Physics 112
Homework 4 (solutions)
(2004 Fall)
Solutions to Homework Questions 4
Chapt18, Problem-1:
A battery having an emf of 9.00 V delivers 117 mA when connected to a
72.0-! load. Determine the internal resistance of the battery.
Solution:
From !V = I ( R + r ) , the internal resistance is
r=
!V
9.00 V
"R =
" 72.0 # = 4.92 !
I
0.117 A
Chapt18, Problem-3:
A lightbulb marked “75 W [at] 120 V” is screwed into a socket at one end
of a long extension cord in which each of the two conductors has a resistance of 0.800 !. The other end of
the extension cord is plugged into a 120-V outlet. Draw a circuit diagram, and find the actual power of the
bulb in this circuit.
Solution:
The ‘rating’ on the bulb is for an ideal case where there is a potential difference of exactly 120V across
the two contacts of the bulb. However in this problem, this is not quite the case since there extension
cord is sufficiently long to have some resistance (and hence there is a small potential difference between
each of the two ends of the cord). So basically we have a circuit with three resistances in series, and a
source of emf (the outlet). We are given the resistances of each ‘arm’ of the cord, and we can find the
resistance of the bulb from its rating:
Rbulb =
(!V )
2
"
=
(120 V )
2
75.0 W
= 192 #
0.800 !
So we can now calculate the current in the circuit:
120 V
!V
120 V
I=
=
= 0.620 A ,
R eq 0.800 " +192 " + 0.800 "
192 !
0.800 !
and so the actual power dissipated in the bulb:
2
! = I 2 Rbulb = ( 0.620 A ) (192 ") = 73.8 W
Chapt18, Problem-5:
(a) Find the equivalent resistance
between points a and b in Figure P18.5. (b) Calculate the current
in each resistor if a potential difference of 34.0 V is applied
between points a and b.
Solution:
(1
(a)
1 %
" 1
+
' = 4.12 !
# 7.00 ! 10.0 ! &
The equivalent resistance of the two parallel resistors is R p = $
Thus,
(b)
Rab = R4 + R p + R9 = ( 4.00 + 4.12 + 9.00 ) ! = 17.1 !
I ab =
(!V )ab
Also, ( !V ) p
Then, I 7 =
34.0 V
= 1.99 A , so I 4 = I 9 = 1.99 A
17.1 "
Rab
= I ab R p = (1.99 A) ( 4.12 " ) = 8.18 V
(!V) p
R7
=
=
8.18 V
= 1.17 A
7.00 "
and
I 10 =
1
(!V )p
R10
=
8.18 V
= 0.818 A
10.0 "
Physics 112
Homework 4 (solutions)
Chapt18, Problem-13:
(2004 Fall)
Find the current
in the 12-! resistor in Figure P18.13.
Solution:
Clearly the first thing we need to determine is
The current in the whole circuit, then we can determine how
it branches. So the resistors in the circuit can be combined
in the stages shown below to yield an equivalent
resistance of Rad = 63 11 ! .
(
)
6.0 !
3.0 !
a 3.0 !
b
I1
c
3.0 !
6.0 !
4.0 !
I2
I12
d
3.0 !
a
I1
2.0 !
12 !
b
d
I2
I
e
3.0 !
c
e
3.0 !
2.0 !
I
18 V
18 V
Figure 1
Figure 2
6.0 !
3.0 ! I1
a
b
d
3!
a
b
30
—
11
!
d
63
—
11
a
!
d
I2
I
I
5.0 !
I
18 V
18 V
Figure 3
18 V
Figure 4
So from Figure 5, a simple application of Ohm’s law gives
Figure 5
I=
(!V)ad
=
18 V
= 3.14 A .
(63 11) "
(!V )bd = I R bd = ( 3.14 A )( 30 11 ") = 8.57 V .
