Answer
The total equivalent resistance is reduced by connecting a second resistor in parallel
Work Step by Step
Consider Resistor R is connected across voltage V.
equivalent resistance $=R_{eq}$
when R' is connected in parallel to R, $I_{1}$ passes through R & $I_{2}$ passes through R'.
By Kirchoff's theorem,
$I=I_{1}+I_{2}$
or, $\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R'}$
In only one resistor case, $\frac{1}{R_{eq}}=\frac{1}{R}$ only.
But in 2 resistor case, $\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R'}$.
that is $\frac{1}{R'}$ more than the first case. So, in the second case $\frac{1}{R_{eq}}$ increases. So, $R_{eq}$ decreases.
So the total equivalent resistance is reduced by connecting a second resistor in parallel.
