Question

Given three resistors of $2\Omega ,4\Omega$ and $6\Omega$, what will be the minimum equivalent resistance that can be obtained through their various permutations?

• A. $1.0909\Omega$
• B. $0.9167\Omega$
• C. $0.8125\Omega$
• D. $1\Omega$

Answer 1

The correct option is A $1.0909\Omega$

Given, $R_1=2\Omega$
$R_2=4\Omega$
$R_3=6\Omega$
Let $R$ be the equivalent resistance.
When you connect a group of resistors in parallel, the value of effective resistance will be smaller than the value of smallest resistor of the group.
We know, for parallel combination,
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$
$⟹\frac{1}{R}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$

$⟹\frac{1}{R}=\frac{6+3+2}{12}$

$⟹R=\frac{12}{11}\Omega =1.0909\Omega$ is the minimum equivalent resistance we can get from the three given resistors.