with a resistance of 0.300  .
Solution Using values from Exercise 20.22,
(a) P  IV  (5.00 A)(0.300 V)  1.50 W
(b) P  IV  (5.00 A)(1.50 V)  7.50 W
45.
Verify that the units of a volt-ampere are watts, as implied by the equation P  IV .
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
Solution Starting with the equation P  IV , we can get an expression for a watt in terms of
current and voltage: P  W , IV   A.V  (C/s)(J/C)  J/s  W , so that a watt is
equal to an ampere-volt.
46.
Show that the units 1 V 2 /   1 W , as implied by the equation P  V 2 / R .
Solution
V2 V2

 AV  (C/s )( J/C )  J/s  W
Ω V
A
47.
Show that the units 1 A 2    1 W , as implied by the equation P  I 2 R .
Solution
V
A 2 Ω  A 2    AV  (C/s )( J/C )  J/s  W
A
48.
Verify the energy unit equivalence that 1 kW  h = 3.60  106 J .
Solution
 60.0 s 
6
1 kW  h  (1.00  103 J/s )(60.0 min )
  3.60  10 J
 1 min 
49.
2
Electrons in an X-ray tube are accelerated through 1.00  10 kV and directed toward
a target to produce X-rays. Calculate the power of the electron beam in this tube if it
has a current of 15.0 mA.
Solution
P  IV  (1.50 10 2 A)(1.00 10 2 kV)  1.50 kW
50.
An electric water heater consumes 5.00 kW for 2.00 h per day. What is the cost of
running it for one year if electricity costs 12.0 cents/kW h? See Figure 20.41.
Solution
51.
(5.00 kW)( 2.00 h/d) 365 d/y $0.120 / kW  h   $438 / y

With a 1200-W toaster, how much electrical energy is needed to make a slice of toast
(cooking time = 1 minute)? At 9.0 cents/kW  h , how much does this cost?
OpenStax College Physics
Solution
P
Instructor Solutions Manual
 60 s
E
 E  Pt  (1200 W)(1 min) 
t
 1 min
Chapter 20

  7.2  10 4 J

 9.0 cent  1 kW  h 
  0.18 cents
(7.2  10 4 J)

6
 kW  h  3.6  10 J 
52.
What would be the maximum cost of a CFL such that the total cost (investment plus
operating) would be the same for both CFL and incandescent 60-W bulbs? Assume the
cost of the incandescent bulb is 25 cents and that electricity costs 10 cents/kW  h .
Calculate the cost for 1000 hours, as in the cost effectiveness of CFL example.
Solution Energy consumption by an incandescent bulb in 1000 hrs is
(60 W)(1000 h)  60 kW  h
 10 cent 
Cost of electricity is (60 kW  h) 
  600 cent  $6.00 ; the total cost for the
 kW  h 
incandescent bulb is then $0.25+$6.00 = $6.25. This is the maximum cost for the CFL.
53.
Solution
Some makes of older cars have 6.00-V electrical systems. (a) What is the hot
resistance of a 30.0-W headlight in such a car? (b) What current flows through it?
(a) P 
V2
V 2 (6.00 V) 2
R

 1.20 Ω
R
P
30.0 W
(b) P  IV  I 
54.
P 30.0 W

 5.00 A
V
6.00 V
Alkaline batteries have the advantage of putting out constant voltage until very
nearly the end of their life. How long will an alkaline battery rated at 1.00 A  h and
1.58 V keep a 1.00-W flashlight bulb burning?
Solution The total energy output of the battery is E  QV  ( It )V or
E  (1.00 A  h )(1.58 V)  1.58 W  h . So at P  1.00 W it can last
E 1.58 W  h
t 
 1.58 h
P
1.00 W
55.
A cauterizer, used to stop bleeding in surgery, puts out 2.00 mA at 15.0 kV. (a) What
is its power output? (b) What is the resistance of the path?
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
Solution (a) P  IV  (2.00  10 3 A)(15.0  103 V)  30.0 W
(b) I 
V
V (1.50 10 4 V)
R 
 7.50 10 6 Ω  7.50 MΩ
3
R
I
2.00 10 A
Note: this assumes the cauterizer obeys Ohm’s law, which will be true for ohmic
materials like good conductors.
56.
The average television is said to be on 6 hours per day. Estimate the yearly cost of
electricity to operate 100 million TVs, assuming their power consumption averages
150 W and the cost of electricity averages 12.0 cents/kW h.
Solution
 365 d 
Total cost  (100  10 6 )(6 h/d )  (150 W)(1.00  10 3 kW/W )($0.0900/kW  h )
 1y 
 $2.96 billion/ye ar
57.
An old lightbulb draws only 50.0 W, rather than its original 60.0 W, due to
evaporative thinning of its filament. By what factor is its diameter reduced, assuming
uniform thinning along its length? Neglect any effects caused by temperature
differences.
Solution
R
L
A

