Current Electricity

JEE Main level, numerical answer, medium

136 of 380

Question 136 of 380

First, a set of nn equal resistors of 10Ω10\Omega each are connected in series to a battery of E.M.F. 20V20V and internal resistance 10Ω.10\Omega . A current II is observed to flow. Then, the nn resistors are connected in parallel to the same battery. It is observed that the current is increased 2020 times, then the value of nn is ___________.

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Answer: 20

When the resistances of each 10Ω10\Omega ​ are connected in series and parallel then,Rs=10n{R}_{s}=10n, and Rp=10n{R}_{p}=\dfrac{10}{n}. Given, ip=20is{i}_{p}=20{i}_{s}, ⇒20(10n+10)=20(2010n+10)⇒20n10+10n=20(2010n+10)\Rightarrow \dfrac{20}{\left(\dfrac{10}{n}+10\right)}=20\left(\dfrac{20}{10n+10}\right)\Rightarrow \dfrac{20n}{10+10n}=20\left(\dfrac{20}{10n+10}\right)
⇒n=20\Rightarrow n=20.

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