Question

Equivalent resistance of n resistor of is X, when connected in series. Equivalent resistance is Y when connected in parallel. Find the ratio of X and Y. Resistance of each is R.

Answer 1

Answer:

n resistors each of resistance R is provided.

First they are connected in series , that is connected end to end such that current through each resistance remains same , but the potential difference supplied by battery gets divided.

Let equivalent resistance be X

$X = R + R + ..... \: n \: times$

$= > X = nR$

Next they are connected in parallel with each other such that potential difference remains same but current gets divided.

Let equivalent resistance be Y .

$\dfrac{ 1}{Y} = \dfrac{1}{R} + \dfrac{1}{R} + .....n \: times$

$= > \dfrac{ 1}{Y} = \dfrac{n}{R}$

$= > Y = \dfrac{R}{n}$

Required ratio will be :

$\boxed{ \huge{ \sf \: X : Y = {n}^{2} : 1}}$

Answer 2

Solution :

✴ Given :-

▪ Eq. resistance of n resistors which are connected in series = X

▪ Eq. resistance of n resistance which are connected in parallel = Y

▪ Resistance of each resistor = R

✴ To Find :-

▪ Ratio of X and Y

✴ Formula :-

• For series connection

$\bigstar\:\boxed{\sf{\pink{\large{R_{s}=R_1+R_2+.....+\infty}}}}$

• For parallel connection

$\bigstar\:\boxed{\sf{\purple{\large{\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+.....+\infty}}}}$

✴ Calculation :-

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• Eq. resistance of series connection

$\implies\sf\:R_s=R+R+.....n\:times\\ \\ \implies\:\boxed{\sf{\red{X=nR}}}$

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• Eq. resistance of parallel connection

$\implies\sf\:\dfrac{1}{R_p}=\dfrac{1}{R}+\dfrac{1}{R}+.....n\:times\\ \\ \implies\sf\:\dfrac{1}{Y}=\dfrac{n}{R}\\ \\ \implies\:\boxed{\sf{\green{Y=\dfrac{R}{n}}}}$

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• Ratio of X and Y

$\mapsto\sf\:\dfrac{X}{Y}=\dfrac{nR}{\frac{R}{n}}\\ \\ \mapsto\:\boxed{\sf{\orange{\large{X:Y=n^2:1}}}}$