The value 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-

The correct option is B 3.875% [ R_1=6 ±0.3kΩ; R_2=10± 0.2 kΩ; R= net resistance in parallel; 1R=1R_ ...

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Standard XII

Physics

Error and Uncertainty

The value of ...

Question

The value 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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2 %

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3.875 %

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7 %

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Solution

The correct option is B $3.875 %$
$\begin{matrix} R_{1} = 6 \pm 0.3 k Ω R_{2} = 10 \pm 0.2 k Ω R = net resistance in parallel \frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} ⟶ (1) \frac{1}{R} = \frac{1}{6} + \frac{1}{10} \end{matrix}$

$\begin{matrix} R = 3.75 k Ω differentiating (1) \begin{matrix} 2 - \frac{d R}{R^{2}} & = - \frac{d R_{1}}{R_{1}^{2}} - \frac{d R_{2}}{R_{2}^{2}} \Rightarrow \frac{d R}{R} & = R (\frac{d R_{1}}{R_{1}^{2}} + \frac{d R_{2}}{R_{2}^{2}}) = 3.75 (\frac{0.3}{6^{2}} + \frac{0.2}{10^{2}}) = 3.75 (0.0083 + 0.002) = 3.75 (0.0103) = 0.038 \end{matrix} \end{matrix}$

$\begin{matrix} % error & = 0.038 \times 100 % = 3.8 % . \end{matrix}$

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The value 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 3.875%R1=6±0.3kΩR2=10±0.2kΩR= net resistance in parallel 1R=1R1+1R2⟶(1)1R=16+110R=3.75kΩ differentiating (1)2−dRR2=−dR1R21−dR2R22⇒dRR=R(dR1R21+dR2R22)=3.75(0.362+0.2102)=3.75(0.0083+0.002)=3.75(0.0103)=0.038% error =0.038×100%=3.8%.

The value 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?