Answer, Key – Homework 11 – David McIntyre This print-out should have 36 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. Chapters 27 and 28 problems. 001 (part 1 of 1) 0 points It is known that about one electron per atom of copper contributes to the current. The atomic mass of copper is 63.54 g and its density is 8.92 g/cm3 . Calculate the average drift speed of electrons traveling through a copper wire with a cross-sectional area of 1 mm2 when carrying a current of 1 A (values similar to those for the electric wire to your study lamp). Correct answer: 7.38624 × 10−5 m/s. Explanation: Given : N = 1 , M = 63.54 g , ρ = 8.92 g/cm3 , A = 1 mm2 = 1 × 10−6 m2 , I = 1 A , and qe = 1.602 × 10−19 C/electron . We first calculate n, the number of currentcarrying electrons per unit volume in copper. Assuming one free conduction electron per NA ρ atom, n = , where NA is Avogadro’s M number and ρ and M are the density and the atomic weight of copper, respectively ¶ µ electron NA ρ . n≡ 1 atom M ¶ electron 6.02 × 1023 atoms n= 1 atom 63.54 g µ 6 ¶ ¡ ¢ 10 cm3 3 · 8.92 g/cm 1 m3 = 8.45112 × 1028 electrons/m3 . µ
=
1
1A 8.45112 × 1028 electrons/m3 1 · −19 (1.602 × 10 C/electron) 1 · (1 × 10−6 m2 )
= 7.38624 × 10−5 m/s .
002 (part 1 of 3) 0 points The damage caused by electric shock depends on the current flowing through the body; 1 mA can be felt and 5 mA is painful. Above 15 mA, a person loses muscle control, and 70 mA can be fatal. A person with dry skin has a resistance from one arm to the other of about 100000 Ω. When skin is wet, the resistance drops to about 5300 Ω. What is the minimum voltage placed across the arms that would produce a current that could be felt by a person with dry skin? Correct answer: 100 V. Explanation: Given : Imin = 1 mA , and Rdry = 100000 Ω . The minimum voltage depends on the minimum current for a given resistance, so Vmin = Imin Rdry µ = (1 mA)
1A 1000 mA
¶
(100000 Ω)
= 100 V .
003 (part 2 of 3) 0 points Using the same electric potential as in Part 1, what would be the current if the person had wet skin? Correct answer: 18.8679 mA. Explanation:
The drift speed is vd , vd =
I n qe A
Given :
Vmin = 100 V , and Rwet = 5300 Ω .
Answer, Key – Homework 11 – David McIntyre For the voltage V from Part 1 and the resistance Rwet , Vmin Rwet µ ¶ 100 V 1000 mA = 5300 Ω 1A = 18.8679 mA .
I=
2
Solution: The charge carrier density n is unaffected. This is simply a property of the material, and is related to the mass density and the number of valence electrons available per atom. 006 (part 2 of 4) 0 points (b) What happens to the current density? 1. It doubles. correct
004 (part 3 of 3) 0 points What would be the minimum voltage that would produce a current that could be felt when the skin is wet? Correct answer: 5.3 V. Explanation: Given :
Imin = 1 mA , and Rwet = 5300 Ω .
V1 = Imin Rwet µ = (1 mA)
1A 1000 mA
¶
(5300 Ω)
= 5.3 V . 005 (part 1 of 4) 0 points If the current carried by a metallic conductor is doubled, (a) What happens to the charge carrier density? Assume that the temperature of the metallic conductor remains constant.
2. It is unchanged. 3. It is quartered. 4. It quadruples. 5. It is halved. Explanation: I The current density J = only depends A on the current I and the cross-sectional area of the conductor A, so if I doubles and A remains the same, J must also double: I 2I = 2 = 2J . J0 = A A 007 (part 3 of 4) 0 points (c) What happens to the electron drift velocity? 1. It is unchanged.
