PHYS208 Fundamentals of Physics II

Answer for Ch. 27, 5E

GIVEN:
Gauge:4681012141618
Diameter (mils)20416212910281645140
Safe Current (A)70503525201563
CALCULATED:
Diameter (mm)5.184.113.282.592.11.61.31.0
Area (mm2)21138.45.33.32.11.30.8
Current Density (A/mm23.33.84.24.86.07.24.63.6

[graph]

Solution to Ch. 27, 27E:

54 ohm

Solution to Ch. 27, 35P:

1750 degrees C.

Solution to Ch. 27, 41P:

a. R = rho L / (pi a b)

b. R = rho L / (pi b2) = rho L/A.

Solution to Ch. 27, 55P:

a. For a 30-day month, cost is $4.32.

b. 144 ohm

c. 0.833 A

d. The lightbulb filament is probably tungsten; the resistivity of metal increases with temperature. Since the temperature of a glowing filament is typically several thousand degrees, the resistance will be less when its turned off.

Solution to Ch. 27, 62P:

a. 8.5% drop. Your may use the formula P=V2/R and directly calculate each power and compute the fractional difference. You may also notice that delta P/P = 2 delta V/V can approximate the power drop as being twice the voltage drop (in percentages).

b. Denominator in power formula above also decreases with lowered voltage, so power drop is not as large as expected in (a).


"http://www.physics.udel.edu/~watson/phys208/exercises/answers/0218.html"
Last updated Feb. 17, 1998.
Copyright George Watson, Univ. of Delaware, 1998.