First, a set of n equal resistors of $10 \Omega$ each are connected in series to a battery of emf 20 V and internal resistance $10 \Omega$. 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 ______

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Current Electricity

First, a set of n equal resistors of

10Ω

each are connected in series to a battery of emf 20 V 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 ______

Answer: 20

Solution

In series the total resistance is

10 n + 10

, so

I = \frac{20}{10 n + 10}

. In parallel the

n

equal resistors give

\frac{10}{n}

, so

I^{'} = \frac{20}{\frac{10}{n} + 10} = \frac{20 n}{10 n + 10}

. The condition

I^{'} = 20 I

becomes

\frac{20 n}{10 n + 10} = \frac{20 \times 20}{10 n + 10} \Rightarrow 20 n = 400 \Rightarrow n = 20.

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Integer answer

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