Question

There are m resistor each of resistance R. First they are all are connected in series and equivalent resistance is x. Now they are connected in parallel and equivalent resistance is y. What is the ratio of x and y

Answer 1

Answer:

m 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

$x = R + R + R + .....m \: times$

$= > x = mR$

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

Let equivalent resistance be y .

$\dfrac{1}{y} = \dfrac{1}{R} + \dfrac{1}{R} + ....m \: times$

$= > \dfrac{1}{y} = \dfrac{m}{R}$

$= > y = \dfrac{R}{m}$

Required ratio will be :

$\boxed{ \red{ \huge{ \bold{x \: : \: y = {m}^{2} : 1 }}}}$

Answer 2

Solution :

✴ Given :-

▪ m resistors of resistance R are first connected in series and then connected in parallel.

▪ Eq. resistance of series connection = x

▪ Eq. resistance of parallel connection = y

✴ To Find :-

▪ Ratio of x and y

✴ Formula :-

✒ Formula of eq. resistance in series connection is given by...

$\boxed{\sf{\pink{\large{R_{eq}=R_1+R_2+R_3+....+\infty}}}}$

✒ Formula of eq. resistance in parallel connection is given by...

$\boxed{\sf{\purple{\large{\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...+\infty}}}}$

✴ Calculation :

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• For series connection

$\implies\sf\:R_{s}=R+R+..... m \:times\\ \\ \implies\:\boxed{\sf{\blue{\large{x=mR}}}}$

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• For parallel connection

$\mapsto\sf\:\dfrac{1}{R_p}=\dfrac{1}{R}+\dfrac{1}{R}+.....m \: times\\ \\ \mapsto\sf\:\dfrac{1}{y}=\dfrac{m}{R}\\ \\ \mapsto\:\boxed{\sf{\green{\large{y=\dfrac{R}{m}}}}}$

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• Ratio of x and y

$\leadsto\sf\:\dfrac{x}{y}=\dfrac{mR}{\frac{R}{m}}\\ \\ \leadsto\:\boxed{\tt{\orange{\large{x : y = m^2 : 1}}}}$

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