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

Answer: (B)

4.33

Answer 2

In parallel combination, 1/𝑅𝑒𝑞 = 1/𝑅1 + 1/𝑅2
⟹ 1/𝑅𝑒𝑞 = 1/10 + 1/15 = 5/30 = 1/6
Now, for error calculation, 𝑑𝑅𝑒𝑞/𝑅𝑒𝑞^2 = 𝑑𝑅1/𝑅1^2 + 𝑑𝑅2/𝑅2^2
⟹ 𝑑𝑅𝑒𝑞 36 = 0.5 100 + 0.5 225 𝑑𝑅𝑒𝑞 = 36 × 0.5 × ( 13/900) = 18 × 13/900 = 26/100 = 0.26
Now , 𝑑𝑅𝑒𝑞/𝑅𝑒𝑞 × 100 = 0.26/6 × 100 = 26/6 = 4.33