Question

Given three resistors of $2\Omega ,4\Omega and6\Omega$, what will be the minimum equivalent resistance one can obtain through their various arrangements?

• A. $\frac{12}{11}\Omega$
• B. $\frac{11}{12}\Omega$
• C. $\frac{13}{16}\Omega$
• D. $1\Omega$

Answer 1

The correct option is A $\frac{12}{11}\Omega$

Always remember that when you connect a bunch of resistors in parallel, the effective resistance will be smaller than the smallest resistor.
And when you connect them in series, the effective resistance will obviously be larger than the largest resistor.
So connecting them in parallel is the right plan of action. Let’s see what we get.
$\frac{1}{R}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$
Once again, we appeal to $3^{rd}$ standard mathematics to add fractions.
$\frac{1}{R}=\frac{6+3+2}{12}$
$R=\frac{12}{11}\Omega$