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
$\frac{\Delta R_p}{R_p^2}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}$;
$\Delta R_1{\left(\frac{R_p}{R_1}\right)}^2+\Delta R_2{\left(\frac{R_p}{R_2}\right)}^2$
$=3{\left(\frac{66.7}{100}\right)}^2+4{\left(\frac{66.7}{200}\right)}^2=1.8\Omega$
Hence, the equivalent resistance along with error in parallel combination is $\left(66.7\pm 1.8\right)\Omega$.