L
V2
; therefore,
;
P

R
r 2
2
2
r 
P2 V 2 / R2 R1 L / r1
r
P  50.0 W 
 2


  2  , or 2  2  

2
P1 V / R1 R2 L / r2
r1
P1  60.0 W 
 r1 
1/ 2
 0.913.
Thus, there is a 91.3% reduction in the diameter of the filament.
58.
Solution
00-gauge copper wire has a diameter of 9.266 mm. Calculate the power loss in a
2
kilometer of such wire when it carries 1.00  10 A .
R
L (1.72 10 8 Ω  m)(1.00 10 3 m)

 0.255 Ω
r 2
π(4.633 10 3 m) 2
so that P  I 2 R  (100 A)(0.255 Ω)  25.5 W
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
59.
Integrated Concepts Cold vaporizers pass a current through water, evaporating it
with only a small increase in temperature. One such home device is rated at 3.50 A
and utilizes 120 V AC with 95.0% efficiency. (a) What is the vaporization rate in grams
per minute? (b) How much water must you put into the vaporizer for 8.00 h of
overnight operation? (See Figure 20.42.)
Solution
(a) P  IV  (3.50 A)(120 V)  420 J/s  0.420 kJ/s and since the vaporizer has an
efficiency of 95.0%, the heat that is capable of vaporizing the water is
Q  (0.950) Pt .
This heat vaporizes the water according to the equation Q  mLv , where
Lv  2256 kJ/kg , from Table 14.2, so that (0.950) Pt  mLv , or
(0.950) Pt (0.950)(0.420 kJ/s )(60.0 s)
m

 0.0106 kg  10.6 g/min
Lv
2256 kJ/kg
(b) If the vaporizer is to run for 8.00 hours, we need to calculate the mass of the
water by converting units:
 60 min 
3
mrequired  (10.6 g/min )(8.0 h )
  5.09  10 g  5.09 kg
 1h 
In other words, it requires 5.09 kg 
m3
L
 3 3  5.09 L of water to run
3
10 kg 10 m
overnight.
60.
Integrated Concepts (a) What energy is dissipated by a lightning bolt having a
2
20,000-A current, a voltage of 1.00  10 MV , and a length of 1.00 ms? (b) What
mass of tree sap could be raised from 18.0C to its boiling point and then evaporated
by this energy, assuming sap has the same thermal characteristics as water?
Solution (a) E  Pt  Ivt  (2.00 10 4 A)(1.00 108 V)(1.00 10 3 s)  2.00 109 J
(b) E  mcΔT  mLv  m( Lv  cT )
E
2.00  10 9 J
m

 769 kg
Lv  cT 2256  10 3 J/kg  (4186 J/kg  C)(82.0 c)
61.
Integrated Concepts What current must be produced by a 12.0-V battery-operated
bottle warmer in order to heat 75.0 g of glass, 250 g of baby formula, and
3.00 102 g of aluminum from 20.0C to 90.0C in 5.00 min?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 20
Q  ( mg cg  mF c F  mAL c AL )T , where Q is heat and T is the change in
temperature. Assume the specific heat of baby formula is equal to that of water.
Then
Q  [(0.075 kg)( 840 J/kg  C)  (0.250 kg)( 4186 J/kg  C)
 (0.300 kg)( 900 J/kg  C)](70.0C)
 9.66  10 4 J
P
Q 9.66  10 4 J
P 3.22  10 2 W

 3.22  10 2 W, and P  IV  I  
 26.8 A
t
300 s
V
12.0 V
62.
Integrated Concepts How much time is needed for a surgical cauterizer to raise the
temperature of 1.00 g of tissue from 37.0C to 100C and then boil away 0.500 g of
water, if it puts out 2.00 mA at 15.0 kV? Ignore heat transfer to the surroundings.
Solution
Q  m1c1T  m2 Lv
 (1.00  10 3 kg)( 300 J/kg  C)(63.0 C)  (0.500  10 3 kg)( 2256  10 3 J/kg)
 1349 J
Q  Pt  IVt  t 
63.
Q
1349 J

