nswer, Key Homework 11 avid Mcntyre 1 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 entral time. hapters 27 and 28 problems. 001 (part 1 of 1) 0 points t 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/cm 3. alculate the average drift speed of electrons traveling through a copper wire with a cross-sectional area of 1 mm 2 when carrying a current of 1 (values similar to those for the electric wire to your study lamp). orrect answer: 7.38624 10 5 m/s. Given : N = 1, M = 63.54 g, ρ = 8.92 g/cm 3, = 1 mm 2 = 1 10 6 m 2, = 1, and q e = 1.602 10 19 /electron. We first calculate n, the number of currentcarrying electrons per unit volume in copper. ssuming one free conduction electron per atom, n = N ρ M, where N is vogadro s number and ρ and M are the density and the atomic weight of copper, respectively n = n ( 1 electron atom ) N ρ M. ( 1 electron ) 6.02 10 23 atoms atom 63.54 g (8.92 g/cm 3) ( 10 6 cm 3 ) 1 m 3 = 8.45112 10 28 electrons/m 3. The drift speed is v d, v d = n q e = 1 8.45112 10 28 electrons/m 3 1 (1.602 10 19 /electron) 1 (1 10 6 m 2 ) = 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 m can be felt and 5 m is painful. bove 15 m, a person loses muscle control, and 70 m can be fatal. 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? orrect answer: 100 V. Given : min = 1 m, and R dry = 100000 Ω. The minimum voltage depends on the minimum current for a given resistance, so V min = min R dry ( ) 1 = (1 m) (100000 Ω) 1000 m = 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? orrect answer: 18.8679 m. Given : V min = 100 V, and R wet = 5300 Ω.
nswer, Key Homework 11 avid Mcntyre 2 For the voltage V from Part 1 and the resistance R wet, = V min R wet = 100 V ( ) 1000 m 5300 Ω 1 = 18.8679 m. 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? orrect answer: 5.3 V. Given : min = 1 m, and R wet = 5300 Ω. V 1 = min R wet ( ) 1 = (1 m) (5300 Ω) 1000 m = 5.3 V. 005 (part 1 of 4) 0 points f the current carried by a metallic conductor is doubled, (a) What happens to the charge carrier density? ssume that the temperature of the metallic conductor remains constant. 1. t quadruples. 2. t is unchanged. correct 3. t is quartered. 4. t is halved. 5. t doubles. asic oncept: J = σ = n q v d v d = q m τ. 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. t doubles. correct 2. t is unchanged. 3. t is quartered. 4. t quadruples. 5. t is halved. The current density J = only depends on the current and the cross-sectional area of the conductor, so if doubles and remains the same, J must also double: J = 2 = 2 = 2 J. 007 (part 3 of 4) 0 points (c) What happens to the electron drift velocity? 1. t is unchanged. 2. t is halved. 3. t is quartered. 4. t quadruples. 5. t doubles. correct The electron drift velocity v d is given by v d = n q, where q is the charge of an electron. So if doubles, v d must also double: v d = 2 n q = 2 n q = 2v d.
nswer, Key Homework 11 avid Mcntyre 3 008 (part 4 of 4) 0 points (d) What happens to the average time between collisions? 1. t doubles. 2. t is halved. 3. t is quartered. 4. t is unchanged. correct 5. t quadruples. The average time τ between collisions is given by τ = v d m q, where m is the mass of an electron and is the applied electric field (determined by Ohm s Law: J = σ ). Substituting for, τ = v d m σ q J. Since v d 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). ll 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. 1. > > > 2. = > = V 3. > = > 4. = > > 5. = > = correct 6. = = = 7. = > = 8. > = > 9. > > > 10. = = > From the figure, the total resistance of the circuit is 2 R + R 2 = 5 R 2. Then = = V 5 R 2 = 2 5 V R = 0.4 V R = = 1 2 = 1 V 5 R = 0.2V R The rankings of the currents are also the corresponding rankings of the brightness. 010 (part 2 of 2) 0 points Suppose a switch has been added to the circuit as shown. The switch is initially closed. V When the switch is opened, what happens to the currents through bulbs,, and? 1. decreases, remains the same, decreases. 2. remains the same, decreases, remains the same 3. increases, increases, increases.
