# First, a set of

Question:

First, a set of $\mathrm{n}$ equal resistors of $10 \Omega$ each are connected in series to a battery of emf $20 \mathrm{~V}$ and internal resistance $10 \Omega$. A current $\mathrm{I}$ is observed to flow. Then, the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of $n$ is

Solution:

In series

$\mathrm{R}_{\mathrm{eq}}=\mathrm{nR}=10 \mathrm{n}$

$\mathrm{i}_{\mathrm{s}}=\frac{20}{10+10 \mathrm{n}}=\frac{2}{1+\mathrm{n}}$

in parallel

$\mathrm{R}_{\mathrm{eq}}=\frac{10}{\mathrm{n}}$

$\mathrm{i}_{\mathrm{p}}=\frac{20}{\frac{10}{\mathrm{n}}+10}=\frac{2 \mathrm{n}}{1+\mathrm{n}}$

$\frac{i_{p}}{i_{s}}=20$

$\frac{\left(\frac{2 n}{1+n}\right)}{\left(\frac{2}{1+n}\right)}=20$

$\mathrm{n}=20$