The values of two resistors are R1- 6 -0-3k- and R2 - 10 - 0-2k- The percentage error in the equivalent resistance when- they are connected in parallel is

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Transversal of Parallel Lines

The values of...

Question

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

5.125%

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10.125%

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Solution

The correct option is B 10.125%
When the two resistors are connected in parallel, the equivalent resistance will be
$R_{P} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$

Differentiating both sides ,
$\frac{Δ R_{P}}{R_{P}} = \frac{Δ R_{1}}{R_{1}} + \frac{Δ R_{2}}{R_{2}} + \frac{Δ (R_{1} + R_{2})}{R_{1} + R_{2}}$
$\frac{Δ R_{P}}{R_{P}} = \frac{0.3}{6} + \frac{0.2}{10} + \frac{0.3 + 0.2}{16} = 0.05 + 0.02 + 0.03125 = 0.10125$

% error $= \frac{Δ R_{P}}{R_{P}} \times 100 = 0.10125 \times 100 = 10.125 %$

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The values of two resistors are R1=(6±0.3)kΩ and R2=(10±0.2)kΩ. The percentage error in the equivalent resistance when, they are connected in parallel is

The correct option is B 10.125%When the two resistors are connected in parallel, the equivalent resistance will beRP=R1R2R1+R2Differentiating both sides ,ΔRPRP=ΔR1R1+ΔR2R2+Δ(R1+R2)R1+R2ΔRPRP=0.36+0.210+0.3+0.216=0.05+0.02+0.03125=0.10125 % error =ΔRPRP×100=0.10125×100=10.125%

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