Problem 21 - A Complete Resource Book in Physics for JEE Main

Question

A set of \(n\) identical resistors, each of resistance \(R \Omega\), when connected in series, has an effective resistance of \(x\) ohm. When the resistors are connected in parallel, the effective resistance is \(y\) ohm. What is the relation between \(R, x\), and \(y ?\) (A) \(R=\frac{x y}{(x+y)}\) (B) \(R=(y-x)\) (C) \(R=\sqrt{x y}\) (D) \(R=(x+y)\)

Step-by-Step Solution

Answer

The relation between \(R, x\), and \(y\) is given by Option A: \(R=\frac{x y}{(x+y)}\).

Step 1: Analyze the series connection

When n identical resistors are connected in series, their effective resistance (x) can be calculated by adding the individual resistances. Since each resistor has a resistance of R, we can write: \(x= nR\)

Step 2: Analyze the parallel connection

When n identical resistors are connected in parallel, their effective resistance (y) can be calculated using the formula: \(\frac{1}{y} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R}\) (n terms)

Step 3: Simplify the parallel resistance formula

Using the parallel resistance formula from Step 2, we can see that the expression has n terms. We can simplify the expression as follows: \(\frac{1}{y} = n \cdot \frac{1}{R}\)

Step 4: Make R the subject of both equations

In order to find the relationship between R, x, and y, we need to rewrite both formulas in terms of R. For the series connection (Step 1): \(x=nR\) \(R = \frac{x}{n}\) For the parallel connection (Step 3): \(\frac{1}{y} = n \cdot \frac{1}{R}\) \(R = \frac{n}{y}\)

Step 5: Equate both expressions for R and solve

Now that we have both expressions as R, we will equate them and solve for a final relation that involves R, x, and y. \(\frac{x}{n} = \frac{n}{y}\) To solve for R, multiply both sides by n and y: \(R =\frac{x y}{(x+y)}\)

Step 6: Answer

We can now see that the correct relation among the options given in the exercise is Option A: \(R=\frac{x y}{(x+y)}\).