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 -
- 2.33
- 4.33
- 5.33
- 6.33
The Correct Option is B
Approach Solution - 1
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%.
Approach Solution -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).
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