Topic: Internal resistance, emf, cells in series and parallel· Current Electricity

A set of 'n' equal resistors, of value 'R' each, 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. Then the current drawn from ba…

A set of 'n' equal resistors, of value 'R' each, 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. Then the current drawn from battery becomes 10 I. The value of 'n' is

1. 10

2. 20

3. 11

4. 9

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Answer: 1. 10

Series: total resistance $nR+R$, so $I=\dfrac{E}{(n+1)R}$.

Parallel: total resistance $\dfrac Rn+R=\dfrac{(n+1)R}{n}$, so $I'=\dfrac{nE}{(n+1)R}$.

$$\frac{I'}{I}=n=10\Rightarrow n=10$$

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A set of 'n' equal resistors, of value 'R' each, 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. Then the current drawn from ba…

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