Physics 1304/1404 - Spring 1999
Homework Assignment #3
Due: Thursday 25 February 1999
Chapters 27 and 28.
Chapter 27 - 3, 4, 15, 17, 20.
Chapter 28 - 4, 5, 6, 8, 9, 11, 13, 16, 17, 22, 28.
Chapter 27 - 4, 16, 19A, 19, 25, 42, 45.
Chapter 28 - 1, 5, 12, 14, 18, 25, 42, 43, 44, 50.
These are the ANSWERS only, not the SOLUTIONS. It is not sufficient to
copy these and turn them in as homework. You must show your work.
Q 27-3) Resistance (R) is the constant of proportionality in most
conductors between voltage and current (V=IR); the MKS unit is the ohm.
Resistance increases with the length of a conductor and decreases with
the cross-sectional area.
Resistivity (rho) is an intrinsic property of the material from which the
conductor is made, without regard to its length (L) or cross-sectional
area (A). The relation is (R = rho L/A); the MKS unit is the ohm.meter.
Q 27-4) AreaA/AreaB = 1/3;
rA/rB = 1/sqrt(3)
Q 27-15) When one electron enters a wire, another electron is
pushed out the other end of the wire. The push travels at nearly the
speed of light.
Q 27-17) (P = V2/R) The smaller resistance will
dissipate more power.
Q 27-20) (R = V2/P) The 25 W bulb has the higher resistance.
(I=V/R) The 100 W bulb carries the greater current.
25 W bulb: R = 484 ohms; I = 0.227 amps.
100 W bulb: R = 121 ohms; I = 0.909 amps.
Q 28-4) Set A is wired with the bulbs (resistors) in parallel;
set B is wired in series.
Q 28-5) Connect the resistors in series.
Q 28-6) Connect the resistors in parallel.
Q 28-8) Current only.
Q 28-9) Potential difference (voltage) only.
Q 28-11) Parallel, because one sees cars with only one headlight on.
Q 28-13) Conservation of Charge (junction rule), and
Conservation of Energy (loop rule).
Q 28-16) The light glows brightly at first (as bright as it would glow
if it were connected across the battery without the capacitor), then grows
dimmer exponentially as the cap charges because the current is decreasing
Q 28-17) The internal resistance of an ideal ammeter is zero, and the
internal resistance of an ideal voltmeter is infinite. Real device only
approximate these values.
Q 28-22) No.
- Lamps A and B glow brighter.
- Lamp C is now unlit because it has been shorted out of the circuit.
- The current increases.
- The voltage drops across both A and B increase; the voltage drop
across C is now zero.
- The power dissipated increases. Even though only two bulbs are now
lit, they use up more power than the three dimmer original bulbs.
(Was P=V2/(3R), now is P=V2/(2R))
- 17 A
- 85,000 A/m2
P 27-16) 0.310 ohms
P 27-19A) R/9
P 27-19) 1.33 ohms
P 27-25) 6.43 A
P 27-45) 36.1%
- 7.67 ohms
- 1.76 W
P 28-5) 12 ohms
P 28-12) There are many ways to do this.
- 17.1 ohms
- current in 4 ohm resistor: 1.99 A
current in 7 ohm resistor: 1.17 A
current in 10 ohm resistor: 0.82 A
current in 9 ohm resistor: 1.99 A
- The 11 ohm resistor uses more power.
- (proof required)
- The 22 ohm resistor uses more power.
- (proof required)
- The parallel configuration uses more power.
parallel: 148.5 W; series: 33 W
- Vb - Va = -10.42 V
- I1 = 0.141 A
I2 = 0.915 A
I3 = 0.774 A
- 5 s
- 150 microcoulombs
- 4.06 microamps
- 1.5 s
- 1 s
- [200 + 100 exp(-t/1s)] microamps
- RC ln(2) = 0.693 RC
- an infinite amount of time
P 28-50) 0.982 s
Please report any corrections to
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