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  4. 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

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Q. 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

NEETNEET 2018Current Electricity

Solution:

$I = \frac{E}{nR + R}$ .....(i)
$ 10\,I = \frac{E}{\frac{R}{n} + R} $ ....(ii)
Dividing (ii) by (i),
$10 = \frac{\left(n+1\right)R}{\left(\frac{1}{n} + 1\right)R} $
After solving the equation, $n = 10$