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
The current from a cell is the emf divided by the total resistance, external plus internal resistance.
In series the external resistance is \( nR \):
\( \displaystyle I = \frac{E}{nR + R} = \frac{E}{(n+1)R} \)
In parallel the external resistance is \( R/n \):
\( \displaystyle I' = \frac{E}{\frac{R}{n} + R} = \frac{nE}{(n+1)R} = nI \)
Given \( I' = 20\,I \):
\( \displaystyle nI = 20\,I \;\Rightarrow\; n = 20 \)