 44.95 s  45.0 s
3
IV (2.00  10 A)(15.0  10 3 V)
Integrated Concepts Hydroelectric generators (see Figure 20.43) at Hoover Dam
3
produce a maximum current of 8.00  10 A at 250 kV. (a) What is the power output?
(b) The water that powers the generators enters and leaves the system at low speed
(thus its kinetic energy does not change) but loses 160 m in altitude. How many cubic
meters per second are needed, assuming 85.0% efficiency?
Solution (a) P  IV  (8.00 103 A)( 2.50 105 V)  (2.00 109 W)
(b) PE  mgh  m  PE
gh


m (2.00  10 9 J) / (9.80 m/s 2 )(160 m)

1s
1s
 1 m3 
  1.276  10 3 m 3 /s
 1.276  10 6 kg/s 
1000
kg


For 85.0% efficiency , we divide the value for m by 0.850  1.50  10 3 m 3 /s
OpenStax College Physics
64.
Solution
Instructor Solutions Manual
Chapter 20
Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical
power by the motor, what current must the 12.0-V batteries of a 750-kg electric car
be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a
2.00  102 -m-high hill in 2.00 min at a constant 25.0-m/s speed while exerting
5.00  102 N of force to overcome air resistance and friction? (c) To travel at a
constant 25.0-m/s speed, exerting a 5.00  102 N force to overcome air resistance and
friction? See Figure 20.44.
2
mv 2
(a) P  E  P  1 / 2 mv  IV  I 
t
0.950 t
2V (0.950)t

(750 kg)( 25.0 m/s ) 2
 343 A
2(12.0 V)(0.950)(60.0 s)
mgh  Fd (750 kg )(9.80 m/s 2 )( 200 m)  (500 N )( 25.0 m/s )(120 s)

0.950
0.950
6
 3.13  10 J (Note : no change in KE).
(b) E 
P
E
E
3.13  10 6 J
 IV  I 

 2.17  10 3 A
t
Vt (12.0 V)(120 s)
(c) Take a time interval of 1 s
P
E
Fd
(500 N )( 25.0 m)


 1.32  10 4 W
t 0.950t
0.950(1 s)
P  IV  I 
65.
P 2.60  10 5 W

 2.17  10 4 A
V
12.0 V
Integrated Concepts A light-rail commuter train draws 630 A of 650-V DC electricity
when accelerating. (a) What is its power consumption rate in kilowatts? (b) How long
does it take to reach 20.0 m/s starting from rest if its loaded mass is 5.30 104 kg ,
assuming 95.0% efficiency and constant power? (c) Find its average acceleration. (d)
Discuss how the acceleration you found for the light-rail train compares to what
might be typical for an automobile.
Solution (a) P  IV  (630 A)(650 V)  4.10  10 5 W  410 kW
(b) Since the efficiency is 95.0% , the effective power is: Peff  (0.950) P  389.0 kW .
Then, W  ( Peff )t. Set that equal to the change in kinetic energy,
OpenStax College Physics
W 
Instructor Solutions Manual
Chapter 20
1
1
2
mv 2  mv0  ( Peff )t , so that
2
2
(1 / 2)mv 2  (1 / 2)mv0
0.5(5.30  10 4 kg)( 20.0 m/s ) 2
t

 27.25 s  27.3 s
Peff
3.890  105 W
2
(c) v  v0  at , so that a 
v-v0 20.0 m/s

 0.734 m/s 2
t
27.25 s
(d) A typical automobile can go from 0 to 60 mph in 10 seconds, so that its
v 60 mi/hr
1 hr
1609m
acceleration is: a  


 2.7 m/s 2
t
10 s
3600 s
mi
Thus, a light-rail train accelerates much slower than a car, but it can reach final
speeds substantially faster than a car can sustain. Typically, light-rail tracks are
very long and straight, to allow them to reach these faster final speeds without
decelerating around sharp turns.
66.
Integrated Concepts (a) An aluminum power transmission line has a resistance of
0.0580  / km . What is its mass per kilometer? (b) What is the mass per kilometer of
a copper line having the same resistance? A lower resistance would shorten the
heating time. Discuss the practical limits to speeding the heating by lowering the
resistance.
Solution (a) Let  R be the resistivity of the transmission line, and  m be the density of the
transmission line.
R
R L
A

