Question

A set of “$n$” identical resistors, each of resistance $R$ ohm, when connected in series have an effective resistance of $X$ ohm and when the resistors are connected in parallel, the effective resistance is $Y$ ohm. Find the relation between $R, X$ and $Y$.
A. ${X^2} = RY$
B. ${Y^2} = XR$
C. ${R^2} = XY$
D. $R = \dfrac{X}{Y}$

Answer 1

Hint: In order to solve this question we need to understand the combination of resistance. So, a large number of resistors can form two combinations, one is series and other is parallel combination. So in this we first define equivalent resistance for both series and parallel and later establish the required relation.

Complete step by step answer:
Consider resistors ${R_1},{R_2},{R_3}........{R_n}$.Now suppose these are connected in series, then the equivalent resistance is given by,
${R_{eq}} = {R_1} + {R_2} + {R_3}....... + {R_n}$
$\Rightarrow {R_{eq}} = \sum\limits_{i = 1}^n {{R_i}} \to (i)$
Similarly, Consider resistors ${R_1},{R_2},{R_3}........{R_n}$ now suppose these are connected in parallel, then the equivalent resistance is given by,
$\dfrac{1}{{R{'_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + \dfrac{1}{{{R_4}}}....... + \dfrac{1}{{{R_n}}}$
$\Rightarrow \dfrac{1}{{R{'_{eq}}}} = \sum\limits_{i = 1}^n {\dfrac{1}{{{R_i}}}} \to (ii)$

According to question, ${R_1} = {R_2} = {R_3} = .... = {R_n} = R\,\Omega $
First consider them in series. According to the problem, equivalent resistance when all resistors are connected in series is $X{\kern 1pt} \Omega $. So from equation (i), we get equivalent resistance as,
$X = {R_1} + {R_2} + {R_3}........ + {R_n}$
Putting values we get, $X = R + R + R......... + R$
$X = nR{\kern 1pt} \Omega \to (iii)$
Similarly, consider them in parallel combination
According to problem, equivalent resistance when all resistors are connected in parallel is $Y{\kern 1pt} {\kern 1pt} \Omega $

So from equation (ii), we get equivalent resistance as,
$\dfrac{1}{Y} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + ...... + \dfrac{1}{{{R_n}}}$
$\Rightarrow \dfrac{1}{Y} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + ...... + \dfrac{1}{R}$
$\Rightarrow \dfrac{1}{Y} = \dfrac{n}{R}$
$\Rightarrow Y = \dfrac{R}{n}{\kern 1pt} \Omega \to (iv)$
Finding the value of “n” from equation (iii) we get,
$n = \dfrac{X}{R}$
Putting the value of “n” in equation (iv) we get,
$Y = \dfrac{R}{{(\dfrac{X}{R})}}$
$\Rightarrow Y = \dfrac{{{R^2}}}{X}$
$\therefore {R^2} = XY$

Therefore, the correct option is C.

Note: It should be remembered that in series resistance, voltage across each resistor is different but the current across each resistor is the same. In a parallel combination, current across each resistor is different but the voltage difference across each resistor is the same. Also in series, voltage divide between two resistances is in the ratio of their resistance while in parallel, current between two resistances divide in ratio of their inverse resistances.