nswer, Key Homework 11 avid Mcntyre 4 4. remains the same, increases, remains the same. 5. decreases, increases, decreases. correct 6. remains the same, remains the same, remains the same. 7. increases, remains the same, decreases. 8. increases, remains the same, increases. 9. increases, decreases, increases. 10. decreases, decreases, decreases. With the switch open, the total resistance of the circuit is 3 R, so = = = V 3 R = 0.33V R This is to be compared with the case that the switch is closed, where from previous part, = = 0.4 V R and = 0.2 V R. So for the present case, and decrease and increases. 011 (part 1 of 4) 0 points battery has an emf of 12 V and an internal resistance of 0.13 Ω. ts terminals are connected to a load resistance of 2 Ω. Find the current in the circuit. orrect answer: 5.6338. Given : = 12 V, R = 2 Ω, and r = 0.13 Ω. The total resistance is R + r, so = R + r 12 V = 2 Ω + 0.13 Ω = 5.6338. 012 (part 2 of 4) 0 points alculate the terminal voltage of the battery. orrect answer: 11.2676 V. The terminal voltage V of the battery is equal to V r = r = 12 V (5.6338 ) (0.13 Ω) = 11.2676 V. 013 (part 3 of 4) 0 points Find the power dissipated in the load resistor. orrect answer: 63.4795 W. The power dissipated in the load resistor is P R = 2 R = (5.6338 ) 2 (2 Ω) = 63.4795 W. 014 (part 4 of 4) 0 points Find the power dissipated in the battery. orrect answer: 4.12617 W. The power dissipated in the battery is P r = 2 r = (5.6338 ) 2 (0.13 Ω) = 4.12617 W. 015 (part 1 of 2) 0 points n the figure below the switch S is initially in position 1. Given, = R 2 = R 3. Neglect the internal resistance of the battery.
nswer, Key Homework 11 avid Mcntyre 5 2 1 3 V S 0 R 2 as when was in the circuit. Since R 2 and R 3 have the same terminal voltage and resistance, the current through R 2 and R 3 must now be the same. 017 (part 1 of 1) 0 points onsider the following circuit containing identical bulbs. What happens to the current through R 3 when the switch is moved to the open position 2. 1. The current through R 3 decreases to twothirds its original value. V 2. The current through R 3 increases to twice its original value. 3. The current through R 3 increases to three-halves its original value. 4. The current through R 3 is reduced to one-half its original value. 5. The current through R 3 remains the same. correct The voltage across R 3 is the of the battery, and is unchanged. The current through R 3 remains the same,. R 3 016 (part 2 of 2) 0 points What happens when switch S is moved to position 3, leaving R 2 and R 3 parallel? 1. The current through R 2 is half what it was with in the circuit. 2. The current through R 2 and R 3 are now the same. correct 3. The current through R 3 increases. 4. The current through R 3 decreases. 5. The current through R 2 remains the same 1. = = = = 2. = = > = 3. = > = = 4. = = > = correct 5. = = > > 6. = > = = 7. = > > = 8. = = > > 9. > > > > 10. = > > = The potential across bulbs, and is V. The potential across bulbs and is V 2. Thus i = i = i = V R i = i = V 2R. 018 (part 1 of 2) 0 points Four identical light bulbs are connected either in series (circuit 1) or parallel (circuit 2)