So now, going back to Figure 4, we know
R ad
(!V )bd
8.57 V
= 1.71 A ,
3.0 " + 2 .0 " 5.0 "
(!V )be = I2 R be = (1.71 A )(3.0 ") = 5.14 V
(!V )be 5.14 V 0.43 A
I 12 =
=
=
12 "
R12
Going back again, this time to Figure 2, we see
I2 =
so
So finally, from Figure 1, we can determine that
2
=
Physics 112
Homework 4 (solutions)
Chapt18, Problem-19:
(2004 Fall)
Figure P18.19 shows a circuit
diagram. Determine (a) the current, (b) the potential of
wire A relative to ground, and (c) the voltage drop across
the 1 500-! resistor.
Solution:
(a) Applying Kirchhoff’s loop rule, as you go clockwise around the loop, gives
+ 20.0 V ! ( 2000) I ! 30.0 V ! ( 2500) I + 25.0 V ! ( 500 ) I = 0
or
I = 3.00 !10 "3 A = 3.00 mA
(b) Start at the grounded point and move up the left side, recording
changes in potential as you go, to obtain
(
)
(
or
VA = +20.0 V ! (2000 " ) 3.00 # 10!3 A ! 30.0 V ! ( 1000 ") 3.00 #10 !3 A
VA = ! 19.0 V
(c)
(!V )1500 = (1500 " )(3.00 # 10$3
)
A = 4.50 V
)
(The upper end is at the higher potential.)
Chapt18, Problem-22:
Find the current in
each of the three resistors of Figure P18.22 (a) by the
rules for resistors in series and parallel and (b) by the
use of Kirchhoff’s rules.
Solution:
(a) The resistors can be combined as shown below to yield
an equivalent of 6.6 W.
I6 6.0 !
3.0 !
3.6 !
3.0 !
a
b
c
a
b
I9
I3
I3
9.0 !
I3
I3
12 V
12 V
Figure 2
Figure 1
(!V)ac
c
6.6 !
a
c
I3
I3
12 V
Figure 3
12 V
= 1.8 A
6.6 "
R ac
Then, from Figure 2,
(!V )ab = I3 Rab = (1.82 A )( 3.6 " ) = 6.6 V
(!V)ab 6.6 V 1.1 A
(!V)ab 0.73 A
From Figure 1,
I6 =
=
=
, and I 9 =
=
6.0 "
6.0 "
9.0 "
(b) Using Figure 1 above, apply Kirchhoff’s junction rule at point a to obtain I 3 = I 6 + I 9
(1)
Next, apply Kirchhoff’s loop rule to the small loop containing the 6.0 - ! and 9.0 - ! resistors to find
(2)
!( 6.0) I 6 + ( 9.0 ) I 9 = 0 , or I 9 = (2 3 ) I 6
From Figure 3,
I3 =
=
Finally, apply Kirchhoff’s loop rule to the outside perimeter of Figure 1 to obtain
!( 6.0) I 6 ! ( 3.0 ) I 3 +12 V = 0 , or 2.0 I 6 + I 3 = 4.0 V (3)
( )
Solving equations (1), (2), and (3) simultaneously yields
I 3 = 1.8 A, I 6 = 1.1 A, I 9 = 0.73 A
3
Physics 112
Homework 4 (solutions)
Chapt18, Problem-30:
(2004 Fall)
Show that ! = RC has units of time.
Solution:
The time constant is ! = R C . Considering the units, we find
" Volts % " Coulombs % " Coulombs %
RC ! ( Ohms )( Farads ) = $
' =$
'$
'
# Amperes & # Volts & # Amperes &
!
$
Coulombs
=#
so indeed ! = R C has units of time.
& = Second
" Coulombs Second %
Chapt18, Problem-32:
An uncharged capacitor and a resistor are connected in series to a source
of emf. If E = 9.00 V, C = 20.0 µF, and R = 100 ", find (a) the time constant of the circuit, (b) the
maximum charge on the capacitor, and (c) the charge on the capacitor after one time constant.
