Physics, asked by ashoksa001, 3 months ago

5
There
are m resistar each of resistance R
they connected in series & equvalent resistence
is X. Now they are conned in parallel the resistance
isY.What is the ratioof Xand Y.

Answers

Answered by Anonymous
73

'm' identical resistors of magnitude R are connected in :

1. Series Connection : The equivalent resistance is X.

\sf R + R + R + \cdots + R = X \\ \\ \longrightarrow \sf mR = X ---------(1)

2. Parallel Connection : The equivalent resistance is Y.

\sf \dfrac{m}{R} + \dfrac{m}{R} + \cdots +  \dfrac{1}{R}  = Y \\ \\ \longrightarrow \sf \dfrac{m}{R} = Y --------(2)

We have to find ratio of X and Y. From equations (1) and (2),

 \sf  \dfrac{X}{Y}  =  \dfrac{mR}{ \dfrac{m}{R} }  \\   \\  \implies \boxed{ \boxed{ \sf X  : Y =  {R}^{2} :  1}}

Thus, the ratio of X and Y is R² : 1.

Answered by Anonymous
120

@ BeBrainliest is here to help you buddy. :)

Thanks for asking your doubts.

⌬ Keep asking .....

⌬ Keep Answering .....

⌬ Keep Contributing .....

\\

\Large{\underline{\underline{\textsf{\maltese\: {\red{Given :-}}}}}}

✏ There are ‘m’ number of resistors each of resistance ‘R’.

✏ Equivalent resistance of the resistors when connected in series is X.

✏ Equivalent resistance of the resistors when connected in parallel is Y.

\\

\Large{\underline{\underline{\textsf{\maltese\: {\red{To Find :-}}}}}}

\sf \dfrac{X}{Y} \; = \; ?

\\

\Large{\underline{\underline{\textsf{\maltese\: {\red{Recalling the concepts :-}}}}}}

✏ Equivalent resistance of resistors when connected in series.

✰ Rₙₑₜ = R₁ + R₂ + R₃ ..... Rₙ

\\

✏ Equivalent resistance of resistors when connected in parallel.

\sf \dfrac{1}{R_{net}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \; .... \; \dfrac{1}{R_n}

\\

\Large{\underline{\underline{\textsf{\maltese\: {\red{Solution :-}}}}}}

\\

✏ Equivalent resistance of resistors when connected in series.

✰ Rₙₑₜ = R₁ + R₂ + R₃ ..... Rₙ

⇒ X = R + R + R ... ‘m’ times

⇒ X = Rm

\\

✏ Equivalent resistance of resistors when connected in parallel.

\sf \dfrac{1}{R_{net}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} \; .... \; \dfrac{1}{R_n}

⇒ Y = \sf \dfrac{1}{R} +  \dfrac{1}{R} +  \dfrac{1}{R} \; .....\; m \: times

⇒ Y = \sf \dfrac{1}{R} \; * \; m

⇒ Y = \sf \dfrac{m}{R}

\\

\sf \implies \dfrac{X}{Y} \; = \; \dfrac{Rm}{\dfrac{m}{R}}

 \implies \sf \dfrac{X}{Y} = Rm \div \dfrac{m}{R}

  \implies \sf \dfrac{X}{Y} = Rm \times \dfrac{R}{m}

 \implies \sf \dfrac{X}{Y} = R\not{m} \times \dfrac{R}{\not{m}}

  \implies \sf \dfrac{X}{Y} = R \times R

  \implies \sf \dfrac{X}{Y} = R^2

\\

{\underline{ \boxed{\therefore \; \sf X : Y = R^2 : 1}}}


BrainlyPotter176: Awesome!
Anonymous: Thanks mate !
Similar questions