Question

Two resistors of resistances $R_1$ = (100 $\pm$ 3) $\Omega$ and $R_2$ = (200 $\pm$ 4) $\Omega$ are connected in parallel. The equivalent resistance of the parallel combination is:

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

Answer 1

The correct option is A (66.7 $\pm$ 1.8) $\Omega$

$\text{Step 1: Calculation of equivalent resistance}$

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$=\frac{1}{100}+\frac{1}{200}$

$\Rightarrow R_{eq}=\frac{200}{3}=66.7\Omega$

$\text{Step 2: Error in equivalent resistance}$

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

$\Rightarrow R_{eq}^{-1}=R_1^{-1}+R_2^{-1}$

Differentiating on both the sides:

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

OR

$\frac{\Delta R}{R_{eq}^2}=\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2}$

$\text{Step 3: Calculations}$

$\Delta R=R_{eq}^2(\frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2})$

Substituting all the values:

$\Delta R_{eq}={\left(\frac{200}{3}\right)}^2(\frac{3}{(100{)}^2}+\frac{4}{(200{)}^2})$

$=\left(\frac{4}{9}\right)(\frac{3}{1}+\frac{4}{4})$ $=\frac{16}{9}$

$\Delta R_{eq}=1.8\Omega$

Therefore, $R_{eq}=66.7\pm 1.8\Omega$

Hence, Option A is correct