Solution:
( a ) ! = R C = (100 " ) 20.0 # 10$6 F = 2.00 #10 $3 s = 2 .00 ms
(
)
(
)
(b) Q max = C ! = 20.0 " 10#6 F ( 9.00 V ) = 1.80 "10 #4 C =
(
)
(
(c) Q = Q max 1 ! e ! t " = Q max 1! e !"
"
)=Q
#
$
max % 1!
180 µC
1 & 114 µC
(=
e'
Chapt18, Problem-37:
An electric heater is rated at 1 300 W, a toaster is rated at 1 000 W, and an
electric grill is rated at 1 500 W. The three appliances are connected in parallel to a common 120-V circuit.
(a) How much current does each appliance draw? (b) Is a 30.0-A circuit breaker sufficient in this situation?
Explain.
Solution:
(a) The current drawn by each appliance is
! 1300 W
=
= 10.8 A
"V
120 V
! 1000 W
Toaster: I =
=
= 8.33 A
"V
120 V
! 1500 W
Grill:
I=
=
= 12.5 A
"V
120 V
Heater:
I=
( b ) If the three appliances are operated simultaneously, they will draw a total current of
I total = 10.8 + 8.33 + 12.5 A = 31.7 A . Therefore, a 30!ampere circuit breaker is
(
)
insufficient to handle the load .
Chapt18, Problem-39:
A heating element in a stove is designed to dissipate 3 000 W when
connected to 240 V. (a) Assuming that the resistance is constant, calculate the current in this element if it is
connected to 120 V. (b) Calculate the power it dissipates at this voltage.
Solution:
From ! = ( "V )
2
R , the resistance of the element is
R=
(!V)
"
When the element is connected to a 120-V source, we find that
(a)
(b)
!V 120 V
=
= 6.25 A , and
R 19.2 "
! = ( "V) I = ( 120 V ) ( 6.25 A ) = 750 W
I=
4
2
=
(240 V )
2
3000 W
= 19.2 #
Physics 112
Homework 4 (solutions)
Chapt18, Problem-41:
(2004 Fall)
Assume that a length of axon
membrane of about 10 cm is excited by an action potential.
(Length excited = nerve speed x pulse duration = 50 m/s x 2.0 ms
= 10 cm.) In the resting state, the outer surface of the axon wall is
charged positively with K+ ions and the inner wall has an equal and
opposite charge of negative organic ions as shown in Figure P18.41.
Model the axon as a parallel plate capacitor and use C=!"0A/d and
Q = C#V to investigate the charge as follows. Use typical values for
a cylindrical axon of cell wall thickness d = 1.0x10–8 m, axon
radius r = 10 µm, and cell wall dielectric constant ! = 3.0.
(a) Calculate the positive charge on the outside of a 10-cm piece
of axon when it is not conducting an electric pulse.
How many K+ ions are on the outside of the axon? Is this a
large charge per unit area? [Hint: Calculate the charge per unit area
in terms of the number of square angstroms (Å2) per electronic charge.
An atom has a cross section of about 1 Å2 (1 Å = 10–10 m).]
(b) How much positive charge must flow through the cell membrane to reach the excited state of +30 mV
from the resting state of –70 mV? How many sodium ions is this?
(c) If it takes 2.0 ms for the Na+ ions to enter the axon, what is the average current in the axon wall in this
process?
(d) How much energy does it take to raise the potential of the inner axon wall to +30 mV starting from the
resting potential of –70 mV?