R L
πr 2
 r2 
R L
πR

(2.65 10 8 Ω  m)(1000 m)
 1.45 10 4 m 2
π (0.0580 Ω)
Now let m be the mass of the transmission line, and V be the volume of the
transmission line.
m   mV   m (r 2 ) L  (2.70  10 3 kg/m 3 ) (1.45  10 4 m 2 )(1000 m)
 1.23  10 3 kg
2
(b) r 
1.72  10

8

Ω  m (1000 m)
 9.44  10 5 m 2
 (0.0580 Ω)


m  8.90  10 3 kg/m 3  9.44  10 5 m 2 (1000 m)  2.64  10 3 kg
OpenStax College Physics
67.
Instructor Solutions Manual
Chapter 20
Integrated Concepts (a) An immersion heater utilizing 120 V can raise the
temperature of a 1.00  102 -g aluminum cup containing 350 g of water from 20.0C
to 95.0C in 2.00 min. Find its resistance, assuming it is constant during the process.
(b) A lower resistance would shorten the heating time. Discuss the practical limits to
speeding the heating by lowering the resistance.
Solution (a) Q  mAl c Al T  m w c w T
 (0.100 kg )(900 J/kg  C)  (0.350 kg )( 4186 J/kg  C)(75.0 C)
 1.17  10 5 J
E  Pt  P 
E Q V2
V 2 t (120 V) 2 (120 s)
 
R

 14.8 Ω
t
t
R
Q
1.17  10 5 J
(b) The resistance of the heater must be substantially larger than the resistance of
the wires carrying the current to the heater. For instance, if the immersion heater
was connected to an extension cord, as discussed in Exercise 20.22(b), the
resistance of the cord is approximately 2% of the resistance of the heater. Heat
will be dissipated through the extension cord and, if large enough, could lead to a
fire hazard.
68.
Integrated Concepts (a) What is the cost of heating a hot tub containing 1500 kg of
water from 10.0C to 40.0C , assuming 75.0% efficiency to account for heat transfer
to the surroundings? The cost of electricity is 12 cents/kW h . (b) What current was
used by the 220-V AC electric heater, if this took 4.00 h?
mcw T (1500 kg )( 4186 J/kg  C)(30.0 C)

 2.51  10 8 J

0.750
0.750
 1 kWh  $0.0900 
Cost  (2.51  10 8 J) 

  $6.28
6
 3.60  10 J  1 kW  h 
Q
(b) P  IV  , where t  4.00 h  14,400 s , so that
t
Q
2.51108 J
I 
 79.2 A
Vt (220 V)(14,400 s)
Solution (a)
69.

Q
Unreasonable Results (a) What current is needed to transmit 1.00  10 2 MW of
power at 480 V? (b) What power is dissipated by the transmission lines if they have a
1.00 -  resistance? (c) What is unreasonable about this result? (d) Which
assumptions are unreasonable, or which premises are inconsistent?
OpenStax College Physics
Instructor Solutions Manual
Solution (a) P  IV  I 
Chapter 20
P 1.00  108 W

 2.08 10 5 A
V
480 V
(b) P  I 2 R  (2.06 105 A)(1.00 )  43,264 MW  4.33 104 MW
(c) The power dissipated is greater than the power transmitted.
(d) The voltage is too low for transmission of power of this magnitude along the
transmission line.
70.
Unreasonable Results (a) What current is needed to transmit 1.00  10 2 MW of
power at 10.0 kV? (b) Find the resistance of 1.00 km of wire that would cause a
0.0100% power loss. (c) What is the diameter of a 1.00-km-long copper wire having
this resistance? (d) What is unreasonable about these results? (e) Which assumptions
are unreasonable, or which premises are inconsistent?
Solution (a) P  IV  I 
P 1.00  108 W

 10.0 kA
V 1.00  10 4 V
4
2
(b) (1.00  10 ) P  I R 
(1.00  10 -4 ) P (1.00  10 -4 )(1.00  10 8 W)
R

 1.00  10 -4 
2
4
2
I
(1.00  10 A)
1
2
1
2
 4(1.72  10   m)(1.00  10 m) 
4 L
 4 L 
d 
 
  0.468 m
2
A d
 (1.00 10 -4 )
 R 


(d) This is an unreasonably thick “wire.”
(e) The assumption of such a small power loss is unreasonable, and the voltage
should be much higher so that a lower current can carry the power.
(c) R 
L

