Question

Resistance of $n$ resistors each of $r\Omega$, when connected in parallel give an equivalent resistance of $R\Omega$. If these resistance's were connected in series, the combination would have a resistance in $\Omega$, equal to

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

Answer 1

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

Resistance in parallel,
$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\cdots +\frac{1}{R_n}$
If $R_1=R_2=\cdots =R_n=r$
$\frac{1}{R}=\frac{1}{r}+\frac{1}{r}+\cdots +n$
$\frac{1}{R}=\frac{n}{r}$
$\begin{array}{}r=nR& \text{(i)}\end{array}$
Now resistance in series,
$R=R_1+R_2+\cdots +R_n$
$R=r+r+\cdots +n$
$R=nr$
From Eq.(i) put $r$ value in above equation
$R=n\left(nR\right)=n^{2}R$