A set of ' n ' equal resistors, of value ' R ' each, are connected in series to a battery of emf ' E ' and int

Question

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 10I . The value of ' n '

Accepted Answer

Case 1: Series combinationEquivalent external resistance

R_{ser i es} = n R

.Total resistance

= n R + R = (n + 1) R

.Current

I = \frac{E}{( n + 1 ) R} \dots (1)

Case 2: Parallel combinationEquivalent external resistance

R_{p a r a ll e l} = \frac{R}{n}

.Total resistance

= \frac{R}{n} + R = R (\frac{1 + n}{n})

.Current

10 I = \frac{E}{R ( \frac{1 + n}{n} )} = \frac{n E}{( n + 1 ) R} \dots (2)

Divide (2) by (1):

\frac{10 I}{I} = \frac{\frac{n E}{( n + 1 ) R}}{\frac{E}{( n + 1 ) R}}

10 = n

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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 battery becomes 10I. The value of 'n ' is: