EET 102
Fall 1996
Prof. Ken Reid
9/16/96

Homework 2 - Answers


ch 3, prob 4) Find the resistance of the given wire

Copper, therefore p (rho) = 10.37
d = 0.01 inch = 10 mils
A (CM) = d squared = (10 mils)(10 mils) = 100 CM
R = p (l/A) = 10.37 (200 ft / 100 CM) = 20.74 Ohm

ch.3, prob 11a) Find the resistance of the copper bar

Copper, so p = 10.37
l = 4 ft
A (CM) = w * h * (4/pi) = (500 mil)(3000 mil)(4/pi)
= 1.9099x10E6 CM

R = p (l/A) = 10.37 (4/1.9099x10E6) = 21.72 micro Ohm

ch 4, #1) V = RI = (6) (2.5) = 15V

ch 4, #2) I = V/R = 12/72 = .1667A = 166.7mA

ch 4, #3) R = V/I = 6/(1.5x10E-3) = 4000 ohm = 4k Ohm

ch 4, #4) I = V/R = 12/.056 = 214.3 A

ch 4, #5) V = RI = (.02M ohm)(3.6 micro A) = .072 V = 72 mV

ch 4, #6) I = V/R = 62/15k = .00413A = 4.13mA

ch 4, #7) R = V/I = 120/2.2 = 54.54 Ohm

ch 4, #8) I = V/R = 120/7.5k = .016A = 16mA