1. It quadruples.
2. It is halved.
2. It is unchanged. correct
3. It is quartered.
3. It is quartered.
4. It quadruples.
4. It is halved.
5. It doubles. correct
5. It doubles. Explanation: Basic Concept: J = σ E = n q vd vd =
qE τ. m
Explanation: The electron drift velocity vd is given by I vd = , nqA where q is the charge of an electron. So if I doubles, vd must also double: 2I I vd0 = =2 = 2vd . nqA nqA
Answer, Key – Homework 11 – David McIntyre 008 (part 4 of 4) 0 points (d) What happens to the average time between collisions?
3
3. D > B = C > A 4. A = C > D > B
1. It doubles.
5. A = D > B = C correct
2. It is halved.
6. A = B = C = D
3. It is quartered.
7. A = B > D = C
4. It is unchanged. correct
8. A > B = C > D
5. It quadruples.
9. A > B > C > D
Explanation: The average time τ between collisions is given by vd m , τ= qE where m is the mass of an electron and E is the applied electric field (determined by Ohm’s Law: J = σ E). Substituting for E, τ=
vd m σ . qJ
Explanation: From the figure, the total resistance of the 5R R . Then circuit is 2 R + = 2 2 IA = I D =
V 2V V = = 0.4 5R 5R R 2
1V V 1 IA = = 0.2 2 5R R The rankings of the currents are also the corresponding rankings of the brightness. IB = I C =
Since vd and J are each multiplied by a factor of two, the factors of two cancel (as long as σ does not change due to heating). All the other variables remain the same (as long as σ does not change due to heating). Therefore τ remains unchanged. 009 (part 1 of 2) 0 points The circuit below shows four identical bulbs connected to an ideal battery, which has negligible internal resistance. Rank the bulbs in order from brightest to dimmest. A V
10. A = B = C > D
B
010 (part 2 of 2) 0 points Suppose a switch has been added to the circuit as shown. The switch is initially closed. A V
B
C
D
When the switch is opened, what happens to the currents through bulbs A, B, and D? C
D
1. IA decreases, IB remains the same, ID decreases.
1. D > C > B > A
2. IA remains the same, IB decreases, ID remains the same
2. B = C > A = D
3. IA increases, IB increases, ID increases.
Answer, Key – Homework 11 – David McIntyre 4. IA remains the same, IB increases, ID remains the same. 5. IA decreases, IB increases, ID decreases. correct 6. IA remains the same, IB remains the same, ID remains the same. 7. IA increases, IB remains the same, ID decreases. 8. IA increases, IB remains the same, ID increases. 9. IA increases, IB decreases, ID increases. 10. IA decreases, IB decreases, ID decreases. Explanation: With the switch open, the total resistance of the circuit is 3 R, so IA = I B = I D =
V V = 0.33 3R R
This is to be compared with the case that the switch is closed, where from previous part, V V and IB = 0.2 . So for IA = ID = 0.4 R R the present case, IA and ID decrease and IB increases. 011 (part 1 of 4) 0 points A battery has an emf of 12 V and an internal resistance of 0.13 Ω. Its terminals are connected to a load resistance of 2 Ω. Find the current in the circuit. Correct answer: 5.6338 A. Explanation: Given : E = 12 V , R = 2 Ω , and r = 0.13 Ω . The total resistance is R + r, so I=
E R+r
4
12 V 2 Ω + 0.13 Ω = 5.6338 A .
=
012 (part 2 of 4) 0 points Calculate the terminal voltage of the battery. Correct answer: 11.2676 V. Explanation: The terminal voltage V of the battery is equal to Vr = E − I r = 12 V − (5.6338 A) (0.13 Ω) = 11.2676 V .
013 (part 3 of 4) 0 points Find the power dissipated in the load resistor. Correct answer: 63.4795 W. Explanation: The power dissipated in the load resistor is PR = I 2 R = (5.6338 A)2 (2 Ω) = 63.4795 W .
014 (part 4 of 4) 0 points Find the power dissipated in the battery. Correct answer: 4.12617 W. Explanation: The power dissipated in the battery is Pr = I 2 r = (5.6338 A)2 (0.13 Ω) = 4.12617 W .
