First, a set of n equal resistors of 10 Ω  each are connected in series to a battery of emf 20V 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 increases 20 times, then the value of n is………

0 2 Views | Posted 2 months ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago

    In series Req = nR = 10n

    l s = 2 0 1 0 + 1 0 n = 2 1 + n

    In parallel R e q = 1 0 n

    l ρ l s = 2 0 = 2 n / 1 + n 2 / 1 + n

    n = 20

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