CUET · PHYSICS · PYQ PAPER 2025
Two wires with resistances \( R_1 \) and \( R_2 \) are stretched n times their original length. After stretching, the two wires are connected in parallel. The equivalent resistance of the combination will be:
- A \( \frac{n^2 R_1 R_2}{R_1 + R_2} \)
- B \( \frac{n R_1 R_2}{R_1 + n R_2} \)
- C \( \frac{R_1 R_2}{n^2 (R_1 + R_2)} \)
- D \( \frac{n R_1 R_2}{n R_1 + R_2} \)
Answer & Solution
Correct Answer
(A) \( \frac{n^2 R_1 R_2}{R_1 + R_2} \)
Step-by-step Solution
Detailed explanation
\(R_1' = n^2 R_1, R_2' = n^2 R_2\) \(R_{eq} = \frac{R_1' R_2'}{R_1' + R_2'} = \frac{(n^2 R_1)(n^2 R_2)}{n^2 R_1 + n^2 R_2}\) \(R_{eq} = \frac{n^4 R_1 R_2}{n^2 (R_1 + R_2)} = \frac{n^2 R_1 R_2}{R_1 + R_2}\)
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