# First a set of n equal resistors of R each are connected in series to a battery of emf E and internal resistance R. 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 10 times. What is ‘n’?

$R_{Series}=nR$

$I_{Series}=\frac{E}{R+nR}=\frac{E}{(n+1)R}$

$R_{Parallel}=\frac{R}{}n$

$I_{Parallel}=\frac{E}{R+\frac{R}{}n}=\frac{E}{}R\times \frac{n}{n+1}$

$I_{Parallel}=10\times I_{Series}$

$\frac{E}{}R\times \frac{n}{n+1}=10\times \frac{E}{}R\times \frac{1}{n+1}$

$n=10$

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