Solution:
(a)
The area of each surface of this axon membrane is
[ (
)]
A = l ( 2!r ) = (0.10 m ) 2! 10" 10#6 m = 2! " 10#6 m 2 , and the capacitance is
C = ! "0
' 2& #10 $6 m 2 *
A
$8
= 3.0 8.85 # 10$12 C2 N % m 2 )
, = 1.67 #10 F
-8
d
( 1.0 # 10 m +
(
)
In the resting state, the charge on the outer surface of the membrane is
(
)(
)
"9
Q i = C ( !V) i = 1.67 "10 #8 F 70" 10#3 V = 1.17 " 10#9 C $ 1.2 !10 C
The number of potassium ions required to produce this charge is
NK + =
Q i 1.17 ! 10"9 C
9
+
=
= 7.3 ! 10 K ions ,
-19
e
1.6 ! 10 C
and the charge per unit area on this surface is
!=
#20
2
1e
Q i 1.17 "10 #9 C %
1e
1e
( % 10 m (
=
=
*
'
'
*
-6
2
-19
2
4
2 =
)
A 2$ "10 m & 1.6 " 10 C & 1 Å ) 8.6 "10 Å
(290 Å )2
This corresponds to a low charge density of one electronic charge per square of side 290 Å, compared to a
normal atomic spacing of one atom per several Å2 .
(b)
In the resting state, the net charge on the inner surface of the membrane is ! Q i = !1.17 " 10!9 C ,
and the net positive charge on this surface in the excited state is
(
)(
)
Q f = C ( !V) f = 1.67 "10 #8 F +30" 10#3 V = +5.0 " 10#10 C
The total positive charge which must pass through the membrane to produce the excited state is
therefore
!Q = Q f " Q i
(
)
"9
= +5.0 ! 10"10 C " "1.17 !10 "9 C = 1.67 ! 10"9 C # 1.7 ! 10 C ,
corresponding to
N Na +
!Q
1.67 " 10#9 C
10
+
=
=
= 1.0 !10 Na ions
e
1.6 "10 -19 C Na+ ion
5
Physics 112
Homework 4 (solutions)
(c)
If the sodium ions enter the axon in a time of !t = 2.0 ms , the average current is
I=
(2004 Fall)
!Q 1.67 "10 #9 C
=
= 8.3 "10 #7 A = 0.83 µ A
!t
2.0 " 10#3 s
(d)
When the membrane becomes permeable to sodium ions, the initial influx of sodium ions neutralizes
the capacitor with no required energy input. The energy input required to charge the now neutral capacitor
to the potential difference of the excited state is
Chapt18, Conceptual-4:
How would you connect resistors so that the equivalent resistance is
larger than the individual resistances? Give an example involving two or three resistors.
Solution:
The resistors should be connected in series. For example, connecting three resistors of 5 ! , 7 ! , and 2 !
in series gives a resultant resistance of 14 ! .
Chapt18, Conceptual-5:
If you have your headlights on while starting your car, why do they
dim while the car is starting?
Solution:
The starter in the car draws a relatively large current from the battery. This large current causes a
significant voltage drop across the internal resistance of the battery. As a result the terminal voltage of
the battery is reduced, and the headlights dim accordingly.
Chapt18, Conceptual-7:
Electrical devices are often rated with a voltage and a current – for
example, 120V, 5A. Batteries, however are only rated witth a voltage – for example, 1.5 V. Why?
Solution:
An electrical appliance has a given resistance. Thus, when it is attached to a power source with a known
potential difference, a definite current will be drawn. The device can be labeled with both the voltage and
the current. Batteries, however, can be applied to a number of devices. Each device will have a different
resistance, so the current from the battery will vary with the device. As a result, only the voltage of the
battery can be specified.
Chapt18, Conceptual-10:
If electrical power is transmitted over long distances, the resistance of
the wores becomes significant. Why? Which mode of transmission would result in less energy loss- high
current and low voltage or low current and high voltage?
Solution:
A wire or cable in a transmission line is thick and made of material with very low resistivity. Only when its
length is very large does its resistance become significant. To transmit power over a long distance it is
most efficient to use low current at high voltage, minimizing the I 2 R power loss in the transmission line.
Chapt18, Conceptual-13:
Why is it possible for a bird to sit on a high-voltage woire without
being electrocuted?
Solution:
The bird is at rest on the wire whose electrical potential is also constant along its length. In order to be
electrocuted, a large potential difference is required between the bird’s feet. The potential difference
between the bird’s feet is too small to harm the bird.
6