If 'n' identical resistors, each of value 'R' are first connected in series to a battery of emf 'E' and internal resistance 'R', the current drawn is 'I'. Now if they are connected in parallel, then the current drawn is '5 I'. The value of n is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5
In each case the current is the emf divided by the total resistance, external plus internal (Ohm's law for a complete circuit). In series the external resistance is \( nR \); in parallel it is \( R/n \).
\( \displaystyle I = \frac{E}{nR + R} = \frac{E}{(n + 1)R} \)
\( \displaystyle 5I = \frac{E}{R/n + R} = \frac{nE}{(n + 1)R} \)
Dividing the second equation by the first:
\( \displaystyle \frac{5I}{I} = \frac{nE}{(n + 1)R} \times \frac{(n + 1)R}{E} = n \;\Rightarrow\; n = 5 \)