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Chapter 12: Problem 1812

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

Short Answer

Expert verified
The minimum possible value of \(n\) is 4. (A)

Step by step solution

01

Recall the formulas for calculating resistance in series and parallel

Let's denote the two resistances as \(R_1\) and \(R_2\). The formula for calculating the total resistance in series (\(S\)) and parallel (\(P\)) are as follows: For series: \(S = R_1 + R_2\) For parallel: \(\displaystyle P = \frac{R_{1} R_{2}}{R_{1} +R_{2}}\) We are given that \(S = nP\) and our task is to find the minimum value of \(n\).
02

Substitute the formulas for series and parallel into the equation

Replace the values of \(S\) and \(P\) with their respective formulas from Step 1 into the equation \(S = nP\): \(R_1 + R_2 = n \cdot \frac {R_1R_2}{R_1+R_2}\)
03

Simplify the equation and eliminate common terms

The equation from Step 2 can be simplified by multiplying both sides by \((R_1+R_2)\): \((R_1 + R_2)^2 = n R_1 R_2\) Now, expand the left side of the equation: \(R_1^2 + 2R_1R_2 + R_2^2 = n R_1 R_2\)
04

Minimize the value of n

To find the minimum value of \(n\), we will look for the minimum possible non-negative difference between the left side and right side of the equation. This occurs when the equation becomes an equality, which means the right term of the left side (\(R_1^2 + R_2^2\)) should be as small as possible. The smallest value for the sum of squares of two positive values (as resistances are always positive) occurs when both \(R_1\) and \(R_2\) are equal. In that case, \(R_1 = R_2\) Now, substitute this condition into the equation from Step 3: \((R_1 + R_1)^2 = n R_1 R_1\) \(4R_1^2 = n R_1^2\)
05

Solve for n

Now divide both sides by \(R_1^2\) to find the minimum possible value of n: \(n = \frac{4R_1^2}{R_1^2}\) \(n = 4\) So, the minimum possible value of \(n\) is 4, which corresponds to option (A).

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Most popular questions from this chapter

Two wires of equal lengths, equal diameters and having resistivities \(\rho_{1}\) and \(\rho_{2}\) are connected in series The equivalent resistivity of the combination is.... (A) \(\left(\rho_{1}+\rho_{2}\right)\) (B) \((1 / 2)\left(\rho_{1}+\rho_{2}\right)\) (C) $\left\\{\left(\rho_{1} \rho_{2}\right) /\left(\rho_{1}+\rho_{2}\right)\right\\}$ (D) \(\left.\sqrt{(} \rho_{1} \rho_{2}\right)\)

An infinite sequence of resistances is shown in the figure. The resultant resistance between \(\mathrm{A}\) and \(\mathrm{B}\) will be, when \(\mathrm{R}_{1}=1 \mathrm{ohm}\) and \(\mathrm{R}_{2}=2 \mathrm{ohm}\) (A) \(3 \Omega\) (B) \(2 \Omega\) (C) \(1 \Omega\) (D) \(1.5 \Omega\)

How many dry cells, each of emf \(1.5 \mathrm{~V}\) and internal resistance $0.5 \Omega\(, much be joined in series with a resistor of \)20 \Omega$ to give a current of \(0.6 \mathrm{~A}\) in the circuit? (A) 2 (B) 8 (C) 10 (D) 12

A parallel combination of three resistors takes a current of \(7.5 \mathrm{~A}\) form a \(30 \mathrm{~V}\) supply, It the two resistors are \(10 \Omega\) and $12 \Omega$ find which is the third one? (A) \(4 \Omega\) (B) \(15 \Omega\) (C) \(12 \Omega\) (D) \(22 \Omega\)

A carbon resistor has a set of coaxial colored rings in the order brown, violet brown and silver. The value of resistance (in ohms) is. (A) \((27 \times 10) \pm 5 \%\) (B) \((27 \times 10)=10 \%\) (C) \((17 \times 10)=5 \%\) (D) \((17 \times 10) \pm 10 \%\)
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