113th Problem 2016

$\text{Resistance in a parallel connection is given by} \\ \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\\ \implies \frac{1}{R_p} = \frac{1}{100} + \frac{1}{300} + \frac{1}{600} \\ = \frac{1}{100} (\frac{1}{1} + \frac{1}{3} + \frac{1}{6}) = \frac{1}{100} ( frac{3}{2}) \\ \implies \frac{1}{R_p} = \frac{3}{200} \\ \implies R_p = \frac{200}{3} = 66.6666\\ \therefore \text{ The answer is } \fbox{66.67}$

The reciprocal of the equivalent resistance of a set of resistors connected in parallel is equal to the sum of reciprocals of the individual resistances:

$\dfrac{1}{R_T}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...$

So we have,

$\dfrac{1}{R_T}=\dfrac{1}{100}+\dfrac{1}{300}+\dfrac{1}{600}=\dfrac{3}{200}$

$R_T=\dfrac{200}{33}=66.67$

Total resistance in a parallel circuit is the reciprocal of the sum of the reciprocals of the individual resistors.
In this question, the individual resistance is ${\left(\dfrac{1}{100} + \dfrac{1}{300} + \dfrac{1}{600}\right)}^{-1}\\ \\ = \dfrac{600}{9}\\ \\ = \color{#3D99F6}{\boxed{66.67 \Omega}}$