A set of n equal resistors, each of resistance R 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. The current drawn from the battery is 20 I. The value of n is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 20
Given:
Each resistor = $ R $
Battery emf = $ E $
Internal resistance = $ R $
Current in series = $ I $
Current in parallel = $ 20I $
Case 1: Series connection
Total resistance, $ R_s = nR + R = R(n + 1) $
$ I = \frac{E}{R_s} = \frac{E}{R(n + 1)} $
Case 2: Parallel connection
Equivalent resistance of n parallel resistors, $ R_p = \frac{R}{n} $
Total resistance, $ R_{total} = R_p + R = \frac{R}{n} + R = R\left(1 + \frac{1}{n}\right) $
Current, $ I' = \frac{E}{R_{total}} = \frac{E}{R\left(1 + \frac{1}{n}\right)} $
Given, $ I' = 20I $
$ \frac{E}{R\left(1 + \frac{1}{n}\right)} = 20 \times \frac{E}{R(n + 1)} $
Cancelling $ \frac{E}{R} $ from both sides:
$ \frac{1}{1 + \frac{1}{n}} = \frac{20}{n + 1} $
$ \frac{n}{n + 1} = \frac{20}{n + 1} $
$ n = 20 $
Therefore, the value of n is $ 20 $.