Question

Two resistances are given as \(R_1=(10 ±0.5) Ω\) and \(R_2= (15±0.5) Ω\). The percentage error in the measurement of equivalent resistance when they are connected in parallel is -

• A. 2.33
• B. 4.33
• C. 5.33
• D. 6.33

Answer 1

The Correct Option is B

Parallel Resistor Combination and Error Calculation 

In parallel combination,

\( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \)

Given \( R_1 = 10 \) and \( R_2 = 15 \),

\( \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6} \)

Therefore, \( R_{eq} = 6 \).

Now, for error calculation,

\( \frac{dR_{eq}}{R_{eq}^2} = \frac{dR_1}{R_1^2} + \frac{dR_2}{R_2^2} \)

Given \( dR_1 = 0.5 \) and \( dR_2 = 0.5 \),

\( \frac{dR_{eq}}{36} = \frac{0.5}{100} + \frac{0.5}{225} \)

\( dR_{eq} = 36 \times 0.5 \times \left(\frac{1}{100} + \frac{1}{225}\right) \)

\( dR_{eq} = 36 \times 0.5 \times \left(\frac{9 + 4}{900}\right) = 18 \times \frac{13}{900} = \frac{26}{100} = 0.26 \)

Now, the percentage error is,

\( \frac{dR_{eq}}{R_{eq}} \times 100 = \frac{0.26}{6} \times 100 = \frac{26}{6} = 4.33 \)

Conclusion:

The percentage error is approximately 4.33%.

Answer 2

Equivalent Resistance and Error Calculation 

Step 1: Equivalent Resistance in Parallel

The formula for equivalent resistance (\( R_{eq} \)) when two resistors are connected in parallel is:

\( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \)

Substituting the given values of \( R_1 = 10 \ \Omega \) and \( R_2 = 15 \ \Omega \):

\( \frac{1}{R_{eq}} = \frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} \)

Therefore, \( R_{eq} = 6 \ \Omega \).

Step 2: Error Propagation in Parallel Combination

The formula for error propagation in parallel combination is:

\( \frac{dR_{eq}}{R_{eq}^2} = \frac{dR_1}{R_1^2} + \frac{dR_2}{R_2^2} \)

Given \( dR_1 = 0.5 \ \Omega \) and \( dR_2 = 0.5 \ \Omega \), we can substitute the values:

\( \frac{dR_{eq}}{6^2} = \frac{0.5}{10^2} + \frac{0.5}{15^2} \)

\( \frac{dR_{eq}}{36} = \frac{0.5}{100} + \frac{0.5}{225} \)

\( \frac{dR_{eq}}{36} = 0.5 \left( \frac{1}{100} + \frac{1}{225} \right) = 0.5 \left( \frac{9 + 4}{900} \right) = 0.5 \times \frac{13}{900} = \frac{13}{1800} \)

\( dR_{eq} = 36 \times \frac{13}{1800} = \frac{2 \times 13}{100} = \frac{26}{100} = 0.26 \ \Omega \)

Step 3: Percentage Error

The percentage error is calculated as:

\( \text{Percentage Error} = \frac{dR_{eq}}{R_{eq}} \times 100 \)

Substituting the calculated values:

\( \text{Percentage Error} = \frac{0.26}{6} \times 100 = \frac{26}{6} = 4.33\% \)

Conclusion:

The percentage error in the measurement of equivalent resistance is 4.33% (Option 2).