What is the minimum resistance which can be made using five resistors each of $\dfrac{1}{5}{\text{ }}\Omega $?
A) $\dfrac{1}{5}{\text{ }}\Omega $
B) $\dfrac{1}{{25}}{\text{ }}\Omega $
C) $\dfrac{1}{{10}}{\text{ }}\Omega $
D) $25{\text{ }}\Omega $

Hint:Use the concept of the series combination and the parallel combination of the resistors. To get the minimum resistance in this question add the resistors in a parallel combination and apply the equivalent resistor formula.

Complete step by step solution:
According to the question it is given that the value of each resistor is $\dfrac{1}{5}{\text{ }}\Omega $.
In a circuit there are two possible combinations, parallel connection and series connection.
If the two or more resistors are connected in series then the equivalent resistance of the given circuit will be the sum of individual resistances.
Let be two resistors ${R_1}$ and ${R_2}$ are in series, then the equivalent resistance is calculated as,
$ \Rightarrow R = {R_1} + {R_2}$

If the two or more resistors are connected in parallel then the reciprocal of equivalent resistance of the given circuit will be the sum of reciprocal of the individual resistances.
Let be two resistors ${R_1}$ and ${R_2}$ are in parallel, then the equivalent resistance is calculated as,
$ \Rightarrow \dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}}$

Let the five resistors each of $\dfrac{1}{5}{\text{ }}\Omega $ are connected in series and the equivalent resistance is calculated as,
$ \Rightarrow R = \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5} + \dfrac{1}{5}$

Simplify the above equation and get the equivalent resistance as,
$
   \Rightarrow R = \dfrac{5}{5} \\
   \Rightarrow R = 1{\text{ }}\Omega \\
 $

Let the five resistors each of $\dfrac{1}{5}{\text{ }}\Omega $ are connected in parallel and the equivalent resistance is calculated as,
\[ \Rightarrow \dfrac{1}{R} = \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}} + \dfrac{1}{{\left( {\dfrac{1}{5}} \right)}}\]

Simplify the above equation and get the equivalent resistance as,
$
   \Rightarrow \dfrac{1}{R} = 5 + 5 + 5 + 5 + 5 \\
   \Rightarrow R = \dfrac{1}{{25}}{\text{ }}\Omega \\
 $

So, from the above explanation it is concluded that if we connect all the five resistors in parallel then the minimum resistor of $\dfrac{1}{{25}}{\text{ }}\Omega $ can be made.

Hence, the correct answer is $\dfrac{1}{{25}}{\text{ }}\Omega $ and the correct option is B.

Note:If the two or more capacitors are connected in series, then the reciprocal of equivalent capacitance of the given circuit will be the sum of reciprocal of the individual capacitance.
$ \Rightarrow \dfrac{1}{C} = \dfrac{1}{{{C_1}}} + \dfrac{1}{{{C_2}}}$