Question

Find the resistance of a silver wire 50 yd long and 0.0045 in. in diameter. T=20 C

Answer 1

Answer:

Heya !

Here's your answer mate !

→Resistance Problem #5: Find the resistance of a silver wire 50 yd long and 0.0045 in. in diameter. (T = 20 o C, ρ silver = 9.9 CM·Ω/ft). Solution: The resistance is 73.33 Ω.

Answer 2

Given: the length of the silver wire, L =50 yd

           the diameter of the silver wire, d = 0.0045 inch

To Find: the resistance of the wire, R.

Solution:

To calculate R, the formula used:

• Resistance = resistivity x ( Length / Area)
• R = ρ x (L/A)

Applying the above formula:

R = ρ x (L/A)

here, ρ of silver = 1.59 x 10⁻⁸ ohm.m

R = 1.59 x 10⁻⁸ x (L/A)                                            ⇒ 1

Convert 50 yds into m:

1 yd = 0.914 m

50 yds = 0.914 x 50

            = 45.7 m                                                   ⇒ 2

Convert 0.0045 in into m:

1 inch = 0.03 m

0.0045 in = 0.03 x 0.0045

                = 0.000135 m

                = 1.35 x 10⁻⁴ m                                            

Now calculate the area of the silver wire:

Area, A = π x r²

and, r, radius of the wire = diameter /2 = d/2

∴ A = π x (d/2)²

As d = 1.35 x 10⁻⁴m

So, A = 3.14 x (1.35 x 10⁻⁴/ 2)²

         = 3.14 x (0.675 x 10⁻⁴)²

         = 3.14 x 0.46 x 10⁻⁸

   A    = 1.44 x 10⁻⁸ m²                                         ⇒ 3

Putting the values 2 and 3 into equation 1:

R = 1.59 x 10⁻⁸ x ( 45.7 /  1.44 x 10⁻⁸ )

  =  1.59 x 45.7 / 1.44

  =  72.66 / 1.44

  =  50.46

R = 50.46 ohm

Hence the resistance of the silver wire is 50.46 ohms.