-8
3
20.5 ALTERNATING CURRENT VERSUS DIRECT CURRENT
72.
Solution
(a) What is the hot resistance of a 25-W light bulb that runs on 120-V AC? (b) If the
bulb’s operating temperature is 2700C , what is its resistance at 2600C ?
(a) P 
V2
V 2 (120 V) 2
R

 576 Ω  5.8  10 2 Ω
R
P
25 W
(b) R  R0 (1  T )  (576 Ω)[1  (4.5 10 3 /C)( 100 C)]
 317 Ω  3.2 10 2 Ω
Where   4.5  10 -3 /C is the temperature coefficient of resistivity for tungsten.
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
73.
Certain heavy industrial equipment uses AC power that has a peak voltage of 679 V.
What is the rms voltage?
Solution
Vrms 
V0
2

679 V
2
 480 V
74.
A certain circuit breaker trips when the rms current is 15.0 A. What is the corresponding
peak current?
Solution
I 0  2I rms  2 (15.0 A)  21.2 A
75.
Military aircraft use 400-Hz AC power, because it is possible to design lighter-weight
equipment at this higher frequency. What is the time for one complete cycle of this
power?
Solution
T
76.
A North American tourist takes his 25.0-W, 120-V AC razor to Europe, finds a special
adapter, and plugs it into 240 V AC. Assuming constant resistance, what power does
the razor consume as it is ruined?
Solution
V2
V 2 (120 V) 2
R

 576 Ω
R
P
25.0 W
(V ' ) 2 (240 V) 2
and P ' 

 100 W
R
576 Ω
77.
In this problem, you will verify statements made at the end of the power losses for
Example 20.10. (a) What current is needed to transmit 100 MW of power at a voltage
of 25.0 kV? (b) Find the power loss in a 1.00 -  transmission line. (c) What percent loss
does this represent?
Solution
1
1

 2.50  10 3 s  2.50 ms
1
f
400 s
P
(a) P  IV  I 
P 1.00 108 W

 4.00 103 A  4.00 kA
4
V 2.50 10 V
(b) P '  I 2 R  (4.00  10 3 A) 2 (1.00 Ω)  1.60  10 7 W  16.0 MW
OpenStax College Physics
(c)
78.
Solution
Instructor Solutions Manual
Chapter 20
P ' 1.60  10 7 W

 0.160  16.0% Loss
P 1.00  10 8 W
A small office-building air conditioner operates on 408-V AC and consumes 50.0 kW. (a)
What is its effective resistance? (b) What is the cost of running the air conditioner
during a hot summer month when it is on 8.00 h per day for 30 days and electricity
costs 9.00 cents/kW  h ?
(a) P 
V2
V2
(408 V) 2
R

 3.33 Ω
R
P
5.00  10 3 W
(b) Total cost  (50.0 kW)(8.00 h/d)(30.0 d)($0.0900/kW  h)  $1080
79.
What is the peak power consumption of a 120-V AC microwave oven that draws 10.0
A?
Solution
Pave  I rms Vrms  (10.0 A)(120 V)  1.20 kW, so that
1
P0  I 0V0  2( I 0V0 )  2 Pave  2.40 kW
2
80.
What is the peak current through a 500-W room heater that operates on 120-V AC
power?
Solution
Pave  I rms Vrms  I rms 
Pave 500 W

 4.17 A, so that
Vrms 120 V
I 0  2 I rms  2 (4.17 A)  5.89 A
81.
Two different electrical devices have the same power consumption, but one is
meant to be operated on 120-V AC and the other on 240-V AC. (a) What is the
ratio of their resistances? (b) What is the ratio of their currents? (c) Assuming its
resistance is unaffected, by what factor will the power increase if a 120-V AC
device is connected to 240-V AC?
OpenStax College Physics
Solution
Instructor Solutions Manual
2
Chapter 20
2
2
V 
V1
V2
R
V2
 220 V 
P



 2   2   
(a)
  3.361  3.36
R1
R1
R2
R1  V1 
 120 V 
R 
I
(b) P  I 1 R1  I 2 R2  2   1 
I 1  R2 
2
1/ 2
2
 1 


 3.361 
2
1/ 2
 0.545
2
2
P V
V2
 220 V 
 2  22  
(c) P 
  3.36
R
P1 V1
 120 V 
82.
Solution
Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the
average power output is to be 1.00 kW utilizing 120-V AC. (b) What length of
Nichrome wire, having a cross-sectional area of 5.00 mm 2 , is needed if the operating
temperature is 500C ? (c) What power will it draw when first switched on?

