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Solution:-The resistance equivalent is smallest in case of parallel and largest in series. It is shown belowParallel\begin{array}{l}\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \\\frac{1}{R_{e q}}=\frac{1}{R}+\frac{1}{R} \Rightarrow \frac{1}{R_{e q}}=\frac{2}{R} \quad \Rightarrow \quad R_{\text {eq }}=\frac{R}{2}\end{array}Serils\begin{aligned}R_{\text {eq }} & =R_{1}+R_{2} \\& =R+R \\R_{\text {eq }} & =2 R\end{aligned}\therefore In case of series equivalent resistance is dowote wirile in case of parallel, it is halved.\therefore Equivalent resistance in parallel than in series ...