nswer, Key Homework 11 avid Mcntyre 6 to a constant voltage battery with negligible internal resistance, as shown. 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 f one of the bulbs in circuit 2 is unscrewed and removed from its socket, the remaining 3 bulbs 1. go out. circuit 1 2. are unaffected. correct 3. become brighter. 4. become dimmer. circuit 2 ompared to the individual bulbs in circuit 1, the individual bulbs in circuit 2 are 1. the same brightness. 2. 1 4 as bright. 3. less than 1 4 as bright. 4. 4 times brighter. 5. more than 4 times brighter. correct Solution: n circuit 1, the voltage across each light bulb is V = R = 4 R R = 4, so the power of each bulb in circuit 1 is P 1 = V 2 R = 2 16 R. n circuit 2, the voltage across each bulb is identical; namely. Hence the power of each bulb in circuit 2 is P 2 = 2 R = 1 16 P 1. 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. 020 (part 1 of 3) 0 points Hint: pply the Kirchhoff s law to the loop. Given: 2 = 2 R 4 = R 2 = R 3, = r. is the current entering and leaving the circuit. ach current shown in the figure is denoted by the same subscript as the resistor through which it flows (e.g., i 1 is the current flowing through ). R 2 i 1 Find the ratio i 1. 1. i 1 = 2 correct 2. i 1 = 1 3 i5 i 3 R 4 R 3 i 4
nswer, Key Homework 11 avid Mcntyre 7 3. i 1 = 3 4. i 1 = 1 5. i 1 = 1 2 6. i 1 = 1 4 7. i 1 = 4 Given : = r, R 2 = 2 r, R 3 = 2 r, R 4 = r. r 2 r and 4. i 5 = 3 correct 5. i 5 = 2 6. i 5 = 5 7. i 5 = 7 8. i 5 = 4 i 1 = 2 = i 1 + = 2 + = 3 = 3 i 1 = 2 3. 2 r i 1 i5 i 3 r i 4 Following a similar analysis, one finds that i 4 = 2, so that i 3 = i 3 3 and i 4 = 2 3. Note: The junction equation at is asic oncept: ircuit. Solution: ased on Kirchhoff s law, the equation for the loop is given by i 1 + R 2 = 0 i 1 = R 2 = 2 r r = 2. 021 (part 2 of 3) 0 points Find the magnitude of the current i 5 which flows from to. 1. i 5 = 0 2. i 5 = 8 3. i 5 = 6 + i 5 = i 4 i 5 = i 4, or = i 3 i 1 = 3 2 3 = 3 i 5 = 3. 022 (part 3 of 3) 0 points Find the resistance R. 1. R = 3 r 2. R = 0 3. R = r 4. R = 2 r 5. R = 4 3 r correct
nswer, Key Homework 11 avid Mcntyre 8 6. R = 1 3 r 7. R = 2 3 r etermine, i.e. points and. 1. 3 correct the current between 8. R = 3 2 r 2. 4 9. R = 5 3 r y inspection, the following circuit is equivalent to the original circuit. 3. 9 4. 5 5. R 3 6. 6 Here R 2 R 4 R = 2 R 2 = 2 2 r2 r + 2 r = 4 3 r 023 (part 1 of 3) 0 points onsider a cubic resistor network shown, where all the resistors are the same. ach resistor has a resistance r. current comes into the network at. The same current leaves the network at. 7. 2 asic oncept: Symmetry in circuits. Solution: Go through the network systematically from to. The current enters at. Since the network is completely symmetric, there can be no difference in resistance for the three possible paths the current can split into at. Thus, each of the three must carry identical current, and they must add up to (since no current is lost at the junction): = 3 024 (part 2 of 3) 0 points etermine, i.e. the current between points and. 1. 9 2. 4 3. 2 4. 5. 5
nswer, Key Homework 11 avid Mcntyre 9 6. 6 correct 7. 3 Now we are at. The current coming in from has two possible choices of path, and again there is complete symmetry between the two. Thus the current going through splits in half: = 2 = 6 025 (part 3 of 3) 0 points etermine R, i.e. the effective resistance between points and. 1. r 3 2. r 6 3. 5r 6 correct 4. r 2 5. r 6. 4r 3 7. 2r 3 8. 3r 2 9. 7r 6 Finally, we calculate the equivalent resistance. We can step through the path to find the potential V between and. Kirchoff s Laws tell us that each time we cross a resistor (moving with the current) we drop a potential V = R. Therefore V V = r + r + r Now note that = by symmetry, so ( V V = V = r 3 + 6 + ) = 5 3 6 r Now V = R, so R = 5 6 r 026 (part 1 of 2) 0 points Two identical light bulbs and are connected in series to a constant voltage source. Suppose a wire is connected across bulb as shown. ulb V 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. The electric power is given by P = 2 R = V 2 R. efore the wire is connected, = = V 2 R so that P = V 2 4 R. fter the wire is connected, = V R and = 0, so = 0 P = V 2 R = 4 P.
