Homework 4
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Due: Thu Feb 17 2005 11:31 PM EST Description
Chapter 21
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1. Walker2 21.P.002. [239003] A flashlight bulb carries a current of 0.13 A for 65 s. How much charge flows through the bulb in this time?
Q = I T
N = Q/e
e=1.6e-19 C
C
How many electrons flow through the bulb in this time?

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2. Walker2 21.P.006. [239004] A television set connected to a 120 V outlet consumes 90 W of power.
(a) How much current flows through the television?
P=IV
A
(b) How long does it take for 10 trillion electrons to pass through the TV?
s
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3. Walker2 21.P.008. [331317] A iron wire is 4.5 m long and 0.35 mm in diameter. What is its resistance?
R = [Resistivity]* Length/ [Cross Sectional Area]
Area = pi times radius squared
Look up Resistivity values in book.
Use dimensional analysis to check that you really get Ohms for your answer.
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4. Walker2 21.P.009. [239006] When a potential difference of 17 V is applied to a given wire, it conducts 0.35 A of current. What is the resistance of the wire?

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5. Walker2 21.P.012. [239007] A bird lands on a bare copper wire carrying a current of 23 A. The wire is 8 gauge, which means that its cross-sectional area is 0.13 cm2.
(a) Find the difference in potential between the bird's feet, assuming they are separated by a distance of 5.2 cm.
V = I R
R = [Resistivity] L / A
L = distance between bird's feet.
mV
(b) Will your answer to part (a) increase or decrease if the separation between the bird's feet increases?

Explain.
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6. Walker2 21.P.014. [239008] A typical cell membrane is 8.0 nm thick and has an electrical resistivity of 1.3 107 · m.
(a) If the potential difference between the inner and outer surfaces of a cell membrane is 60 mV, how much current flows through a square area of membrane 1.0 µm on a side?
A
(b) Suppose the thickness of the membrane is halved, but the resistivity and potential difference remain the same. Does the current increase or decrease?

By what factor?
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7. Walker2 21.P.022. [239011] The current in a 120 V reading lamp is 2.5 A. If the cost of electrical energy is $0.060 per kilowatt-hour, how much does it cost to operate the light for an hour? (Do not round to the nearest cent.)
Energy = power times time
1 Watt*1sec = 1 J = 1 V * 1 A * 1 sec
1 kW-hour = 1000 W (3600 sec) = 3.6e6 J
1 = (1 kW-hour) / (3.6e6 J)
P = IV
Cost = [Energy in kilowatt hours]* [ Cost per kilowatt hour]]
$
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8. Walker2 21.P.024. [329838] A 60 W light bulb operates on a potential difference of 95 V.
(a) Find the current in the bulb.
A
(b) Find the resistance of the bulb.

(c) If this bulb is replaced with one whose resistance is double the value found in part (b), how would the new power rating compare?
(new power rating / 60 W) =

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9. Walker2 21.P.026. [239014] Find the equivalent resistance between points A and B for the group of resistors shown in Figure 21-29, where R1 = 79 and R2 = 40 .
R1 and 35 Ohm are in series.  Find the equivalent resistance of these two first.
This combined resistance is in parallel with R2.
Find the equivalent resistance of this parallel combination.


Figure 21-29
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10. Walker2 21.P.029. [239015] Your toaster has a power cord with a resistance of 0.017 connected in series with a 11.3 nichrome heating element. The potential difference between the terminals of the toaster is 120 V.
Find total resistance R of series combination cord + nichrome.
Current in circuit I = V/R
Power in each separate resistance is I2 R1 and I2 R2
(a) How much power is dissipated in the power cord?
W
(b) How much power is dissipated in the heating element?
kW
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11. Walker2 21.P.031. [239016] A circuit consists of a 9.0 V battery connected to three resistors (25 , 17 , and 160 ) in series.
(a) Find the current that flows through the battery.
mA
(b) Find the potential difference across each resistor.
V25 = V
V17 = V
V160 = V

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12. Walker2 21.P.033. [239017] A circuit consists of a battery connected to three resistors (81 , 25 , and 170 ) in parallel. The total current through the resistors is 1.9 A.
(a) Find the emf of the battery.
V
(b) Find the current through each resistor.
I81 = A
I25 = A
I170 = A

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13. Walker2 21.P.036. [239019] The equivalent resistance between points A and B of the resistors shown in Figure 21-30 is 40 . Find the value of resistance R.
Work Backwards:  Given total equivalent resistance between A &B, find
the equivalent resistance of 55 Ohm and R in parallel.


Figure 21-30
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14. Walker2 21.P.037. [239020] Find the equivalent resistance between points A and B shown in Figure 21-31, where R1 = 1.5 and R2 = 3.3 .
4.8 Ohm, R2, 8.1 Ohm in parallel = R3_eq
R3_eq in series with 6.3 Ohm
Now you have three resistances in parallel.


Figure 21-31
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15. Walker2 21.P.041. [239022] The terminals A and B in Figure 21-31 are connected to a 9.0 V battery, where R1 = 1.3 and R2 = 3.9 .


