# resistivity

1.Compute the voltage drop along a 28

length of household no. 14 copper wire (used in

15-

circuits). The wire has diameter

1.628

current.

figures.

2.The filament of a light-bulb has a resistance of

when hot.

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

is

,

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

?

=___%

### 2 個解答

• 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.

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• i have no idea what you wrote on the 2nd question part b

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