A man has five resistors each of value $\dfrac{1}{5}\Omega $. What is the minimum resistance he can obtain by connecting them?
A. $\dfrac{1}{{10}}\Omega $
B. $\dfrac{1}{5}\Omega $
C. $\dfrac{1}{{50}}\Omega $
D. $\dfrac{1}{{25}}\Omega $

Hint: A man is given five resistors with equal resistance, each has resistance $\dfrac{1}{5}\Omega $. Connect these five resistors both in series and in parallel configurations and find the minimum value of these. When resistors are connected in series, then the equivalent resistance will be the sum of all the resistances. When the resistors are connected in parallel, then the reciprocal of the equivalent resistance is the sum of reciprocals of all the resistances. Use this info to solve the question.

Complete step by step answer:
We are given that a man has five resistors each of value $\dfrac{1}{5}\Omega $.
We have to find the minimum resistance he can obtain by connecting them.
Let the five resistors be ${R_1},{R_2},{R_3},{R_4},{R_5}$
And the resistance of these five resistors is the same, ${R_1} = {R_2} = {R_3} = {R_4} = {R_5} = \dfrac{1}{5}$
When these resistors are connected in series, then the equivalent resistance will be
$
  {R_{eq}} = {R_1} + {R_2} + {R_3} + {R_4} + {R_5} \\
  {R_1} = {R_2} = {R_3} = {R_4} = {R_5} = \dfrac{1}{5} \\
   \implies {R_{eq}} = \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} = 1\Omega \\
 $
When the resistors are connected in series configuration, the equivalent resistance will be $1\Omega $
When these resistors are connected in parallel, then the equivalent resistance will be
$
  \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + \dfrac{1}{{{R_4}}} + \dfrac{1}{{{R_5}}} \\
  {R_1} = {R_2} = {R_3} = {R_4} = {R_5} = \dfrac{1}{5} \\
   \implies \dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} \\
   \implies \dfrac{1}{{{R_{eq}}}} = \dfrac{5}{1} + \dfrac{5}{1} + \dfrac{5}{1} + \dfrac{5}{1} + \dfrac{5}{1} \\
   \implies \dfrac{1}{{{R_{eq}}}} = 25 \\
  \therefore {R_{eq}} = \dfrac{1}{{25}}\Omega \\
 $
When these resistors are connected in parallel, the equivalent resistance will be $\dfrac{1}{{25}}\Omega $
As we can see $\dfrac{1}{{25}}\Omega $ is less than $1\Omega $
So when the resistors are connected in a parallel configuration, the man can obtain minimum resistance which is $\dfrac{1}{{25}}\Omega $.

So, the correct answer is “Option D”.

Note:
Be careful while calculating the equivalent resistance when the resistors are connected together by a configuration. You may get confused by the equivalent resistance and equivalent capacitance because the formula of series equivalent resistance will be the formula of parallel equivalent capacitance and the formula of parallel equivalent resistance will be the formula of series equivalent capacitance.