Figure 21-31

(a) Find the current flowing through each resistor.
1.3 resistor
A		 2.5 resistor
A
6.3 resistor
A		 4.8 resistor
A
3.9 resistor
A		 8.1 resistor
A

(b) Is the potential difference across the 6.3 resistor greater than, less than, or the same as the potential difference across the 1.3 resistor?

Explain.
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16. Walker2 21.P.043. [239024] Consider the group of resistors in Figure 21-33, where R1 = 15.2 and R2 = 9.49 . The current flowing through the 9.49 resistor is 1.22 A.
Let I1 = current flowing to right through R1 (value given)
Let I2 = current flowing to right through R2
Let I3 = curent flowing down through 13.8 Ohm
Let I4 = current flowing down through 17.2 Ohm
Junction Equation (two junctions combined):
I1 -I2 - I3 - I4 = 0
Outer Loop Clockwise:
I1 (15.0 Ohm + R1) + I2(R2 + 4.11 Ohm) - EMF = 0
Innermost loop Clockwise:
I4 17.2 Ohm - I3 13.8 Ohm = 0 (why the minus sign?)
Write down one more loop equation.
Now you have four equations and four unknowns: (I2, I3, I4, EMF)
Solve one equation symbolically for one variable, substitute into next equation.
Repeat until you have answer for one variable.
Back substitute to get answers for other variables..


Figure 21-33

(a) What is the voltage of the battery?
V
(b) If the 17.2 resistor is increased in value, will the current provided by the battery increase, decrease or stay the same?

Explain.
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17. Walker2 21.P.050. [239029] Suppose point A is grounded (V = 0) in Figure 21-36, in which = 13 V and R = 13 . Find the potential at points B and C.
Ohm's Law V =IR gives the voltage DROP across resistor, in direction of positive current flow.
Battery EMF = Voltage INCREASE from - terminal to + terminal.
If V=0 at A, and current flows from right to left through 11 Ohm, them
V<0 at C.
V (at B)
V (at C)


Figure 21-36
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18. Walker2 21.P.051. [239030] Consider the group of resistors shown in Figure 21-37, where R1 = 2.3 and = 7.9 V.
Use one Junction Equation and two Loop Equations.


Figure 21-37

(a) Find the current in each resistor in Figure 21-37.
I2.3
A		 I9.8
A
I1.2
A		 I6.7
A

(b) Is the potential at point A greater than, less than, or equal to the potential at point B?

Explain.

(c) Determine the potential difference between the points A and B.
V
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19. Walker2 21.P.053. [239031] Find the equivalent capacitance between points A and B for the group of capacitors shown in Figure 21-39, where C1 = 18 µF and C2 = 9.6 µF.
Add C2 and 22 micro-F in series first.
Remember  rules for series and  parallel addition of Capacitance are switched compared to formulae for  Resistors.
µF


Figure 21-39
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20. Walker2 21.P.055. [239032] Consider the group of capacitors shown in Figure 21-39, where C1 = 18 µF and C2 = 8.5 µF. Terminals A and B are connected to a 9.0 V battery. Find the energy stored in each capacitor.

Voltage across C1 is 9.0 V (given)
V2 = Voltage on C2
V(22) = Voltage on 22 micro F capacitor.
C2 and 22micro F have the same CHARGE
 (look at the wire joining C2 to 22muF,
this wire must have zero net charge.  Whatever charge is on the right hand
plate of C2 is equal and opposite the charge on left hand plate of 22mu F).

J
8.5 µF capacitor
J
22 µF capacitor
J


Figure 21-39
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21. Walker2 21.P.062. [239037] The capacitor in an RC circuit (R = 120 , C = 31 µF) is initially uncharged.
(a) Find the charge on the capacitor one time constant ( = RC) after the circuit is connected to a 9.0 V battery.
Q = (V C) [ 1 - exp(-t/RC)].  Subsitutue in t=RC
Q(t=RC) = VC [ 1-exp(-1)]
C
(b) Find the current in the circuit one time constant ( = RC) after the circuit is connected to a 9.0 V battery.
V = IR + Q/C
I = V/R-Q/(RC)
mA
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22. Walker2 21.P.063. [239038] Consider an RC circuit with = 12.0 V, R = 107 , and C = 62.9 µF.
(a) Find the time constant for the circuit.
ms
(b) Find the maximum charge on the capacitor.
At a time t >> RC, the capacitor is fully charged, meaning I = 0.
Use loop equation IR + Q/C - EMF = 0
with I=0:
Q/C = EMF
mC
(c) Find the initial current in the circuit.
Initially, Q=0.  Use same loop equation to find I at initial time.
mA
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23. Walker2 21.P.064. [239039] The resistor in an RC circuit has a resistance of 193 .
(a) What capacitance must be used in this circuit if the time constant is to be 2.7 ms?
µF
(b) Using the capacitance determined in part (a), calculate the current in the circuit 5.0 ms after the switch is closed. Assume that the capacitor is uncharged initially and that the EMF of the battery is 9.0 V.
mA