The values of two resistors are R 1 = 6 ± 0.3 k Ω and R 2 = 10 ± 0.2 k Ω . The percentage error in the equivalent resistance when they are connected in parallel is

R 1 = 6 ± 0.3 k Ω , R 2 = 10 ± 0.2 k Ω R p a r a l l e l = R 1 R 2 R 1 + R 2 Let R 1 + R 2 = x ⇒ R P = R 1 R 2 x Taking log of both sides l n R P = l n R 1 + ln ⁡ R 2 - ln ⁡ x Differentiating, ∆ R P R P = ∆ R 1 R 1 + ∆ R 2 R 2 + - ∆ x x ∆ x m e a n = 0.3 + 0.2 2 = 0.25 Ω R m e a n = 6 + 10 2 = 8 Ω ∴ x = 6 + 10 2 = 8 Ω ⇒ ∆ x x = 0.25 8 ∴ T o t a l e r r o r = 0.3 6 + 0.2 10 + 0.25 8 = 0.05 + 0.02 + 0.03125 = 0.10125 ∴ ∆ R P R P = 10.125 %

Physics - Error Question with Solution | TestHub

PhysicsErrorMiscellaneousEasy2 minPYQ_2020

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The values of two resistors are $R_{1} = (6 \pm 0.3) k Ω$ and $R_{2} = (10 \pm 0.2) k Ω .$ The percentage error in the equivalent resistance when they are connected in parallel is

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Answer:

Solution:

$R_{1} = (6 \pm 0.3) k Ω, R_{2} = (10 \pm 0.2) k Ω$

$R_{p a r a l l e l} = \frac{R_{1} R_{2}}{(R_{1} + R_{2})}$

Let $(R_{1} + R_{2}) = x$

$\Rightarrow R_{P} = \frac{R_{1} R_{2}}{x}$

Taking log of both sides

$l n R_{P} = l n R_{1} + \ln R_{2} - \ln x$

Differentiating,

$\frac{∆ R_{P}}{R_{P}} = \frac{∆ R_{1}}{R_{1}} + \frac{∆ R_{2}}{R_{2}} + (- \frac{∆ x}{x})$

$∆ x_{m e a n} = \frac{0.3 + 0.2}{2} = 0.25 Ω$

$R_{m e a n} = \frac{6 + 10}{2} = 8 Ω$

$∴ x = \frac{6 + 10}{2} = 8 Ω$

$\Rightarrow \frac{∆ x}{x} = \frac{0.25}{8}$

$∴ T o t a l e r r o r = \frac{0.3}{6} + \frac{0.2}{10} + \frac{0.25}{8}$

$= 0.05 + 0.02 + 0.03125 = 0.10125$

$∴ \frac{∆ R_{P}}{R_{P}} = 10.125 %$

Stream:NTA_ABHYASSubject:PhysicsTopic:ErrorSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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