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# Help me please, A set of 'n' equal resistors, of value 'R' each, areconnected in series to a battery of emf 'E' and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors are connected in paral

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

• Option 1)

20

• Option 2)

11

• Option 3)

10

• Option 4)

9

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As we have learned

Main current / current from each cell -

$i=\frac{nE}{R+nr}$

- wherein

$n-$ identical cells which connected in series

Main current -

$i= \frac{E}{R+\frac{r}{n}}$

-

$I= \frac{E}{R+nR}$    And $10I= \frac{E}{R/n+R}$

$10I= \frac{nE}{R+nR}= nI$

$n= 10$

Option 1)

20

This is incorrect

Option 2)

11

This is incorrect

Option 3)

10

This is correct

Option 4)

9

This is incorrect

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