Question

If two resistors of resistances have values \(R_1 = (350\pm 3) \Omega\) and \(R_2 = (140\pm 4) \Omega\). The percentage error for the sum and difference of \(R_1\) and \(R_2\) are respectively

• A. \(1.43\%\) , \(3.33\%\)
• B. \(7\%\) , \(1\%\)
• C. \(1\%\) , \(7\%\)
• D. \(3.33\%\) , \(1.43\%\)

Hint:

For a sum or difference, absolute errors add; then divide by the result.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When two measured quantities are added or subtracted, their absolute errors add. The percentage error is then found from the final value.

Step 2: Key Formula or Approach:
\(R = R_1 \pm R_2\) has error \(\Delta R = \Delta R_1 + \Delta R_2\), and percentage error \(= \frac{\Delta R}{R} \times 100\).

Step 3: Detailed Explanation:
\(\Delta R = 3 + 4 = 7\ \Omega\) for both cases.
Sum: \(R = 350 + 140 = 490\ \Omega\), so the percentage error \(= \frac{7}{490} \times 100 = 1.43\%\).
Difference: \(R = 350 - 140 = 210\ \Omega\), so the percentage error \(= \frac{7}{210} \times 100 = 3.33\%\).
The error for the difference is larger because the final value is smaller while the error is the same.

Final Answer:
The percentage errors are \(1.43\%\) and \(3.33\%\), option (A). \[ \boxed{1.43\%,\ 3.33\%} \]