Question

We are given n resistors, each of resistance R. The ratio of the maximum to minimum resistance that can be obtained by combining them is:
A. $n^n$
B. $n$
C. $n^2$
D. $log^n$

Answer 1

Hint: The maximum resistance can be obtained by connecting the resistors in series and the minimum resistance can be obtained by connecting the resistors in parallel. Use this to determine the sum of n resistors in series and the inverse sum of n resistances in parallel and divide the two resulting expressions to obtain the appropriate ratio.

Formula used: Effective resistance of two resistors in series: $R_{eff} = R_1 +R_2$
Effective resistance of two resistors in parallel: $\dfrac{1}{R_{eff}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}$

Complete step by step answer:
We know that a maximum effective resistance can be obtained by taking an additive sum of resistances of all the constituent resistors. This is nothing but the resistances connected in series.
The minimum effective resistance, however, is obtained by taking an additive sum of the inverse of resistances of all the constituent resistors. This is when the resistances are connected in parallel.
Therefore, for n resistors each with a resistance R:
$R_{max} = R_{series} = R + R +R + ……n$ times $=nR$.
$R_{min} = R_{parallel} \Rightarrow \dfrac{1}{R_{parallel}}=\dfrac{1}{R} + \dfrac{1}{R} +\dfrac{1}{R} + …n$ times $ = \dfrac{n}{R} \Rightarrow R_{parallel} = \dfrac{R}{n}$
Therefore, the ratio of the maximum to minimum resistance is given as:
$\dfrac{R_{max}}{R_{min}}=\dfrac{R_{series}}{R_{parallel}} = \dfrac{nR}{\dfrac{R}{n}} = \dfrac{nR \times n}{R} = n^2$

So, the correct answer is “Option C”.

Note: Remember that a series circuit is a voltage divider whereas a parallel circuit is a current divider. We add up all the resistances in a series circuit because the resistances are arranged in such a way that the current has only one path to take. However, when resistances are connected in parallel, there are multiple paths for the current to pass through but have the same potential difference across them. This results in a net resistance that is lower than any of the individual resistances.