nswer, Key Homework 11 avid Mcntyre 10 027 (part 2 of 2) 0 points and bulb 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. Since there is no potential difference between the two ends of bulb, it goes off. 028 (part 1 of 3) 0 points onsider the circuit shown below. 1 r 1 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. f a resistor is traversed in the direction of the current, the change in potential is R. f a emf source is traversed from the to + terminals, the change in potential is +. pply the opposite sign for traversing the elements in the opposite direction. Hence, by inspection : 1 2 + r 2 i 1 r 1 = 0. 029 (part 2 of 3) 0 points What equation does the loop F yield? 1. 2 r 2 + i R = 0 2. 2 + r 2 i R = 0 3. 2 r 2 i R = 0 4. 2 r 2 i R = 0 correct i 1 2 r 2 F R i pply Kirchhoff s rules. does the loop yield? What equation 5. 2 + r 2 i R = 0 6. 2 + r 2 + i R = 0 7. 2 r 2 + i R = 0 F : 2 i R r 2 = 0. 1. 1 2 r 2 + i 1 r 1 = 0 2. 1 2 + r 2 i 1 r 1 = 0 3. 1 2 r 2 i 1 r 1 = 0 4. 1 + 2 + r 2 + i 1 r 1 = 0 5. 1 + 2 r 2 + i 1 r 1 = 0 6. 1 2 + r 2 i 1 r 1 = 0 correct 7. 1 + 2 + r 2 i 1 r 1 = 0 030 (part 3 of 3) 0 points Let: 1 = 2 =, and r 1 = r 2 = r, where = 9 V and r = 2.4 Ω. lso R = 4.1 Ω. Hint: From symmetry, one expects i 1 =. Find the current i. orrect answer: 1.69811. Given : 1 = 2 = = 9 V, r 1 = r 2 = 2.4 Ω, and R = 4.1 Ω. We are given that 1 = 2 and r 1 = r 2. This implies that i 1 =. (Why? Look at the loops
nswer, Key Homework 11 avid Mcntyre 11 F and F and see if there are any similarities.) Hence the junction rule yields i 1 + = 2 = i =. Substituting this into the loop equation F, Solving for i yields 2 i R r 2 = 0 i = 2 R + r 2 2 9 V = 4.1 Ω + 2.4 Ω 2 = 1.69811. 031 (part 1 of 3) 0 points onsider the circuit shown below, the capacitor is initially uncharged. R 2 S b fter S is switched to position a, the initial current through is 1. t=0 = R 2 V o 2. t=0 = V o 3. t=0 = V o 4. t=0 = R 2 V o 5. t=0 = R 2 6. t=0 = V o R 2 V o a 7. t=0 = ( ) V o 8. t=0 = V o correct 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 R 2, so the entire potential drop, V o, is across the resistor. From Ohm s law, = V o. 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? 1. τ = 2. τ = 3. τ = R 2 4. τ = R 2 5. τ = R 2 ( ) 6. τ = ( ) 7. τ = R 2 8. τ = R 2 9. τ = R 2 10. τ = ( ) correct When the switch is in position b, and R 2 are now in series so the equivalent resistance is R =. y definition, the time constant is τ = R = ( ).
nswer, Key Homework 11 avid Mcntyre 12 033 (part 3 of 3) 0 points Let: V o = 9 V, = 1 µf, = 7 Ω, and R 2 = 20 Ω. t a time 3 τ after S has been switched to 2 position b, what is the power consumption of the circuit? orrect answer: 0.149361 W. The switch on the circuit is closed at t = 0. Find the charge on the capacitor at 4.73 s. orrect answer: 19.626 µ. Given : t = 4.73 s, R = 1.1 MΩ = 1.1 10 6 Ω, = 1.4 µf = 1.4 10 6 F, = 14.7 V. and Given : V o = 9 V, = 1 µf, = 7 Ω, R 2 = 20 Ω. and While the capacitor is discharging, the magnitude of the current decreases as a function of time as R S (t) = V o e t/τ. Noting that R =, the power consumed by the circuit at t = 3 2 τ is P = 2 ( ) [ ] 2 V o = e 3/2 ( ) Vo 2 = e 3 (9 V)2 = 7 Ω + 20 Ω e 3 = 0.149361 W. 034 (part 1 of 3) 0 points Given: The R circuit in the figure below. 1.1 MΩ 14.7 V 1.4 µf S t t = 4.73 s, t/(r q = (1 e )) = (1.4 10 6 F)(14.7 V) { [ ]} 4.73 s 1 exp (1.1 10 6 Ω) (1.4 10 6 F) = 1.9626 10 5 = 19.626 µ. 035 (part 2 of 3) 0 points Find the current in the resistor at 4.73 s. orrect answer: 0.61947 µ. t t = 4.73 s, = ) e t/(r R = 14.7 V 1.1 10 6 Ω [ ] 4.73 s exp (1.1 10 6 Ω) (1.4 10 6 F) = 6.1947 10 7 = 0.61947 µ. 036 (part 3 of 3) 0 points
nswer, Key Homework 11 avid Mcntyre 13 t 4.73 s the current in the resistor is (Part 2) and the charge on the capacitor is q (Part 1). What is the power delivered by the battery? orrect answer: 9.10621 µw. n the time interval t, work done by the battery in pushing charge q across the battery is W battery = q. orrespondingly, the power is d W battery dt = dq dt =. The power dissipated in a resistor is d W resistor dt = 2 R. The power to create the electric field in a capacitor is d W capacitor dt = q. Thus the total power dissipated in the capacitor and resistor, that is the power delivered by the battery is d W battery dt ( = R + q ) = (6.1947 10 7 ) [ (6.1947 10 7 ) (1.1 10 6 Ω) + (1.9626 10 5 ) (1.4 10 6 F) = 9.10621 10 6 W = 9.10621 µw. ]