015 (part 1 of 2) 0 points In the figure below the switch S is initially in position “1”. Given, R1 = R2 = R3 . Neglect the internal resistance of the battery.
Answer, Key – Homework 11 – David McIntyre
5
as when R1 was in the circuit. R1 V
2
1
3 S
0
R1
R2
What happens to the current through R3 when the switch is moved to the open position “2”.
Explanation: Since R2 and R3 have the same terminal voltage and resistance, the current through R2 and R3 must now be the same. 017 (part 1 of 1) 0 points Consider the following circuit containing identical bulbs.
V
A
B
C
D E
1. The current through R3 decreases to twothirds its original value. 2. The current through R3 increases to twice its original value. 3. The current through R3 increases to three-halves its original value. 4. The current through R3 is reduced to one-half its original value. 5. The current through R3 remains the same. correct
1. A = B = C = D = E 2. A = D = E > B = C 3. B = C > A = D = E 4. A = B = C > D = E correct 5. A = B = C > D > E 6. A = B > A = D = E
Explanation: The voltage across R3 is the E of the battery, and is unchanged. The current through E R3 remains the same, . R3
8. A = D = E > B > C
016 (part 2 of 2) 0 points What happens when switch S is moved to position “3”, leaving R2 and R3 parallel?
10. E = D > A > B = C
1. The current through R2 is half what it was with R1 in the circuit. 2. The current through R2 and R3 are now the same. correct 3. The current through R3 increases. 4. The current through R3 decreases. 5. The current through R2 remains the same
7. A = C > B > D = E
9. C > B > A > D > E
Explanation: The potential across bulbs A, B and C is V V . The potential across bulbs D and E is . 2 Thus V iA = i B = i C = R V iD = i E = . 2R 018 (part 1 of 2) 0 points Four identical light bulbs are connected either in series (circuit 1) or parallel (circuit 2)
Answer, Key – Homework 11 – David McIntyre to a constant voltage battery with negligible internal resistance, as shown.
6
We can see that the bulbs in circuit 2 are more than 4 times brighter than the bulbs in circuit 1. 019 (part 2 of 2) 0 points If one of the bulbs in circuit 2 is unscrewed and removed from its socket, the remaining 3 bulbs 1. go out. 2. are unaffected. correct
circuit 1
3. become brighter. 4. become dimmer.
Compared to the individual bulbs in circuit 1, the individual bulbs in circuit 2 are 1. the same brightness. 1 as bright. 4 1 3. less than as bright. 4 2.
4. 4 times brighter. 5. more than 4 times brighter. correct Explanation: Solution: In circuit 1, the voltage across each light bulb is
020 (part 1 of 3) 0 points Hint: Apply the Kirchhoff’s law to the loop ACDA. Given: 2 R1 = 2 R4 = R2 = R3 , R1 = r. I is the current entering and leaving the circuit. Each current shown in the figure is denoted by the same subscript as the resistor through which it flows (e.g., i1 is the current flowing through R1 ). C R R1 3
V2 E2 = . R 16 R In circuit 2, the voltage across each bulb is identical; namely E. Hence the power of each bulb in circuit 2 is
i1 R
A
E E V =IR= R= , 4R 4 so the power of each bulb in circuit 1 is
I
2
i2 Find the ratio
i1 . i2
i1 = 2 correct i2 1 i1 = 2. i2 3 1.
B
R4
I
i4 D
P1 =
1 E2 = P1 . P2 = R 16
i3 i5
circuit 2
Explanation: Since the bulbs are parallel, after one of the bulbs is unscrewed, the voltage across each remaining bulb is unchanged, and the brightness is unaffected.
Answer, Key – Homework 11 – David McIntyre i1 =3 i2 i1 4. =1 i2 1 i1 = 5. i2 2 i1 1 6. = i2 4 i1 7. =4 i2 Explanation:
I correct 3 I 5. i5 = 2 I 6. i5 = 5 I 7. i5 = 7 I 8. i5 = 4 Explanation: 4. i5 =
3.
