The conductance of a wire varies directly as the square of the wire's diameter and inversely as its length. Fi

Question

The conductance of a wire varies directly as the square of the wire's diameter and inversely as its length. Fifty meters of wire with a diameter 2 mm has a conductance of 0.12 ohms. If a wire of the same material has length of 75 m and diameter of 2.5 mm, what is its conductance?

Answer 1

let conductance = c

diameter = d

lenght = l



Therefore, c = kd^2/l



Substitute the parameters given to get k - the constant of variation:

0.12 = k * 2^2/50 to give

k = 1.5



The formula connecting all these c, d and l becomes



c = 1.5 * d^2/l



To find solution to the problem, substitute the last parameter given to find the conductance:



c = 1.5 *2.5^2/75

c = 0.125



The conductnce is 0.125 ohms

Answer 2

Firstly, make sure you understand the difference between conductance and resistance, ohms and mhos. Conductance as units of mhos and resistance has units of ohms.

http://en.wikipedia.org/wiki/Electrical_resistance

http://en.wikipedia.org/wiki/Electrical_conduction



So, are you paraphrasing the problem? Or are you indeed given that exact wording? If so, shame on your instructor.



The resistance is proportional to length and inversely proportional to area.:

0.12 ohms*(75/50)*(2/2.5)^2



The "conductance" is the inverse of the resistance

Answer 3

R = ok * (l / (r^2)) the place R =resistance, l = length, r = diameter and ok is a relentless via our meusurements i.e. ohms, meters millimeters. to locate ok plug in you inital condition for whilst R = 32 to get the linked fee ok =4/3 now you will locate the resistance of the twine of 1500m length and 10mm by using putting those because of the fact the values of l and r respectively so R = (4/3)*( 1500 / (10^2)) = 20 ohms

Answer 4

the given conductance C1=0.12 mho when length L1=50m and diameter D1=2mm, Now L1=50 X 1000 =50000mm



Let the conductance to be find is C2= ? when length L2= 75m i.e L2=75 X 1000 = 75000mm and diameter D2 = 2.5mm



Now square of D1= 4 sqmm and D2= 6.25 sqmm



As per your given condition C1 =(k X sq of D1)/L1 where K is a constant, so K = (C1 X L1)/ sq of D1



and similarly C2 =( k X sq of D2)/L2 where K is a constant so



K = (C2 X L2)/sq of D2



as the material is same so we can equalize both K



hence (C1 x L1 )/sq D1=(C2 X L2)/ sq D2



or C2 = (C1 X L1) X (sq D2) / (sq D1) X L2



=( 0.12 X 50000 X 6.25)/ (4 X 75000)



= 0.125 mho(ans)

Answer 5

conductance = k (diameter)^2/length



a) 0.12 = k (2e-3)^2 /(50)

= k(4e-6)/(50)

= k*(400e-8)/50

= k*8e-8



So k = 0.12/(8e-8)

= (12/8) (e-2)(e8)

= 1.5 e6



b) So now with length = 75 and diameter = 2.5e-3

conductance = k (2.5e-3)^2/(75)

= (1.5e6) (25^2)(e-8)/75

= (1.5e6)*25*(e-8)/(3)

= (15e5)(25)(e-8)/3

= 5*25 (e-3)

= 125 e-3

=0.125

Answer 6

c = k d ² / l

Work in mm:-

0.12 = k x 2 ² / (50 x 1000)

k = 50000 x 0.12 / 4

k = 1500



c = ( 1500 x 2.5 ² ) / (75 x 1000)

c = (1.5 x 6.25) / 75

c = 0.125 ohms