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For two resistors R1 and R2, connected in parallel, the relative error in their equivalent resistance is (where R1 = (10.0 ± 0.1) Ω and R2 = (20.0 ± 0.4) Ω) (a) 0.08 (b) 0.05 (c) 0.01 (d) 0.04 |
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Answer» Correct option is (c) 0.01 Given, \(R_1 = (10.0 \pm0.1)\Omega\) \(R_2 = (20.0 \pm 0.4 )\Omega\) \(\frac1{R_eq} = \frac1{R_1} + \frac1{R_2}\) .....(1) Differentiate equation (1) \(+ \frac1{{R_e}^2} \Delta R_{eq} = + \frac1{{R_1}^2} \Delta R_1 + \frac1{{R_2}^2} \Delta R_2\) ....(2) Both side multiply Re, so \(\frac{\Delta R_{eq}}{R_{eq}} = R_{eg} \left[\frac{\Delta R_1}{{R_1}}^2 + \frac{\Delta R_2}{{R_2}^2}\right]\) .....(3) ∵ \(R_{eq} = \frac{R_1 + R_2}{R_1 + R_2}= \frac{10 \times 20}{30} = \frac{200}{30}\) \(R_{eq} = \frac{20}3\) Put all value in equation (3) \(\frac{\Delta R_{eq}}{R_{eq}} = \frac{20}3 \left[\frac{0.1}{(10)^2} + \frac{0.4}{(20)^2}\right]\) \(\frac{\Delta R_{eq}}{R_{eq}} = \frac{20}3 \left[\frac{0.1}{100} + \frac{0.4}{400}\right]\) \(\frac{\Delta R_{eq}}{R_{eq}} = \frac{20}3 [0.001 + 0.001]\) \(\frac{\Delta R_{eq}}{R_{eq}} = 0.01\) |
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