Question:

The maximum percentage error in the equivalent resistance of two parallel connected resistors of 100 \( \Omega \) and 900 \( \Omega \), with each having a maximum 5% error, is: \[ \text{(round off to nearest integer value).} \]

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For resistors in parallel, the maximum percentage error in equivalent resistance is {less than or equal to} the individual percentage errors and depends on the relative values of the resistors.
Updated On: Apr 16, 2025
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Solution and Explanation

Let the two resistors be \( R_1 = 100\,\Omega \) and \( R_2 = 900\,\Omega \). The equivalent resistance of resistors in parallel is: \[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{100 \times 900}{100 + 900} = \frac{90000}{1000} = 90\,\Omega \] For small errors, the percentage error in parallel resistance is approximately: \[ \delta R_{eq} \approx \frac{R_2^2}{(R_1 + R_2)^2} \delta R_1 + \frac{R_1^2}{(R_1 + R_2)^2} \delta R_2 \] With \( \delta R_1 = \delta R_2 = 5\% \), we get: \[ \delta R_{eq} = \left( \frac{900^2}{(1000)^2} + \frac{100^2}{(1000)^2} \right) \times 5 = \left( \frac{810000 + 10000}{1000000} \right) \times 5 = \frac{820000}{1000000} \times 5 = 0.82 \times 5 = 4.1\% \] Rounding off to the nearest integer gives: \[ \boxed{5\%} \]
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