A set of n equal resistances, of value R each, are connected in series to a battery of emf E and internal resistances R . The current drawn is I . Now, the n resistance are connected in parallel to the same battery. Then the current drawn from battery becomes 10 I . The value of n is

Question

Aicharya · Accepted Answer

In the first case, it is given that 'n' resistances are arranged in series, each having a resistance R.

In that case, the net reistance of the external circuit is 'nR'. The internal resistance itself is R. Hence, the resultant resistance of the circuit is $n R + R = (n + 1) R$ .

$∵ I = \frac{ε}{n R + R}$
⇒ $I = \frac{E}{(n + 1) R}$ .... (i)

In the case of parallel combination, these external n resistances are arranged in parallel,thereby their nett external resistance is $\frac{R}{n}$ . This is in series with the internal resistance R.

Hence,
$10 I$ $= \frac{E}{\frac{R}{n} + R} = \frac{n E}{R + n R}$ .... (ii)
From (i) and (ii);
$n \frac{E}{R + n R} = 10 (\frac{E}{n R + R})$
$n = 10$

A set of n equal resistances, of value R each, are connected in series to a battery of emf E and internal…

Solution