Question

$\begin{array}{c}\text{The resultant resistance of n wires each of resistance}\\ \text{r ohms is R, when they are connected in parallel.}\\ \text{When these n resistances are connected in series,}\\ \text{the resultant resistance will be :-}\end{array}$

• A. $\frac{R}{n}$
• B. $\frac{R}{n^2}$
• C. $nR$
• D. $n^{2}R$

Answer 1

The correct option is D $n^{2}R$

In total there area ${}^{\prime }{n}^{\prime }$ wires, resistance of each wirre $=r\Omega$ (given)

Also given,

resultant resistance of those wire when connected in parallel is $R$

i.e. $r_{eq}\left(inparallel\right)=R$

$R_{eq}$ in parallel :-

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+........+\frac{1}{R_n}$ (in parallel )

$\Rightarrow \frac{1}{R}=\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+.......$ ($n$ times )

[ $\because$ There area ${}^{\prime }{n}^{\prime }$ total wires ]

$\Rightarrow \frac{1}{R}=\frac{n}{r}$

$\Rightarrow r=nR$ ------- $\left(1\right)$

When they area connected in series,

$R_{eq}=R_1+R_2+R_3+.....+R_n$ (in series )

$\Rightarrow R_{eq}=r+r+r+....$ ( $n$ times )

$\Rightarrow R_{eq}$ ( in series )$=nr$

from $\left(1\right)$ , we get,

$\Rightarrow R_{eq}$ (in series ) $=n\left(nR\right)$

$\Rightarrow R_{eq}$ (in series ) $=n^{2}R\Omega$