Given :
R1 R2 R3 R4
= r, = 2r, = 2r, = r.
i1 = 2 i 2 and
I = i 1 + i2 = 2 i 2 + i2 = 3 i 2 I ⇒ i2 = 3 2I . ⇒ i1 = 3
C
2r
r i3
2r
I
i2
i5
i1
A
B
r
I
i4
D Basic Concept: DC Circuit. Solution: Based on Kirchhoff’s law, the equation for the loop ACDA is given by −i1 R1 + i2 R2 = 0 i1 R2 ⇒ = i2 R1 2r = r = 2. 021 (part 2 of 3) 0 points Find the magnitude of the current i5 which flows from C to D. 1. i5 = 0 I 8 I 3. i5 = 6
2. i5 =
7
Following a similar analysis, one finds that i4 I 2I = 2, so that i3 = and i4 = . i3 3 3 Note: The junction equation at D is i2 + i 5 = i 4 ⇒ i5 = i4 − i2 , or = i3 − i1 2I I I =− = − 3 3 3 I |i5 | = . 3 022 (part 3 of 3) 0 points Find the resistance RAB . 1. RAB = 3 r 2. RAB = 0 3. RAB = r 4. RAB = 2 r 5. RAB =
4 r correct 3
Answer, Key – Homework 11 – David McIntyre 1 6. RAB = r 3 2 7. RAB = r 3 3 8. RAB = r 2 5 9. RAB = r 3 Explanation: By inspection, the following circuit is equivalent to the original circuit.
8
Determine IAC , i.e. the current between points A and C. I correct 3 I 2. 4 I 3. 9 I 4. 5 1.
5. I I 6 I 7. 2 6.
R1
R3
A
B
R2
R4
Here R1 R2 R1 + R 2 2 r2 =2 r + 2r 4 = r 3
RAB = 2
023 (part 1 of 3) 0 points Consider a cubic resistor network shown, where all the resistors are the same. Each resistor has a resistance r. A current I comes into the network at A. The same current leaves the network at B. I B
C
D
Explanation: Basic Concept: Symmetry in circuits. Solution: Go through the network systematically from A to B. The current enters at A. Since the network is completely symmetric, there can be no difference in resistance for the three possible paths the current can split into at A. Thus, each of the three must carry identical current, and they must add up to I (since no current is lost at the junction):
IAC =
024 (part 2 of 3) 0 points Determine ICD , i.e. the current between points C and D. I 9 I 2. 4 I 3. 2
1.
4. I A
I 3
5.
I 5
Answer, Key – Homework 11 – David McIntyre I 6. correct 6 I 7. 3 Explanation: Now we are at C. The current coming in from A has two possible choices of path, and again there is complete symmetry between the two. Thus the current going through C splits in half: I IAC = ICD = 2 6
9
Now |VAB | = RAB I, so 5 RAB = r 6 026 (part 1 of 2) 0 points Two identical light bulbs A and B are connected in series to a constant voltage source. Suppose a wire is connected across bulb B as shown.
A
B
025 (part 3 of 3) 0 points Determine RAB , i.e. the effective resistance between points A and B. r 3 r 2. 6 5r 3. correct 6 r 4. 2
V
1.
5. r 4r 3 2r 7. 3 3r 8. 2 7r 9. 6 Explanation: Finally, we calculate the equivalent resistance. We can step through the path ACDB to find the potential VAB between A and B. Kirchoff’s Laws tell us that each time we cross a resistor (moving with the current) we drop a potential V = RI. Therefore 6.
