Question

If $R_1=600\Omega \pm 1$ and $R_2=600\Omega \pm 2$. Find % error in the calculation of equivalent resistance when these two are connected in parallel.

• A. 0.25%
• B. 0.8%
• C. 1.6 %
• D. 3.2%

Answer 1

The correct option is A 0.25%

$\begin{array}{c}\text{Given:-}{R}_1=600\Omega \pm 1R_2=600\Omega \pm 2\\ \text{for parallel:}\\ \frac{1}{R_{eq}}{v}_1=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{600}+\frac{1}{600}=\frac{2}{600}=\frac{1}{300}-\left(i\right)\\ 8\text{ifferentiating both side}\\ \frac{d{R}_{eq}}{R_{eq}^2}=-\frac{d{R}_1}{R_1^2}+\left(\frac{d{R}_2}{R_2^2}\right)\end{array}$

$\begin{array}{c}\text{Error is always added}\\ \begin{array}{cc}\frac{d{R}_{0}q}{R_{eq}^2}& =\frac{d{R}_1}{R_1^2}+\frac{d{R}_2}{R_2^2}\\ \frac{dRe\gamma }{R_{eq}}& =(\frac{d{R}_1}{R_1^2}+\frac{d{R}_2}{R_2^2})R_{2}V\\ & =(\frac{1}{(600{)}^2}+\frac{2}{(600{)}^2})300\\ & =\frac{(1+2)}{600\times 600}\times 300\end{array}\end{array}$

$\begin{array}{cc}\frac{d{R}_{eq}\times }{\mathrm{Req}}\mathrm{100}\%& =\frac{1}{400}\times 100\%\\ & =0.25\%\end{array}$