Question

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

Answer 1

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`

`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`