Equivalent of battery

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 10*I*. The value of *n* is

Moderate

Solution

Current drawn from a battery when *n* resistors are connected in series is

$I=\frac{E}{nR+R}.......\left(i\right)$

Current drawn from same battery when *n* resistors are connected in parallel is

$10I=\frac{E}{\frac{R}{n}+R}.........\left(ii\right)$

$\text{On dividing eqn. (ii) by (i),}10=\frac{(n+1)R}{\left(\frac{1}{n}+1\right)R}$

After solving the equation, *n=*10

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