VA − VB = rIAC + rICD + rIDB Now note that IAC = IDB by symmetry, so µ ¶ I I I 5 VA − VB = |VAB | = r + + = rI 3 6 3 6
Bulb A 1. will burn half as brightly as before. 2. will burn as brightly as before. 3. will burn nearly four times as brightly as before. correct 4. will burn twice as brightly as before. 5. will go out. Explanation: The electric power is given by V2 . R Before the wire is connected, P = I2 R =
IA = I B =
V 2R
so that
V2 . 4R After the wire is connected, PA =
V and IA0 = 0 , so R IB0 = 0 V2 0 = 4 PA . PA = R IA0 =
Answer, Key – Homework 11 – David McIntyre 027 (part 2 of 2) 0 points and bulb B 1. will burn nearly four times as brightly as before. 2. will burn half as brightly as before. 3. will go out. correct 4. will burn as brightly as before. 5. will burn twice as brightly as before. Explanation: Since there is no potential difference between the two ends of bulb B, it goes off. 028 (part 1 of 3) 0 points Consider the circuit shown below. E1 r1 B A E2 C
i1 r2
R
029 (part 2 of 3) 0 points What equation does the loop DCFED yield? 1. −E2 − i2 r2 + i R = 0 2. −E2 + i2 r2 − i R = 0 3. −E2 − i2 r2 − i R = 0 4. E2 − i2 r2 − i R = 0 correct
6. E2 + i2 r2 + i R = 0 E
1. −E1 − E2 − i2 r2 + i1 r1 = 0 2. −E1 − E2 + i2 r2 − i1 r1 = 0 3. E1 − E2 − i2 r2 − i1 r1 = 0 4. −E1 + E2 + i2 r2 + i1 r1 = 0 5. E1 + E2 − i2 r2 + i1 r1 = 0 6. E1 − E2 + i2 r2 − i1 r1 = 0 correct
Explanation:
ABCDA : E1 − E2 + i2 r2 − i1 r1 = 0 .
D
i Apply Kirchhoff’s rules. What equation does the loop ABCDA yield?
7. E1 + E2 + i2 r2 − i1 r1 = 0
Recall that Kirchhoff’s loop rule states that the sum of the potential differences across all the elements around a closed circuit loop is zero. If a resistor is traversed in the direction of the current, the change in potential is −I R. If a emf source is traversed from the − to + terminals, the change in potential is +E. Apply the opposite sign for traversing the elements in the opposite direction. Hence, by inspection
5. E2 + i2 r2 − i R = 0
i2 F
10
7. E2 − i2 r2 + i R = 0 Explanation: DCF ED : E2 − i R − i2 r2 = 0 . 030 (part 3 of 3) 0 points Let: E1 = E2 = E, and r1 = r2 = r, where E = 9 V and r = 2.4 Ω. Also R = 4.1 Ω. Hint: From symmetry, one expects i1 = i2 . Find the current i. Correct answer: 1.69811 A. Explanation: Given : E1 = E2 = E = 9 V , r1 = r2 = 2.4 Ω , and R = 4.1 Ω . We are given that E1 = E2 and r1 = r2 . This implies that i1 = i2 . (Why? Look at the loops
Answer, Key – Homework 11 – David McIntyre DCFED and ABFEA and see if there are any similarities.) Hence the junction rule yields i1 + i2 = 2i2 = i i ⇒ i2 = . 2
7. It=0 = (R1 + R2 ) Vo Vo correct R1 Explanation: When the switch is in position “a” at t = 0, there is no potential drop across the capacitor. Note: There is no current flowing through R2 , so the entire potential drop, Vo , is across the resistor R1 . From Ohm’s law, 8. It=0 =
Substituting this into the loop equation DCFED, i E2 − i R − r2 = 0 2 Solving for i yields i=
E2
I=
r2 R+ 2 9V = 2.4 Ω 4.1 Ω + 2 = 1.69811 A .
1. τ =
3. τ =
E
S b a
4. τ = 5. τ = 6. τ =
After S is switched to position “a”, the initial current through R1 is 1. It=0 =
R1 R2 Vo R1 + R 2
2. It=0 = R1 Vo 3. It=0 =
Vo R1 + R 2
4. It=0 = R2 Vo R1 + R 2 Vo R1 R2 Vo = R2
5. It=0 = 6. It=0
R1 C
2. τ = R1 C
R2 C
Vo . R1
032 (part 2 of 3) 0 points Leave the switch at position “a” for a long time, then move the switch from “a” to “b”. When the switch is in position “b”, what is the time constant, τ , of the circuit?
