When three resistors, each having resistance r, are connected in paral

When three resistors, each having resistance r, are connected in parallel, their resultant resistance is x. If these three resistances are connected in series, the total resistance will be

3x
3rx
9x
3/x
This question was previously asked in
UPSC NDA-2 – 2016
When three resistors, each with resistance $r$, are connected in parallel, their resultant resistance $x$ is given by:
$\frac{1}{x} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{3}{r}$
So, $x = \frac{r}{3}$, which means $r = 3x$.
When these three resistors are connected in series, the total resistance is the sum of individual resistances:
Total resistance $R_{series} = r + r + r = 3r$.
Substituting the expression for $r$ in terms of $x$ ($r=3x$):
$R_{series} = 3 \times (3x) = 9x$.
Resistors in parallel: $\frac{1}{R_{parallel}} = \sum \frac{1}{R_i}$. Resistors in series: $R_{series} = \sum R_i$.
Parallel connections decrease the total resistance, while series connections increase the total resistance. For identical resistors, the parallel resistance is $R/n$ and the series resistance is $nR$, where $R$ is the individual resistance and $n$ is the number of resistors.