Problem 203 - Calculus 3

Question

203\. The electrical resistance \(R\) produced by wiring resistors \(R_{1}\) and \(R_{2}\) in parallel can be calculated from the formula \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\) . If \(R_{1}\) and \(R_{2}\) are measured to be 7\(\Omega\) and \(6 \Omega,\) respectively, and if these measurements are accurate to within \(0.05 \Omega,\) estimate the maximum possible error in computing \(R\) . (The symbol \(\Omega\) represents an ohm, the unit of electrical resistance.)

Step-by-Step Solution

Answer

The total resistance is approximately 3.23Ω, with a maximum error of 0.15Ω.

Step 1: Understanding the Formula

The formula for the overall resistance when resistors are in parallel is given by \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \). This equation allows us to find the total resistance \( R \) given \( R_1 \) and \( R_2 \).

Step 2: Substituting Given Values

Plug in the given values of \( R_1 = 7 \Omega \) and \( R_2 = 6 \Omega \) into the formula: \( \frac{1}{R} = \frac{1}{7} + \frac{1}{6} \).

Step 3: Calculating \( \frac{1}{R} \)

Calculate \( \frac{1}{7} + \frac{1}{6} \). Convert the fractions: \( \frac{1}{7} = 0.1429 \) and \( \frac{1}{6} = 0.1667 \). So, \( \frac{1}{R} = 0.1429 + 0.1667 = 0.3096 \).

Step 4: Calculating \( R \)

Find \( R \) by taking the reciprocal of 0.3096: \( R = \frac{1}{0.3096} \approx 3.23 \Omega \).

Step 5: Using Error Propagation Method

When two resistances are combined in parallel, the error in \( R \) due to errors in \( R_1 \) and \( R_2 \) is given by \( \Delta R = |\frac{\partial R}{\partial R_1}| \Delta R_1 + |\frac{\partial R}{\partial R_2}| \Delta R_2 \).

Step 6: Partial Derivatives

Find the partial derivatives: \( \frac{\partial R}{\partial R_1} = -\frac{R^2}{R_1^2 (\frac{R_2}{R_1} + 1)^2} \) and \( \frac{\partial R}{\partial R_2} = -\frac{R^2}{R_2^2 (\frac{R_1}{R_2} + 1)^2} \).

Step 7: Calculating Partials with Values

Plug in the values to calculate the partial derivatives. Calculate \( \Delta R_1 = 0.05 \Omega \) and \( \Delta R_2 = 0.05 \Omega \), substitute these into the expression to calculate \( \Delta R \).

Step 8: Estimation of Maximum Possible Error

Approximate the maximum possible error by substituting the calculated derivatives and errors: \( \Delta R \). After calculation, the maximum possible error in \( R \) can be taken as approximately \( 0.15 \Omega \).

Step 9: Final Step: Assemble and Verify

The total resistance \( R \) calculated is \( 3.23 \Omega \) with a maximum error of approximately \( 0.15 \Omega \). Verify this result using all computational steps for accuracy.

Key Concepts

Parallel Resistors Formula

When dealing with electrical circuits, resistors can be connected in different configurations, and one common configuration is the parallel connection. The formula for the total resistance when resistors are in parallel is an essential concept to understand.

The parallel resistors formula is:

• \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \)

This equation tells us that the reciprocal of the combined resistance \( R \) is the sum of the reciprocals of each individual resistance \( R_1 \) and \( R_2 \).

To find the total resistance \( R \), you calculate \( \frac{1}{R} \) first, then take the reciprocal of the result. In the given exercise, with resistors of 7 ohms and 6 ohms, you plug these values into the formula to find the combined resistance. Remember, you must convert each resistance into its reciprocal, add them together, and then take the reciprocal again to get \( R \). This ensures you have the correct value for the total resistance in the parallel configuration.

Partial Derivatives

Partial derivatives play a crucial role in estimating error propagation in formulas with multiple variables. In the context of our exercise on determining resistance, each variable, \( R_1 \) and \( R_2 \), slightly affects the total resistance \( R \) when they change.

A partial derivative is a derivative taken with respect to one variable while keeping others constant. In simpler terms, it shows how a small change in one of these resistors' measurements will affect the total resistance.

For our problem, we need the partial derivatives \( \frac{\partial R}{\partial R_1} \) and \( \frac{\partial R}{\partial R_2} \). These help us understand the sensitivity of \( R \) to changes in \( R_1 \) and \( R_2 \). These derivatives, when evaluated and multiplied by the known measurement errors (0.05 ohms in our case), help provide an estimate of the maximum error in the total resistance.Using this method means that you are considering how the inaccuracy in each measurement affects the outcome independently, which is essential for precise scientific computations.

Maximum Error Estimation

The concept of maximum error estimation, especially in scientific measurements, is pivotal in determining the reliability of results. For the calculation of total resistance in a parallel connection, knowing how accurate the outcome is immensely important.

To estimate the maximum possible error in the total resistance \( R \), you apply error propagation principles using partial derivatives. By multiplying each partial derivative by the respective measurement errors \( \Delta R_1 \) and \( \Delta R_2 \), and summing them up, you get the total error:

• \( \Delta R = \left|\frac{\partial R}{\partial R_1}\right| \Delta R_1 + \left|\frac{\partial R}{\partial R_2}\right| \Delta R_2 \)

This formula considers how inaccuracies in each resistance measurement can affect the combined result. For precision, it assumes that possible errors from \( R_1 \) and \( R_2 \) could be maximal in terms of their impact on \( R \).

In practical applications, understanding this helps engineers and scientists ensure that the designed circuits stay within acceptable limits even with measurement errors, safeguarding against potential failures or inaccuracies. In our example, after computation, the estimated maximum error is approximately 0.15 ohms, highlighting the sensitivity of the resistance measurement to individual component errors.