A set of n equal resistors, of value R each are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same…

Question

A set of n equal resistors, of value R each are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then, the current drawn from battery becomes 10I. The value of n is [NEET 2018]

Accepted Answer

Correct answer: C. (c) When n equal resistors of resistance R are connected in series, then the current drawn is given as I= E nR+r where, nR= equivalent resistance of n resistors in series and r = internal resistance of battery. Given, r=R ⇒I= E nR+R= ER (n+1 ) …(i) Similarly, when n equal resistors are connected in parallel, then the current drawn is given as I '= E Rn+R where, Rn= equivalent resistance of n resistors in parallel. Given, I^ '=10I ⇒10I= E Rn+R= nE (n+1 )R …(ii) Substituting the value of I from Eq. (i) in Eq. (ii), we get 10 ( ER (n+1 ) )= nER (n+1 ) ⇒n=10

$A set of n equal resistors, of value R each are connected in series to a battery of emf E and internal resistance R. The current drawn is I. Now, the n resistors are connected in parallel to the same battery. Then, the current drawn from battery becomes 10I. The value of n is [NEET 2018]$

Solution

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