1.Compute the voltage drop along a 28
length of household no. 14 copper wire (used in
circuits). The wire has diameter
Express your answer using two significant
2.The filament of a light-bulb has a resistance of
a.Calculate the temperature of the filament when it is hot, and take into
account the change in length and area of the filament due to thermal
expansion (assume tungsten for which the thermal expansion coefficient
assuming an average
temperature coefficient of resistivity
b.In this temperature range, what is the percentage change in resistance
due to thermal expansion?
c.What is the percentage change in resistance due solely to the change in
- 天同Lv 71 十年前最愛解答
1. Resistance of copper wire R = (rho).L/A
where (rho) is the resistivity of copper = 1.68 x 10^-8 ohm-m
L is the length of wire (= 28 m)
A is the cross-sectional area of wire (= pi.x (0.001628^2)/4 m^2 ), where pi = 3.14159....
Thus, R = [1.68 x 10^-8 ] x 28/[pi.x (0.001628^2)/4 ] ohms
Hence, voltage drop, by Ohm's Law = 12 x R
2. (a) Usin R = (rho).L/A
hence, 12 = (rho),L/A
and 140 = (rho)',L'/A'
i.e. 140/12 = [(rho)'/(rho)].[L'/L].[A/A'] ------------------------ (1)
Since (rho)'/(rho) = (1+ 0.0045(T-293))
L'/L = (1+ 5.5x10^-6(T-294))
A'/A = (1+2x5.5x10^-6(T-293))
substitute these values into equation (1) and solve for T
(b) assume (rho) remains constant
Ro = (rho).L/A
R(expansion) = (rho).L'/A'
hence, R(expansion)/Ro = [L'/L].[A/A']
using the ratio of L'/L and A'/A is (a), calculate R(expansion)/Ro
and then find the percentage change in resistance
(c) R(change in rho) = (rho)',L/A
hence, R(change in rho)/Ro = [(rho)'/(rho)] = (1+ 0.0045(T-293))
claculate first the ratio of R(change in rho)/Ro then calculate the percentage change.
- 1 十年前
i have no idea what you wrote on the 2nd question part b