Question

R₁ = (15 ± 0.5) Ω
R₂ = (10 ± 0.5) Ω
Find % error in equivalent resistance if resistors R₁ and R₂ are connected in parallel ? 

• A. 5%
• B. 4.33%
• C. 3%
• D. 3.33%

Answer 1

The Correct Option is B

To find the percentage error in the equivalent resistance of two resistors connected in parallel, we'll follow these steps:

1. Understand that for resistors connected in parallel, the formula for equivalent resistance \( R \) is given by: \(R = \frac{R_1 \cdot R_2}{R_1 + R_2}\)
2. Given:
   • \( R_1 = 15 \pm 0.5 \, \Omega \)
   • \( R_2 = 10 \pm 0.5 \, \Omega \)
3. Calculate the equivalent resistance \( R \) without considering the errors: \(R = \frac{15 \cdot 10}{15 + 10} = \frac{150}{25} = 6 \, \Omega\)
4. To find the percentage error, use the formula for relative error in parallel resistances: \(\left( \frac{\Delta R}{R} \right) \times 100\% \approx \frac{\left( R_2^2 (\Delta R_1) + R_1^2 (\Delta R_2) \right)}{(R_1 + R_2)^2} \times 100\%\) where \( \Delta R_1 \) and \( \Delta R_2 \) are the absolute errors in \( R_1 \) and \( R_2 \), respectively.
5. Substitute the given values: \(\Delta R_1 = 0.5 \, \Omega,\; \Delta R_2 = 0.5 \, \Omega\) 
   \(R_1 = 15 \, \Omega,\; R_2 = 10 \, \Omega\) 
   \(\left( \frac{\Delta R}{R} \right) \times 100\% = \frac{(10^2 \cdot 0.5) + (15^2 \cdot 0.5)}{(15 + 10)^2} \times 100\%\)
6. Simplify the expression: \(= \frac{(100 \cdot 0.5) + (225 \cdot 0.5)}{625} \times 100\%\) 
   \(= \frac{50 + 112.5}{625} \times 100\%\) 
   \(= \frac{162.5}{625} \times 100\%\) 
   \(= 0.26 \times 100\% = 4.33\%\)
7. Conclusively, the percentage error in the equivalent resistance is 4.33%. Thus, the correct answer is 4.33%.