If 'n' identical resistors, each of value 'R' are first connected in series to a battery of emf 'E' and internal resistance 'R', the current drawn is 'I'. Now if they are connected in parallel, then the current drawn is '5 I'. The value of n is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5
$\text{Series: Total resistance } = nR + R = (n+1)R$
$I = \frac{E}{(n+1)R}$
$\text{Parallel: Equivalent resistance } = \frac{R}{n}$
$\text{Total resistance } = \frac{R}{n} + R = R\left(1 + \frac{1}{n}\right)$
$I' = \frac{E}{R\left(1 + \frac{1}{n}\right)}$
$I' = \frac{En}{R(n+1)}$
$\frac{I'}{I} = \frac{\frac{En}{R(n+1)}}{\frac{E}{R(n+1)}} = n$
$I' = nI$
$nI = 5I$
$n = 5$
Final Answer: $5$