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Chapter 3: Problem 74

The resistance of a metal is given by \(R=\frac{V}{l}\), where \(\mathrm{V}\) is potential difference and \(l\) is the current \(. \ln\) a circuit the potential difference across resistance is \(V=(8 \pm 0.5)\) \(V\) and current in resistance, \(I=(4 \pm 0.2) \AA\). What is the value of resistance with its percentage error? a. \((2 \pm 5.6 \%) \Omega\) b. \((2 \pm 0.7 \%) \Omega\) c. \((2 \pm 35 \%) \Omega\) d. \((2 \pm 11.25 \%) \Omega\)

Short Answer

Expert verified
The resistance is \((2 \pm 11.25\%) \ \Omega\), so the correct option is d.

Step by step solution

01

Calculate the Resistance

The resistance of a metal is given by the formula \( R = \frac{V}{I} \). We are given that the potential difference \( V \) is \( 8 \pm 0.5 \) volts and the current \( I \) is \( 4 \pm 0.2 \) amperes. First, let's calculate the nominal value of the resistance: \[ R = \frac{8}{4} = 2 \ \Omega \].
02

Calculate Absolute Errors

Next, we calculate the absolute errors. For division, the relative errors add. The relative error in \( V \) is \( \frac{0.5}{8} \) and the relative error in \( I \) is \( \frac{0.2}{4} \). Thus, the total relative error in \( R \) is: \[ \frac{0.5}{8} + \frac{0.2}{4} = 0.0625 + 0.05 = 0.1125. \] The absolute error in \( R \) is \( 2 \times 0.1125 = 0.225 \ \Omega \).
03

Calculate Percentage Error

Convert the total relative error into percentage error. Since the relative error is 0.1125, the percentage error is: \[ 0.1125 \times 100\% = 11.25\%. \]
04

Identify the Correct Option

The calculated resistance is \( 2 \ \Omega \) with a percentage error of 11.25%. Therefore, the correct representation of the resistance with its percentage error, from the given options, is \((2 \pm 11.25\%) \ \Omega\). Option d is the correct answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resistance Calculation
The resistance in a circuit is an essential factor that determines how much current can flow through a component. In this case, the formula used is \[ R = \frac{V}{I} \]where \( R \) is the resistance, \( V \) is the potential difference (voltage), and \( I \) is the current.To find the resistance, we divide the voltage by current. For example, given \( V = 8 \) volts and \( I = 4 \) amperes produces a resistance of \[ R = \frac{8}{4} = 2 \ \Omega \].This basic calculation gives a nominal value without considering any uncertainties. However, realizing that these measurements can have errors is crucial, which leads to the need for error propagation.
Relative Error
Relative error is the measure of the uncertainty of a measurement relative to the size of the measurement itself. It is calculated by dividing the absolute error by the measured value.For instance, in the given exercise, the relative error helps us understand how potential variations in voltage \( V \) and current \( I \) affect the resistance.
  • The relative error of \( V \), \( \frac{0.5}{8} \), results from the fact that voltage is \( 8 \pm 0.5 \; V \) and
  • The relative error of \( I \), \( \frac{0.2}{4} \), reflects the current being \( 4 \pm 0.2 \; A \).
Adding these relative errors gives us a comprehensive picture of the potential deviations from the calculated resistance.
Percentage Error Calculation
Percentage error provides a way to express the uncertainty of a measurement as a percentage of the measured value, which makes it easier to interpret and compare with other measurements.To calculate the percentage error, the total relative error obtained from the previous section is multiplied by 100%. In this problem, the total relative error is 0.1125, translating into a percentage error as follows:\[ 0.1125 \times 100\% = 11.25\% \].This result tells us that the error in the measured resistance is 11.25% of its calculated value of \( 2 \ \Omega \), providing an understanding of the measurement's accuracy and reliability. Understanding these errors can help in assessing the potential performance and efficiency of electrical components used in circuits.

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