Question

There are $n$ exactly identical resistors each having resistance $R$. If the resultant resistance when joined in parallel is $\lambda$, then in series the resistance will come out to be

• A. $\frac{\lambda }{n^2}$
• B. $n^2\lambda$
• C. $\frac{\lambda }{n^3}$
• D. $n^3\lambda$

Answer 1

The correct option is B

$n^2\lambda$

Explanation

Step 1: Given

We have $\frac{1}{R}+\frac{1}{R}+\dots ntimes=\frac{1}{R_{eq}}$ for parallel connection for the same resistors.

We are given that the resultant resistance when joined in parallel,

$\frac{1}{R}+\frac{1}{R}+\dots ntimes=\frac{n}{R}=\frac{1}{\lambda }=\frac{1}{R_{eq}}$

So, $\frac{R}{n}=\lambda$ or, $R=n\lambda$

Step 2: In series connection we have,

$R+R+\dots ntimes=R_{eq}=nR$

Putting, $R=n\lambda$ in above equation we have the series equivalent resistance to be

$R_{eq}=nR=n^2\lambda$

So, the correct option is (B) $n^2\lambda$.