Three resistors of $2\Omega$, $3\Omega$, and $6\Omega$ are connected in parallel. The equivalent resistance of the combination is

Solution

1. The formula for parallel combination is $1/R_{eq}=1/R_1+1/R_2+1/R_3$.
2. Substituting the values, $1/R_{eq}=1/2+1/3+1/6$.
3. Finding a common denominator, $$\begin{array}{rl}1/R_{eq}& =(3+2+1)/6\\ & =6/6\end{array}$$.
4. Thus $1/R_{eq}=1$, which means $R_{eq}=1\Omega$.
5. Hence the answer is (A).