Problem 206 - Calculus 3

Question

Electrical power \(P\) is given by \(P=\frac{V^{2}}{R},\) where \(V\) is the voltage and \(R\) is the resistance. Approximate the maximum percentage error in calculating power if 120 \(\mathrm{V}\) is applied to a \(2000-\Omega\) resistor and the possible percent errors in measuring \(V\) and \(R\) are 3\(\%\) and 4\(\%\) respectively.

Step-by-Step Solution

Answer

The maximum percentage error in calculating power is 10\(\%\).

Step 1: Identify the given values

We have a voltage \( V = 120 \text{ V} \), resistance \( R = 2000 \Omega \), and the percentage errors in measuring \( V \) and \( R \) are 3\(\%\) and 4\(\%\) respectively.

Step 2: Use the power formula

The formula for power is given by \( P = \frac{V^2}{R} \). We need to find how the errors in \( V \) and \( R \) affect the error in \( P \).

Step 3: Determine relative error expressions

The relative error of a derived quantity \( P \) that is calculated using division and powers can be expressed using the formula \( \frac{\Delta P}{P} \approx 2\frac{\Delta V}{V} + \frac{\Delta R}{R} \).

Step 4: Substitute percentage errors into the relative error formula

Insert the percentage errors into the relative error expression: \( \frac{\Delta V}{V} = 0.03 \) (3\(\%\)) and \( \frac{\Delta R}{R} = 0.04 \) (4\(\%\)). Thus,\[ \frac{\Delta P}{P} \approx 2(0.03) + 0.04 \].

Step 5: Calculate the maximum percentage error

Compute the maximum percentage error by performing the arithmetic:\[ \frac{\Delta P}{P} \approx 2(0.03) + 0.04 = 0.06 + 0.04 = 0.10 \].This result indicates a maximum percentage error of 10\(\%\).

Key Concepts

Percentage Error

Percentage error is a way to describe how much error exists in a measurement relative to its true value. It's calculated by taking the absolute error, dividing it by the true value, and then multiplying by 100 to convert it into a percentage.
The formula can be expressed as:\[ \text{Percentage Error} = \left( \frac{\text{Absolute Error}}{\text{True Value}} \right) \times 100\% \]Percentage errors are useful because they give a sense of proportion to measurement inaccuracies. Rather than just stating an error in units, percentage errors communicate how significant the inaccuracy is compared to the true value.
For example, a 5e thing as 20 units of error in 400 units, but the percentage gives a clearer context on the accuracy of the measurement. In our problem, percentage errors help in calculating the overall potential error in power calculation.

Voltage Measurement Error

Voltage measurement error refers to the possible inaccuracy or deviation of measured voltage from its actual value. In typical scenarios, some fluctuation is inevitable due to limitations of the measuring equipment.
In the given problem, we're told that the voltage measurement error is 3\%.
This means that the measured voltage might be off by up to 3\% of the actual value. When dealing with electrical power calculations, voltage errors can have a significant effect because the power is directly dependent on the voltage squared.To understand the impact, consider that if a voltage measurement is systematically higher or lower, the calculated power will be disproportionately affected due to the squared term in the power formula \( P = \frac{V^2}{R} \). Hence, accurately assessing and accounting for such measurement errors is crucial for maintaining reliability in electrical calculations.

Resistance Measurement Error

Resistance measurement error represents the potential deviation between a measured resistance value and its factual resistance. The inaccuracy could stem from instrument calibration, environmental conditions, or interpreting the reading itself.
In our given exercise, there is a 4\% resistance measurement error, implying that the resistance value could vary by this percentage from its true value.
Unlike voltage, where the effect is squared, resistance impacts power calculations inversely. Any error in resistance measurement will affect the denominator in the power formula \( P = \frac{V^2}{R} \). This means that a higher measured resistance will lead to a lower calculated power, and vice versa. Thus, understanding the resistance error and its impact on derived quantities like power helps in error analysis of electrical calculations.

Relative Error Expression

The relative error expression is a way to describe how errors in measurements propagate into errors in any calculated quantity. In our situation, we've calculated electrical power \( P \) as a function of voltage \( V \) and resistance \( R \). The formula for relative error of power \( P \) is:\[ \frac{\Delta P}{P} \approx 2\frac{\Delta V}{V} + \frac{\Delta R}{R} \]This equation communicates how the errors in voltage and resistance measurements cumulatively affect the power calculation. The factor of 2 before the voltage error term signifies that voltage has a squared relationship in the original power equation.
By substituting the known percentage errors (3\% for voltage and 4\% for resistance), we determine the maximum percentage error in power to be 10\%. This computation is critical for error analysis, providing insight into the reliability and precision of power calculations and guiding efforts to minimize overall errors in practical applications.