/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 There are \(n\) similar resistor... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 10

There are \(n\) similar resistors each of resistance \(R\). The equivalent resistance comes out to be \(x\) when connected in parallel. If they are connected in series, the resistance comes out to be (A) \(x / n^{2}\) (B) \(n^{2} x\) (C) \(x / n\) (D) \(n x\)

Short Answer

Expert verified
None of the given options match the result we obtained, which is \( R_\text{series} = x \). There might be an error in the given options or it may not be a multiple choice question with a single correct answer.

Step by step solution

01

1. Formula for Resistance in Parallel

The formula for calculating the equivalent resistance (R_parallel) when resistors are connected in parallel is: \( \frac{1}{R_\text{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n} \) Since all the resistors are identical, we have: \( \frac{1}{R_\text{parallel}} = n \cdot \frac{1}{R} \) Given that the equivalent resistance when connected in parallel is x, we get: \( \frac{1}{x} = n \cdot \frac{1}{R} \)
02

2. Formula for Resistance in Series

The formula for calculating the equivalent resistance (R_series) when resistors are connected in series is: \( R_\text{series} = R_1 + R_2 + \cdots + R_n \) Since the resistors are identical, we have: \( R_\text{series} = nR \)
03

3. Substitute and Find the Relation

Now we substitute the value of R from the equation for resistors in parallel into the equation for resistors in series: \( R = \frac{x}{n} \) Therefore: \( R_\text{series} = n \cdot \frac{x}{n} = x \) Comparing this with the given expressions: (A) \( x / n^{2} \) (B) \( n^{2} x \) (C) \( x / n \) (D) \( n x \) We find that none of the options match the result we obtained, which is \( R_\text{series} = x \). So, either there is an error in the given options or it is not a multiple choice type question with a single correct answer.

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Most popular questions from this chapter

A wire \(l=8 \mathrm{~m}\) long of uniform cross-sectional area \(A=8 \mathrm{~mm}^{2}\) has a conductance of \(G=2.45 \Omega^{-1}\). The resistivity of material of the wire will be (A) \(2.1 \times 10^{-7} \Omega \mathrm{m}\) (B) \(3.1 \times 10^{-7} \Omega \mathrm{m}\) (C) \(4.1 \times 10^{-7} \Omega \mathrm{m}\) (D) \(5.1 \times 10^{-7} \Omega \mathrm{m}\)

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A potentiometer is a device used for measuring EMF and internal resistance of a cell. It consists of two circuits, one is main circuit in which there is a cell of given emf \(\varepsilon^{\prime}\) and given resistance \(R\) which is connected across a wire of length \(100 \mathrm{~cm}\) and having resistance \(r\) and another circuit having unknown EMF \(\varepsilon\) and galvanometer. For a given potentiometer, if \(\varepsilon^{\prime}=30 \mathrm{~V}, r=1 \Omega\), and resistance \(R\) varies with time \(t\) given by \(R=2 t\). The jockey can move on wire with constant velocity \(10 \mathrm{~cm} / \mathrm{s}\) and switch \(S\) is closed at \(t=0\)If jockey starts moving from \(A\) at \(t=0\) and balancing point found at \(t=1 \mathrm{~s}\) then the value of \(\varepsilon\) is (A) \(1 \mathrm{~V}\) (B) \(2 \mathrm{~V}\) (C) \(3 \mathrm{C}\) (D) \(4 \mathrm{~V}\)

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