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

10

11

9

20

Solution

In series grouping equivalent resistance R_series = nR
In parallel grouping equivalent resistance R_parallel = R/n

$I = \frac{E}{n R + R} . . . . . (i) 10 I = \frac{E}{\frac{R}{n} + R} . . . (i i) D i v i d i n g e q . (i i) b y (i), 10 = \frac{(n + 1) R}{(\frac{1}{n} + 1) R} s o l v i n g w e g e t, n = 10$

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