Question

A wire of resistance r is cut into n equal parts these parts are then connected in parallel the equivalent resistance of combination will be

Answer 1

Answer:

$\huge\sf {\frac{R}{n^2}}$

Explanation:

Given resistance = R

It is cut into n equal parts. So the new resistance R' would be R/n

(Since, R = ρ × L/A

R' = ρ ×  (L/n)/A

⇒ R' = R/n)

Now, these are connected in parallel. For parallel connection of resistor,

$\sf \frac{1}{R_{eq}} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}......+\frac{1}{R_n}$

Here, R₁ = R₂ = R₃ ... = Rn

So, equivalent resistance

$\sf \frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} .......\frac{1}{R'} \ n \ times \\\\\implies \frac{1}{R_{eq}} = \frac{n}{R'}\\\\\implies R_{eq} = \frac{R'}{n}$

Now, R' = R/n So,

$\sf R_{eq} = \frac{R'}{n}\\\\\implies R_{eq} = \frac{\frac{R}{n}}{n}\\\\\implies R_{eq} = \frac{R}{n} \div n \\\\\implies R_{eq} = \frac{R}{n} \times \frac{1}{n}\\\\\implies R_{eq} = \frac{R}{n^2}$

Answer 2

Answer :-

R / n²

Solution :-

Refer to the Attachment.

Hope it Helps !!!!