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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 10 I. The value of 'n' is
Explanation
The ratio of the two currents simplifies to n, so n = 10.
Detailed Solution
$I = \frac{E}{nR + R}$ ...(i)
$10I = \frac{E}{\frac{R}{n} + R}$ ...(ii)
Dividing (ii) by (i), $10 = \frac{(n+1)R}{\left(\frac{1}{n}+1\right)R} = n$
After solving the equation, $n = 10$