First, a set of ${n}$ equal resistors of $10\; \Omega$ each are connected in series to a battery of emf $20\; {V}$ and internal resistance $10\; \Omega .$ A current $I$ 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 .... .
JEE MAIN 2021, Diffcult
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In series
${R}_{{eq}}={nR}=10 {n}$
${i}_{{s}}=\frac{20}{10+10 {n}}=\frac{2}{1+{n}}$
in parallel
${R}_{{eq}}=\frac{10}{{n}}$
${i}_{{p}}=\frac{20}{\frac{10}{{n}}+10}=\frac{2 {n}}{1+{n}}$
$\frac{{i}_{{p}}}{{i}_{{s}}}=20$
$\frac{\left(\frac{2 {n}}{1+{n}}\right)}{\left(\frac{2}{1+{n}}\right)}=20$
${n}=20$
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