Question

The measured value of resistance of two wires are $R_1=(3\pm 0.09)\Omega$ and $R_2=(6\pm 0.36)\Omega$. If they are connected in parallel then the percentage error in equivalent resistance is

• A. $2$
• B. $8$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$\begin{array}{c}\text{Given}\\ R_1=(3\pm 0.09)\Omega R_2=(6\pm 0.36)\Omega R=\text{Resistance}\\ \text{For parallel:}\\ \frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}-\text{(i)}\text{Req}=\begin{array}{c}\text{Equivalent}\\ \text{resistance}\end{array}\end{array}$

$\text{Differentiating both side}$

$-\frac{d{R}_{eq}}{R_{eq}^2}=\frac{-d{R}_1}{R_1^2}+\frac{C-d{R}_2}{R_2^2}$

$\text{Error is always added}$

$\begin{array}{cc}\frac{d{R}_{c\psi }}{R_{eq}^2}& =\frac{d{R}_1}{R_1^2}+\frac{d{R}_2}{R_2^2}\\ \frac{d{R}_{eq}}{R_{eq}}& =(\frac{d{R}_1}{R_1^2}+\frac{d{R}_2}{R_2^2})Req\end{array}$

$\begin{array}{c}{\text{Calculating Req from eq}}^n\text{(i)}\\ \frac{1}{R_{eq}}=\frac{1}{3}+\frac{1}{6}\\ \frac{1}{R_{eq}}=\frac{6+3}{18}\\ \text{Req}=2\end{array}$

$\begin{array}{cc}\frac{dReq}{Req}\times 100& =(0.01+0.01)2\times 100\%\\ & =\left(0.02\right)\times 2\times 100\%\end{array}$

$=4\%$