The answer is given below someone show the calculation pleases?! I am stuck at this question

enter image source here

1 Answer
Feb 15, 2018

See the image and explanation.

Explanation:

Given:
Thin film of nichrome with resistivity rho =1.0 xx 10^-6 Omega m.
Thickness of film = 1xx 10^-6 m
Surface area = 1 mm^2 ------ constant.

To find: resistance between opposite edges of film.

(a)Shape of film = square=> area 1 mm xx 1mm = side x side

=> length of sheet across the edges( terminals). = 1mm

=> l = 1 xx 10^-3 m

Cross section area = thickness of film x width of film:

=>A=1xx 10^-6 m xx 1mm = 1xx 10^-9m
enter image source here

R= rho xx l/A

=>R = 1.0 xx 10^-6 xx (1xx 10^-3) / (1 xx 10^-9 m^2)

R = 1 Omega

b) Shape of rectangle with length 20 times width.
Case 1 : terminals across longer edge:
Surface area = 1 xx 10^-6 m^2
Given : l= 20 xx w

Area= l xx w =>20 w xx w = 1xx 10^-6 m^2

=> 20 w^2 = 1 xx10^-6 m^2

w= sqrt ((1 xx10^-6 )/20)

Now the cross sectional area will be width xx thickness:

A =sqrt ((1 xx10^-6 )/20) xx 1 xx10^-6

and Length across terminals:

l = 20 w = 20 xx sqrt ((1 xx10^-6 )/20)

R= rho xx l/A

=> R = 1 xx10^-6 xx [20 xx sqrt ((1 xx10^-6 )/20)]/[sqrt ((1 xx10^-6 )/20) xx 1 xx10^-6]

=> R = 20 Omega

Case 2 : terminals across shorter edge:
from case 1 : w= sqrt ((1 xx10^-6 )/20)
Now the cross sectional area will be length xx thickness:

A =20w xx 1 xx10^-6

and Length across terminals will be :

l = w

R= rho xx l/A = rho xx w/ (20w xx 1 xx10^-6)

=> R = 1 xx10^-6 xx (sqrt ((1 xx10^-6 )/20))/[ 20 xx sqrt ((1 xx10^-6 )/20) xx 1 xx 10^-6]

=> R = 1/20 = 0.05 Omega