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

20

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 $.