When the resistances of each 10Ω â€‹ are connected in series and parallel then,Rs=10n, and Rp=n10. Given, ip=20is, ⇒(n10+10)20=20(10n+1020)⇒10+10n20n=20(10n+1020)
⇒n=20.
JEE Main 2021 — Physics Electromagnetism
First, a set of n equal resistors of 10Ω each are connected in series to a battery of E.M.F. 20V and internal resistance 10Ω. 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 ___________.
Held on 27 Aug 2021 · Verified 6 Jul 2026.
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