Problem 34 - A Complete Resource Book in Physics for JEE Main

Question

The resistance of the series combination of two resistances is \(S\). When they are joined in parallel, the total resistance is \(P\). If \(S=n P\), then the minimum possible value of \(n\) is (A) 4 (B) 3 (C) 2 (D) 1

Step-by-Step Solution

Answer

The minimum possible value of \(n\) is (C) 2.

Step 1: For two resistances \(R_1\) and \(R_2\) in series, the total series resistance is given by: \(S = R_1 + R_2\) For two resistances \(R_1\) and \(R_2\) in parallel, the total parallel resistance is given by: \(\frac{1}{P} = \frac{1}{R_1} + \frac{1}{R_2}\) We are given the relation between \(S\) and \(P\) as: \(S = n P\) #Step 2: Solve for one resistance in terms of the other and the total resistance#

First, let's solve the series resistance equation for \(R_1\): \(R_1 = S - R_2\) Now, let's substitute this expression into the parallel resistance equation: \(\frac{1}{P} = \frac{1}{S - R_2} + \frac{1}{R_2}\) #Step 3: Setup inequality and solve for R_2#

Step 2: Since we are looking for the minimum value for \(n\), we need to find the smallest possible value of \(\frac{S}{P}\). Therefore, we will set an inequality with respect to \(R_2\): \(\frac{1}{P} > \frac{1}{S - R_2} + \frac{1}{R_2}\) Now, cross-multiply and simplify: \(R_2(S - R_2) > (S - R_2)R_2\) Notice that both sides are equal. Since we are looking for the smallest possible value of \(n\), we can take the equality case where: \(R_2(S - R_2) = (S - R_2)R_2\) #Step 4: Solve for n#

Now we have: \(S - 2R_2 = 0\) From the relation between \(S\) and \(P\): \(S = n P\) Substituting the value of \(S\) from the previous relation: \(n P - 2R_2 = 0\) Since the resistances are positive, in this case, the smallest possible value for \(n\) is when \(R_1 = R_2\). We can use the above equation to find the value of \(n\): \(n = 2\) So, the minimum possible value of \(n\) is: (C) 2