First a set of n equal resistors of resistance R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then these n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is the value of n ?

Question

First a set of n equal resistors of resistance R each are connected in series to a battery of emf E and internal resistance R. A current I is observed to flow. Then these n resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is the value of n ? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

I=(E/R+n R) ; I' =(E/R+(R)n)=10 I (1+n/1+(1)n)=10 ⇒ (1+n/n+1) × n=10 ∴ n=10

I=R+nRE​;I′ =R+nR​E​=10I 1+n1​1+n​=10 ⇒n+11+n​×n=10 ∴n=10

$I = \frac{E}{R + n R}; I^{'}$
$= \frac{E}{R + \frac{R}{n}} = 10 I$
$\frac{1 + n}{1 + \frac{1}{n}} = 10$
$\Rightarrow \frac{1 + n}{n + 1} \times n = 10$
$∴ n = 10$