Problem 39 - Calculus Early Transcendentals

Question

If \(R\) is the total resistance of three resistors, connected in parallel, with resistances \(R_{1}, R_{2}, R_{3},\) then $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$$ If the resistances are measured in ohms as \(R_{1}=25 \Omega\) , \(R_{2}=40 \Omega,\) and \(R_{3}=50 \Omega,\) with a possible error of 0.5\(\%\) in each case, estimate the maximum error in the calculated value of \(R .\)

Step-by-Step Solution

Answer

The maximum error in the computed resistance is approximately 0.0587 Ω.

Step 1: Understanding the Problem

We need to calculate the total resistance \( R \) for resistors in parallel using the formula \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \) and estimate the maximum error given the possible errors in \( R_1, R_2, \) and \( R_3 \).

Step 2: Calculate the Total Resistance

First, compute the individual contributions: \( \frac{1}{25} = 0.04 \), \( \frac{1}{40} = 0.025 \), \( \frac{1}{50} = 0.02 \). Add these to get \( \frac{1}{R} = 0.04 + 0.025 + 0.02 = 0.085 \). The resistance \( R \) is the reciprocal, \( R = \frac{1}{0.085} = 11.76 \Omega \).

Step 3: Evaluate Possible Errors in Resistances

Each resistor could vary by 0.5% of its value, which means the errors are: \( 0.5\% \times 25 = 0.125 \Omega \), \( 0.5\% \times 40 = 0.2 \Omega \), \( 0.5\% \times 50 = 0.25 \Omega \).

Step 4: Apply the Error Formula for Parallel Resistances

The error in the total resistance \( \Delta R \) in a parallel circuit can be approximated by: \[ \Delta R \approx R^2 \left( \frac{|\Delta R_1|}{R_1^2} + \frac{|\Delta R_2|}{R_2^2} + \frac{|\Delta R_3|}{R_3^2} \right) \].

Step 5: Calculate Error Contribution

Calculate each term: \( \frac{|\Delta R_1|}{R_1^2} = \frac{0.125}{625} \approx 0.0002 \), \( \frac{|\Delta R_2|}{R_2^2} = \frac{0.2}{1600} \approx 0.000125 \), \( \frac{|\Delta R_3|}{R_3^2} = \frac{0.25}{2500} = 0.0001 \). The sum is \( 0.0002 + 0.000125 + 0.0001 = 0.000425 \).

Step 6: Recalculate Total Maximum Error

Substitute back: \( \Delta R = 11.76^2 \times 0.000425 \approx 0.0587 \Omega \). This is the maximum error in the total resistance.

Key Concepts

Total Resistance Calculation

When dealing with resistors connected in parallel, calculating the total resistance can seem tricky at first, but it's quite straightforward once you understand the concept. In parallel circuits, the total resistance is not the sum of the individual resistances. Instead, you use their reciprocals.

This is because each additional resistor provides another path for current to flow, effectively reducing the total resistance. The formula used is:

• \( \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} \)

Here, \( R \) is the total resistance, and \( R_1, R_2, \) and \( R_3 \) are the individual resistances.

By computing the reciprocal of each resistor's resistance and summing them up, you find the reciprocal of the total resistance. Finally, take the reciprocal of this sum to find \( R \). For instance, with resistances of 25, 40, and 50 ohms, the total resistance is approximately 11.76 ohms, as the calculations in the solution demonstrate.

Error Estimation in Resistances

Error estimation plays a crucial role in accurately assessing the reliability of your measurements in parallel circuits. In this context, consider any possible variance or error that can occur with each resistor. Each resistor has a margin of error, such as 0.5% as given in our problem scenario, which affects the overall calculation of the total resistance.

To find the maximum error in the calculated total resistance, you need to determine how each individual error contributes to it.

• Each error is calculated as a percentage of its respective resistance.
• For instance, 0.5% of a 25 ohm resistor results in a possible error of 0.125 ohms.
• Similarly, for the 40 and 50 ohm resistors, the errors are 0.2 and 0.25 ohms, respectively.

These small discrepancies add up and can affect the total resistance calculated.

Resistors in Parallel

Understanding resistors in parallel is key to successfully managing electrical circuits. In this setup, all resistors are connected across the same two points, leading to multiple paths for the current to take.

This arrangement decreases the overall resistance, unlike series circuits where resistance simply adds up. This behavior results in a lower total resistance than any of the individual resistors. Point to remember is:

• The current must divide among the parallel resistors, which is what lowers resistance.
• If one path is disrupted, the other paths allow the circuit to continue functioning, albeit at a reduced current flow.

This characteristic is why parallel circuits are often used in household wiring to ensure continued operation even when one part fails.

Reciprocal Formula in Electrical Circuits

The reciprocal formula is pivotal when working with parallel resistors. It involves taking the reciprocal of each individual resistance value and summing them up. This method:

• Reflects how each additional resistor reduces total resistance.
• Is expressed as: \( \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} \).

Once summed, the reciprocal of this result gives you the total resistance.

It is a crucial concept because it shows that adding resistors in parallel will always lower the overall resistance due to the multiple pathways for current. Let's recap by applying this to our earlier numbers: for resistances of 25, 40, and 50 ohms, their contributions are added as reciprocals (0.04, 0.025, 0.02).

This ends with a final total resistance of approximately 11.76 ohms when reversed, showcasing the utility of understanding and applying the reciprocal relationships in electrical circuits.