How can three resistors of resistances $2\, \Omega$,$3\, \Omega$,and $6\, \Omega$ be connected to give a total resistance of $1\, \Omega$?

(N/A) To obtain a total resistance of $1\, \Omega$ from resistors of $2\, \Omega$,$3\, \Omega$,and $6\, \Omega$,they must be connected in parallel.
The formula for the equivalent resistance $(R_p)$ of resistors connected in parallel is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
Substituting the given values:
$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$
Taking the least common multiple $(LCM)$ of $2, 3,$ and $6$,which is $6$:
$\frac{1}{R_p} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1\, \Omega^{-1}$
Therefore,$R_p = 1\, \Omega$.
Thus,connecting all three resistors in parallel results in a total resistance of $1\, \Omega$.