031 (part 1 of 3) 0 points Consider the circuit shown below, the capacitor is initially uncharged.
R1
11
7. τ = 8. τ =
R2 C R1 R2 C R1 + R 2 R1 R2 (R1 + R2 ) C (R1 + R2 ) C R1 + R 2 R1 R2 C R1 + R 2 C R1 R2
9. τ = R2 C 10. τ = (R1 + R2 ) C correct Explanation: When the switch is in position “b”, R1 and R2 are now in series so the equivalent resistance is R = R1 + R2 . By definition, the time constant is τ = R C = (R1 + R2 ) C .
Answer, Key – Homework 11 – David McIntyre 033 (part 3 of 3) 0 points Let: Vo = 9 V, C = 1 µF, R1 = 7 Ω, and R2 = 20 Ω. 3 At a time τ after S has been switched to 2 position “b”, what is the power consumption of the circuit? Correct answer: 0.149361 W. Explanation: Given : Vo C R1 R2
The switch on the circuit is closed at t = 0. Find the charge on the capacitor at 4.73 s. Correct answer: 19.626 µC. Explanation: Given : t = 4.73 s , R = 1.1 MΩ = 1.1 × 106 Ω , C = 1.4 µF = 1.4 × 10−6 F , E = 14.7 V .
= 9 V, = 1 µF , = 7 Ω , and = 20 Ω .
R E
While the capacitor is discharging, the magnitude of the current decreases as a function of time as I(t) =
Vo e−t/τ . R1 + R 2
Noting that R = R1 + R2 , the power con3 sumed by the circuit at t = τ is 2 P = I 2 (R1 + R2 ) ¸2 · Vo −3/2 e (R1 + R2 ) = R1 + R 2 Vo2 = e−3 R1 + R 2 (9 V)2 e−3 = 7 Ω + 20 Ω = 0.149361 W . 034 (part 1 of 3) 0 points Given: The RC circuit in the figure below.
1.1 MΩ 14.7 V
1.4 µF
S
12
and
C
S
At t = 4.73 s, ³ ´ q = CE 1 − e−t/(R C)
= (1.4 × 10−6 F)(14.7 V)× ¸¾ ½ · 4.73 s 1 − exp − (1.1 × 106 Ω) (1.4 × 10−6 F) = 1.9626 × 10−5 C = 19.626 µC . 035 (part 2 of 3) 0 points Find the current in the resistor at 4.73 s. Correct answer: 0.61947 µA. Explanation: At t = 4.73 s, E −t/(R C) e R 14.7 V × = 6 Ω 1.1 × 10 ·
I=
4.73 s exp − 6 (1.1 × 10 Ω) (1.4 × 10−6 F) = 6.1947 × 10−7 A = 0.61947 µA . 036 (part 3 of 3) 0 points
¸
Answer, Key – Homework 11 – David McIntyre At 4.73 s the current in the resistor is I (Part 2) and the charge on the capacitor is q (Part 1). What is the power delivered by the battery? Correct answer: 9.10621 µW. Explanation: In the time interval ∆t, work done by the battery in pushing charge ∆q across the battery is ∆Wbattery = ∆q · E . Correspondingly, the power is d Wbattery dq =E =IE. dt dt The power dissipated in a resistor is d Wresistor = I2 R . dt The power to create the electric field in a capacitor is d Wcapacitor q =I . dt C Thus the total power dissipated in the capacitor and resistor, that is the power delivered by the battery is ³ d Wbattery q´ =I IR+ dt C = (6.1947 × 10−7 A) h × (6.1947 × 10−7 A) (1.1 × 106 Ω) (1.9626 × 10−5 C) i + (1.4 × 10−6 F) = 9.10621 × 10−6 W = 9.10621 µW .
13
