Question

The formula \(1 / R=1 / R_{1}+1 / R_{2}\) determines the combined resistance \(R\) when resistors of resistance \(R_{1}\) and \(R_{2}\) are connected in parallel. Suppose that \(R_{1}\) and \(R_{2}\) were measured at 25 and 100 ohms, respectively, with possible errors in each measurement of \(0.5\) ohm. Calculate \(R\) and give an estimate for the maximum error in this value.

Short answer

R = 20 ohms, with a maximum error of 0.34 ohms.

Step-by-step solution

Identify the Given Values

The resistances measured are \( R_1 = 25 \) ohms and \( R_2 = 100 \) ohms. The possible error in each measurement is \( 0.5 \) ohms.

Substitute and Calculate Combined Resistance

Using the formula for combined resistance: \[ \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \]Substitute \( R_1 = 25 \) and \( R_2 = 100 \):\[ \frac{1}{R} = \frac{1}{25} + \frac{1}{100} = \frac{4}{100} + \frac{1}{100} = \frac{5}{100} \]Solving gives \( R = \frac{100}{5} = 20 \text{ ohms} \).

Determine the Maximum Error in the Resistance Measurement

The maximum error can be found using the formula for the derivative: \[ \Delta R = \left| \frac{dR}{dR_1} \Delta R_1 \right| + \left| \frac{dR}{dR_2} \Delta R_2 \right| \]First, calculate \( \frac{dR}{dR_1} \) and \( \frac{dR}{dR_2} \) by implicit differentiation:The reciprocal derivative: \[ \frac{d\left( \frac{1}{R} \right)}{dR_1} = -\frac{1}{R_1^2}, \quad \frac{d\left( \frac{1}{R} \right)}{dR_2} = -\frac{1}{R_2^2} \]Convert it to \( R \):\[ \frac{dR}{dR_1} = \frac{R^2}{R_1^2}, \quad \frac{dR}{dR_2} = \frac{R^2}{R_2^2} \]Substitute \( \Delta R_1 = \Delta R_2 = 0.5\):\[ \Delta R = \left| \frac{20^2}{25^2} \times 0.5 \right| + \left| \frac{20^2}{100^2} \times 0.5 \right| \]Calculate each term:\[ \Delta R = \left| \frac{400}{625} \times 0.5 \right| + \left| \frac{400}{10000} \times 0.5 \right| = 0.32 + 0.02 = 0.34 \text{ ohms} \]

Present the Final Result

The combined resistance \( R \) is 20 ohms, with a maximum error estimate of 0.34 ohms due to measurement errors.

Key concepts

Combined Resistance in Parallel Circuits

When two resistors are connected in parallel, their combined resistance is calculated using a special formula: \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \). This formula results in a combined resistance \( R \) that is always less than the smallest individual resistance. This is because the pathways for the current increase, making it easier for the current to flow, thus reducing overall resistance.

To find the combined resistance, you first take the reciprocal of each resistance, add them up, and then take the reciprocal of that result.

• Example: If \( R_1 = 25 \) ohms and \( R_2 = 100 \) ohms, then the calculation becomes \( \frac{1}{R} = \frac{1}{25} + \frac{1}{100} = \frac{5}{100} \).
• This simplifies to \( R = \frac{100}{5} = 20 \) ohms.

This formula is crucial in electronic circuit design where specific resistances are needed, allowing for flexibility and control over the circuit behavior.

Measurement Error in Resistance Calculations

Measurement errors are common and inevitable in practical environments. In resistance calculations, understanding the potential errors aids in assessing the reliability of the values obtained.

In this scenario, errors of \(0.5\) ohms were noted for both resistors \(R_1\) and \(R_2\). It’s vital to consider these errors to ensure that the measured resistance is as accurate as possible. They can arise due to:

• The quality of the measuring instrument.
• The precision with which the measurement is made.
• Variations in the environmental conditions affecting resistance values.

Understanding measurement errors helps to provide a maximum error estimate, which is crucial especially when high precision is necessary. This understanding allows engineers and technicians to design more reliable circuits.

Derivatives in Calculus and Error Estimation

Derivatives in calculus come in handy when estimating the maximum error in measurements. They allow us to understand how sensitive a function is to changes in its inputs.

For the combined resistance calculation, derivatives are used to determine how small changes in \(R_1\) or \(R_2\) could affect \( R \). By applying implicit differentiation to the combined resistance formula, we can estimate the potential variation.

• The partial derivatives \( \frac{dR}{dR_1} \) and \( \frac{dR}{dR_2} \) represent the rate of change of \( R \) with respect to \( R_1 \) and \( R_2 \), respectively.
• These are calculated as \( \frac{dR}{dR_1} = \frac{R^2}{R_1^2} \) and \( \frac{dR}{dR_2} = \frac{R^2}{R_2^2} \).

This allows us to compute the total error \( \Delta R \) using the formula: \( \Delta R = \left| \frac{dR}{dR_1} \Delta R_1 \right| + \left| \frac{dR}{dR_2} \Delta R_2 \right| \). In this exercise, the total maximum error was estimated to be \( 0.34 \) ohms. Utilizing calculus in this way provides a powerful tool to predict and estimate the effects of small changes in a system, ensuring careful control over electronic design and components.