2
(a) Pave
V
V2
(120 V) 2
 rms  R 

 14.4 Ω
R
Pave 1.00  10 3 W
(b)    0 (1  T )  (1.00  10 6 Ω  m)[1  (4  10 4 / C)( 480 C)]
 1.192  10 6 Ω  m
R
L
A
L
AR


(5.00  10 6 m 2 )(14.4 Ω)
 60.4 m  60 m
1.192  10 6 Ω  m
(c) R  R0 (1  T )  R0 
P
83.
R
14.4 Ω

 12.1 Ω
1  T 1  (4 10 4 / C)( 480 C)
V 2 (120 V) 2

 1190 W  1.2 kW
R0
12.1 Ω
Find the time after t  0 when the instantaneous voltage of 60-Hz AC first reaches the
following values: (a) V0 / 2 (b) V0 (c) 0.
Solution (a) From the equation V  V0 sin 2ft , we know how the voltage changes with time
V
for an alternating current (AC). If we want the voltage to be equal to 0 , we
2
1
V0
1
sin (0.5)
know that
 V0 sin 2ft , so that: sin 2ft  , or t 
. Since we have
2
2
2f
a frequency of 60 Hz, we can solve for the time that this first occurs
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
(remembering to have your calculator in radians!):
0.5236 rad
t
 1.39  10 3 s  1.39 ms
2 (60 Hz )
(b) Similarly, for V  V0 : V  V0 sin 2ft  V0 , so that
t
sin 1 1
 / 2 rad

 4.17 103 s  4.17 ms
2f
2 (60 Hz)
(c) Finally, for V  0 : V  V0 sin 2ft  0 , so that 2ft  0,  , 2 ,....., or for the first
1
time after t  0 : 2ft   , or t 
 8.33 10 3 s  8.33 ms
2(60 Hz )
84.
(a) At what two times in the first period following t  0 does the instantaneous
voltage in 60-Hz AC equal Vrms ? (b)  Vrms ?
Solution (a)
V
1
 3
Vrms  V0 sin 2ft  0  V0 sin 2ft  sin 2ft 
 2ft  ,
4 4
2
2
1 3
1
3
t
,

,
 2.08  10 3 s, 6.25  10 3 s
8 f 8 f 8(60 Hz ) 8(60 Hz )
(b)  Vrms  V0 sin 2ft  sin 2ft   1  2ft  5 , 7
4 4
2
5
7
t
,
 1.04  10  2 s, 1.46  10  2 s
8(60 Hz ) 8(60 Hz )
20.6 ELECTRIC HAZARDS AND THE HUMAN BODY
85.
Solution
86.
(a) How much power is dissipated in a short circuit of 240-V AC through a resistance
of 0.250  ? (b) What current flows?
(a) P 
V 2 (220 V) 2

 1.936 105 W  1.94 105 W  194 kW
R
0.250 Ω
(b) I 
V
220 V

 880 A
R 0.250 Ω
What voltage is involved in a 1.44-kW short circuit through a 0.100 -  resistance?
OpenStax College Physics
Solution
P
Instructor Solutions Manual
Chapter 20
V2
 V  ( PR)1/ 2  [(1.44  103 W)(0.100 Ω)]1/ 2  12.0 V
R
87.
Find the current through a person and identify the likely effect on her if she touches a
120-V AC source: (a) if she is standing on a rubber mat and offers a total resistance of
300 k ; (b) if she is standing barefoot on wet grass and has a resistance of only
4000 k .
Solution
(a) I 
V
120 V

 4.00 10 4 A  0.400 mA
5
R 3.00 10 Ω
This is a small current and will have no effect, since it is below the threshold of
sensation.
(b) I 
V
120 V

