Question

How many times will the resistance of 'n' identical conductors be increases if the parallel resistance be charged in series one

Answer 1

Topic

Electricity

Given

A parallel connection of 'n' identical resistors is changed into series connection of 'n' identical resistors.

To Find

How many times equivalent resistance changes after performing given steps.

Formula to be used

Equivalent Resistance in Parallel Connection,

$\frac{1}{\sf R_eq}$ = $\frac{1}{\sf R_1}$+$\frac{1}{\sf R_2}$ +............

Equivalent Resistance in Series Connection,

$\sf R_eq$ = $\sf R_1$ + $\sf R_2$ + .............

Solution

It is given that 'n' identical resistance is connected in parallel connection.

So, Equivalent Resistance will be,

$\frac{1}{\sf R_eq}$ = $\frac{1}{\sf R_1}$+$\frac{1}{\sf R_2}$ +............

We need to apply formula for 'n' identical resistors.

$\frac{1}{\sf R_eq}$ = $\frac{1}{R}$+$\frac{1}{R}$ +......

$\frac{1}{\sf R_eq}$ = $\frac{n}{R}$

So,

$\sf R_eq$ = $\frac{R}{n}$

Now,

Equivalent Resistance of resistors when connected in series will be

$\sf R_eq$ = $\sf R_1$ + $\sf R_2$ + .............

We need to apply formula for 'n' identical resistors.

$\sf R_eq$ = R + R + R +.......

$\sf R_eq$ = nR

So, ratio of equivalent resistance in series connection and parallel connection is

$\frac{\sf R_eq \:in\:series}{\sf R_eq \:in\:parallel}$ = $\frac{nR}{\frac{R}{n}}$

$\frac{\sf R_eq \:in\:series}{\sf R_eq \:in\:parallel}$ = n²

Answer

So, after performing given steps, resistance of connection is increased by n².