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Chapter 14: Problem 175

Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are \(3 \%\) each, then error in the value of resistance of the wire is (A) \(6 \%\) (B) Zero (C) \(1 \%\) (D) \(3 \%\)

Short Answer

Expert verified
The error in the value of resistance of the wire is \(6\%\). The correct answer is (A) \(6\%\).

Step by step solution

01

Write Ohm's Law formula

The relationship between voltage (V), current (I), and resistance (R) is given by Ohm's Law: \[ V = I R \]
02

Rearrange to find the resistance formula

In order to find the resistance, rearrange the formula as: \[R=\frac{V}{I}\]
03

Calculate the percentage error in resistance

The percentage error in a product or a quotient (in our case, a quotient) is equal to the sum of the percentage errors of all the variables. In this scenario, the percentage error in the measurement of current and voltage is \(3 \%\) each. Therefore, percentage error in R = percentage error in V + percentage error in I \[ \% R = \% V + \% I\]
04

Substitute the values and find the percentage error in resistance

Given the percentage error in the measurement of current and voltage, we plug the values into the formula we established in Step 3: \[\% R = 3\% + 3\%\] \[\% R = 6\%\] So, the error in the value of resistance of the wire is \(6\%\).\ The correct answer is: (A) \(6\%\).

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Most popular questions from this chapter

A galvanometer of resistance \(400 \Omega\) can measure a current of \(1 \mathrm{~mA}\). To convert it into a voltmeter of range \(8 \mathrm{~V}\), the required resistance is (A) \(4600 \Omega\) (B) \(5600 \Omega\) (C) \(6600 \Omega\) (D) \(7600 \Omega\)

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