First, a set of n equal resistors of 10 Ω each are connected in series to a battery of emf 20V and internal resistance 10Ω. A current I is observed to flow. Then, the n resistors are connected in parallel to the same battery. It is observed that the current is increased 20 times, then the value of n is ............... .

Solution

Physics sir · 4 steps

Write both currents with the 10 Ω internal resistance included, then take their ratio.

1. Series circuit. The n resistors in series give 10n Ω. Add the internal resistance, so the current is I = 20/(10n + 10).
2. Parallel circuit. The n resistors in parallel give 10/n Ω. With the internal resistance, the current is I' = 20/(10/n + 10).

   I'/I = (10n + 10)/(10/n + 10) = 10(n+1)/(10(n+1)/n) = n

3. Check. For n = 20: I = 20/210 A and I' = 20/10.5 A. Their ratio is 210/10.5 = 20.
4. So. The current goes up by a factor of n. It goes up 20 times, so n = 20.

The idea to keep. Never drop the internal resistance from the total. Here it equals one external resistor, and that is why the ratio of the two currents comes out to exactly n.

Answer from the NTA final answer key. Solution by PrepWiser.