DW's Homework Answers
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The additional problems handed out, calling the circuits (A), (B) and (C) clockwise from the top
- 3: Series gives the max Rs=786 Ω, parallel gives the min,Rp=20.3 Ω
- 6: a) I=0.400 A, P24=3.84 W, P96=15.4 W, b) Rp=19.2 Ω, I=2.50 A, 2.0 A through R24 and 0.5 A through R96
- 16: 2.9994 V
- 23: a) 200 A, b) 10.0 V, c) 2000 W, d) 80 A, Vmot = 4.0 V, Pmot = 320 W
- 26: b) 1.67 A, c) 27.9 kW, d) 20 MΩ
(A) Req=8.75 Ω total power P=9.26 W, I=1.03 A
Note: R2 and R3 are parallel, then R23 is series with R1
(B) Req=3.58 Ω total power P=22.6 W, I=2.52 A
Note: R3, R2, R4 are parallel , then R324is series with R2 ...
(C) Req=3.27 Ω total power P=24.8 W, I=2.75 A
If instead they are capacitors
(A) Ceq=3.81 μF, Q=34.3 μC, E=1.54 x 10-4 J
(B) Ceq=12.06 μF, Q=108 μC, E=4.88 x 10-4 J
(C) Ceq=12.8 μF, Q=115 μC, E=5.2 x 10-4 J
- 4: 4.00 mA
- 9: 0.12 C, 7.5 x 1017 electrons
- 23: 0.20 mA
- 25: 0.322 Ω
- 27: the Al will be 1.24 times the diameter of the Cu wire
- 40: P = 2 x 1012 W. But not for very long.
- 49: beam power is 1500 W
- 56: $3.94 billion
- 69: a) = 2.08 x 105 A, for that current, b) the waste heat will be I2R= 4.3 x 1010 W. c) but that is greater than the power output of the generator! So it cannot happen. d) Don't confuse the idea of how much power you want to get to the destination with how much power you need to produce at the source.
- 73: 480 V
- 76: R = 576 Ω, power during melt down is 100 W.
- 2: eV=1/2 mv2, so v=1.2 x 108 m/s
- 5: 1940 V
- 7: a) 2 GJ, b) 765 kg of water can evaporate (mcΔT + mLf... remember?)
- 13: yup, they are equivalent
- 17: a) E=2.5 MV, so no breakdown (no spark), b) at d=1.7 mm you get a big enough E-field.
- 29: Q = +0.833 μC
- 38: sketch so equipotentials are perp to E-field. Higher potential is in direction opposite the arrows of E-field
- 50: 20,000 V
- 61: Cnet = 11.4 μF. (a couple of intermediate combinations give 2.06 and 1.37 μF)
- 66: series: 14.2 μC and 63.8 μJ, parallel: 84.6 μC and 381 μJ
- 1: a) 1.25 x 1010 electrons b) 3.13 x 1012 electrons
- 6: a) 1.88 x 106 electrons short of a full load, b) 1.88 x 10-10
- 16: 14.4 N away from Qa, (or in 1 sig fig, 10 N)
- 17: 3.45 x 1016 m/s2
- 20: 1.04 nC
- 25: 85.9 cm right of negative charge, or, if both positive 10.9 cm left of +3 μC
- 30: 69.4 nC and 6.25 N/C
- 43: a) 4 x 107N/C, b) 7 cm, in the middle, c)No, there is no suitable place 0-8 cm, d) The net charge in fig (a) is zero, but the net charge in fig (b) is +1, so E goes to zero faster in the case of fig. (a), and e) Add the electric fields due to the three charges for some point right of 11 cm. This is ugly algebra, which is why they suggest looking at the graph of the sum of the E-fields (if you have some way to make a graph of a function), then the graph of the sum of the E-fields crosses the zero at x=30.6 cm
- 48: Vector addition! All the x-components cancel, and the net vector is 0.102 N in the negative y-direction
- 57: remember circular motion? F=ma=mv2/r, and v=rω, but F=kqQ/r2, put it all together and get ω = 4.12 x 1016 rad/sec (and from v=rω get v=2.2 x 106 m/s, if you are interested)
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