Question

Two resistors of resistances $R_{1}=\left(100 \pm 3\right)\,\Omega$ and $R_{2}=\left(200 \pm 4\right)\,\Omega$ are connected in parallel. The equivalent resistance of the parallel combination is

• A. $\left(66.7 \pm 1.8\right)\,\Omega$
• B. $\left(66.7 \pm 4.0\right)\,\Omega$
• C. $\left(66.7 \pm 3.0\right)\,\Omega$
• D. $\left(66.7 \pm 7.0\right)\,\Omega$

Answer 1

Answer: (A)

Here, $R_{1}=\left(100 \pm 3\right)\,\Omega$;
$R_{2}=\left(200 \pm 4\right)\,\Omega$
The equivalent resistance in parallel combination is
$\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}, \frac{1}{R_{p}}=\frac{1}{100}+\frac{1}{200}=\frac{3}{200}$, $R_{p}=\frac{200}{3}=66.7\,\Omega$
The error in equivalent resistance is given by
$\frac{\Delta R_{p}}{R^{2}_{p}}=\frac{\Delta R_{1}}{R^{2}_{1}}+\frac{\Delta R_{2}}{R_{2}^{2}}$;
$\Delta R_{1}\left(\frac{R_{p}}{R_{1}}\right)^{2}+\Delta R_{2}\left(\frac{R_{p}}{R_{2}}\right)^{2}$
$=3\left(\frac{66.7}{100}\right)^{2}+4\left(\frac{66.7}{200}\right)^{2}=1.8\,\Omega$
Hence, the equivalent resistance along with error in parallel combination is $\left(66.7 \pm1.8\right)\,\Omega$.