Question

The maximum resistance made by four resistor of 1/2 ohm is

Answer 1

Dear Student,
When in series, resistances add up like
$$\begin{gathered} R_{eq}=R+R+R+R=4R \\ forR=\frac{1}{2}ohm \\ {R}_{eq}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2ohms \end{gathered}$$

When in parallel, their equivalent resistance would be
$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R} \\ \frac{1}{R_{eq}}=\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=2+2+2+2=8 \\ {R}_{eq}=\frac{1}{8}ohms \end{gathered}$$
So, we can get a maximum resistance of 2 ohms (in this case) only by series combination.

Regards