 2.67  10  2 A  26.7 mA
R 4500 Ω
Such a current will cause a muscular contraction for the duration of the shock.
88.
While taking a bath, a person touches the metal case of a radio. The path through the
person to the drainpipe and ground has a resistance of 4000  . What is the smallest
voltage on the case of the radio that could cause ventricular fibrillation?
Solution Ventricular fibrillation can initiate at I  100 mA (Table 20.3). Thus, a voltage of
V  IR  (1.00  10 1 A)(4000 Ω)  400 V could cause ventricular fibrillation.
89.
Foolishly trying to fish a burning piece of bread from a toaster with a metal butter
knife, a man comes into contact with 120-V AC. He does not even feel it since, luckily,
he is wearing rubber-soled shoes. What is the minimum resistance of the path the
current follows through the person?
Solution From Table 20.3, we know that the threshold of sensation is I  1.00 mA . So, since
V
V
120 V
I  R 
 1.20  10 5 Ω
R
I 1.00  10 3 A
90.
(a) During surgery, a current as small as 20.0 μA applied directly to the heart may
cause ventricular fibrillation. If the resistance of the exposed heart is 300  , what is
the smallest voltage that poses this danger? (b) Does your answer imply that special
electrical safety precautions are needed?
OpenStax College Physics
Instructor Solutions Manual
Chapter 20
Solution (a) V  IR  (2.00 10 5 A)(300 Ω)  6.00 10 3 V  6.00 mV
(b) It would not be necessary to take extra precautions regarding the power coming
from the wall. However, it is possible to generate voltages of approximately this
value from static charge built up on gloves, for instance, so some precautions are
necessary.
91.
Solution
(a) What is the resistance of a 220-V AC short circuit that generates a peak power of
96.8 kW? (b) What would the average power be if the voltage was 120 V AC?
(a) Pave 
P0
V 2 rms
V 2 rms
(220 V) 2
 4.84 103 W 
R

 1.00 Ω
2
R
P
4.84 103 W
2
2
V 
P V 
 120 V 
(b) 2   2   P2  P1  2   (4.84 kW)
  14.4 kW
P1  V1 
 220 V 
 V1 
2
92.
A heart defibrillator passes 10.0 A through a patient’s torso for 5.00 ms in an attempt
to restore normal beating. (a) How much charge passed? (b) What voltage was
applied if 500 J of energy was dissipated? (c) What was the path’s resistance? (d) Find
the temperature increase caused in the 8.00 kg of affected tissue.
Solution
(a) I 
Q
 Q  It  (10.0 A)(5.00 10 3 s)  5.00 10 2 C
t
(b) E  Pt  IVt  V 
(c) V  IR  R 
V 10,000 V

 1.00 kΩ
I
10.0 A
(d) Q  mcT  T 
93.
E
500 J

 10,000 V  10.0 kV
It (10.0 A)(5.00  10 3 s)
Q
500 J

 1.79  10 2 C
mc (8.00 kg)(3500J/kg  C)
Integrated Concepts A short circuit in a 120-V appliance cord has a 0.500 - 
resistance. Calculate the temperature rise of the 2.00 g of surrounding materials,
assuming their specific heat capacity is 0.200 cal/g  C and that it takes 0.0500 s for a
circuit breaker to interrupt the current. Is this likely to be damaging?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 20
 V 2   (120 V ) 2 
3
t  
E  Pt  
 (0.0500 s)  1.44  10 J
 R   0.500 Ω 
E  Q  mcT 
E
1.44  10 3 J
T 

 860 C
mc (2.00 g )(0.200 cal/g  C)( 4.186 J/cal )
Yes, this would be damaging, as it would likely raise the temperature to above the
melting point, or ignition of the materials.
20.7 NERVE CONDUCTION–ELECTROCARDIOGRAMS
95.
Integrated Concepts Use the ECG in Figure 20.34 to determine the heart rate in beats
per minute assuming a constant time between beats.
Solution The time from diastolic to diastolic on the upper graph is 0.75 s, so the heart rate is:
heart rate =
96.
1
1
60 s


 80 beats/min
time/beat 0.75 s/beat min
Integrated Concepts (a) Referring to Figure 20.34, find the time systolic pressure lags
behind the middle of the QRS complex. (b) Discuss the reasons for the time lag.
Solution (a) The lag is approximately 0.25 s , or 1 / 3 of a heart cycle.
(b) The QRS complex is created by the depolarization of the ventricles, which causes
it to contract. With this contraction comes a decrease in pressure, leading to the
diastolic pressure. After contraction the heart is repolarized, raising the pressure
to its maximum (systolic) value, readying the heart for its next beat. The fact that
the maximum pressure is approximately 1/3 of a heart cycle after the QRS
complex says that during 2/3 of the cycle the heart is doing work; during the last
1/3 it is resetting to start again.
This file is copyright 2016, Rice University. All Rights Reserved.