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

Feb 15, 2018

See the image and explanation.

#### Explanation:

Given:
Thin film of nichrome with resistivity $\rho = 1.0 \times {10}^{-} 6 \Omega m$.
Thickness of film = $1 \times {10}^{-} 6 m$
Surface area = $1 m {m}^{2}$ ------ constant.

To find: resistance between opposite edges of film.

(a)Shape of film = square=> area $1 m m \times 1 m m$= side x side

$\implies$ length of sheet across the edges( terminals). = 1mm

$\implies l = 1 \times {10}^{-} 3 m$

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

$\implies A = 1 \times {10}^{-} 6 m \times 1 m m = 1 \times {10}^{-} 9 m$

$R = \rho \times \frac{l}{A}$

$\implies R = 1.0 \times {10}^{-} 6 \times \frac{1 \times {10}^{-} 3}{1 \times {10}^{-} 9 {m}^{2}}$

$R = 1 \Omega$

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

$A r e a = l \times w \implies 20 w \times w = 1 \times {10}^{-} 6 {m}^{2}$

$\implies 20 {w}^{2} = 1 \times {10}^{-} 6 {m}^{2}$

$w = \sqrt{\frac{1 \times {10}^{-} 6}{20}}$

Now the cross sectional area will be width xx thickness:

$A = \sqrt{\frac{1 \times {10}^{-} 6}{20}} \times 1 \times {10}^{-} 6$

and Length across terminals:

$l = 20 w = 20 \times \sqrt{\frac{1 \times {10}^{-} 6}{20}}$

$R = \rho \times \frac{l}{A}$

$\implies R = 1 \times {10}^{-} 6 \times \frac{20 \times \sqrt{\frac{1 \times {10}^{-} 6}{20}}}{\sqrt{\frac{1 \times {10}^{-} 6}{20}} \times 1 \times {10}^{-} 6}$

$\implies R = 20 \Omega$

Case 2 : terminals across shorter edge:
from case 1 : $w = \sqrt{\frac{1 \times {10}^{-} 6}{20}}$
Now the cross sectional area will be length xx thickness:

$A = 20 w \times 1 \times {10}^{-} 6$

and Length across terminals will be :

$l = w$

$R = \rho \times \frac{l}{A} = \rho \times \frac{w}{20 w \times 1 \times {10}^{-} 6}$

$\implies R = 1 \times {10}^{-} 6 \times \frac{\sqrt{\frac{1 \times {10}^{-} 6}{20}}}{20 \times \sqrt{\frac{1 \times {10}^{-} 6}{20}} \times 1 \times {10}^{-} 6}$

$\implies R = \frac{1}{20} = 0.05 \Omega$