Two resistances are given as and . The percentage error in the measurement of equivalent resistance when they are connected in parallel is [2023]

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Solution

Q.
Two resistances are given as $R_{1} = (10 \pm 0.5) Ω$ and $R_{2} = (15 \pm 0.5) Ω$ . The percentage error in the measurement of equivalent resistance when they are connected in parallel is [2023]

1 6.33

2 5.33

3 2.33

4 4.33

Ans.
(4)

$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$

Differentiating both sides, we get

$\frac{△ R}{R^{2}} = \frac{△ R_{1}}{R_{1}^{2}} + \frac{△ R_{2}}{R_{2}^{2}} [R = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = \frac{10 \times 15}{10 + 15} = 6]$

$\Rightarrow \frac{△ R}{R} = (\frac{△ R_{1}}{R_{1}^{2}} + \frac{△ R_{2}}{R_{2}^{2}}) R$

$\Rightarrow (\frac{0.5}{100} + \frac{0.5}{225}) 6 = (\frac{6 \times 0.5}{25}) (\frac{1}{4} + \frac{1}{9}) = \frac{13}{100}$

$\frac{△ R}{R} \times 100 = \frac{13}{3} = 4.33 %$

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