First, a set of n equal resistors of 10   Ω each are connected in series to a battery of E.M.F. 20…

First, a set of n equal resistors of 10 Ω each are connected in series to a battery of E.M.F. 20 V and internal resistance 10 Ω. 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 ___________.

Solution

When the resistances of each 10 Ω ​ are connected in series and parallel then,Rs=10n, and Rp=10n. Given, ip=20is, 

⇒2010n+10=202010n+10⇒20n10+10n=202010n+10

⇒n=20.

Asked in: JEE Main 2021 (27 Aug Shift 1)

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