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 1. Now, the ‘$n$’ resistors are connected in parallel to the same battery. The current drawn from the battery becomes 10.1. The value of ‘$n$’ is

Answer 1

Hint: In this question, use the concept of combination of resistor and also its equivalent combination of resistance when ‘$n$’ numbers of resistance are given in the circuit.

Complete step by step solution:
First, we are going to solve the series combination for ‘$n$’ resistors.
Now, we have to use an equivalent resistance formula because there are ‘$n$’ numbers of resistors present in the question.
When ‘$n$’ resistance connected in the series then
${R_{eq}} = {R_1} + {R_2} + {R_3} + .... + {R_n}$
or
${R_{eq}} = n \times R$ (Say equation 1)
Current in the series combination is given by
$1 = \dfrac{E}{{n \times R + R}}$ (Say equation 2)
On the other hand, for the parallel combination we used equivalent resistance formula
$\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + .... + \dfrac{1}{{{R_n}}}$
or
${R_{eq}} = \dfrac{R}{n}$ (Say equation 3)
Current in the parallel combination is given by
$10.1 = \dfrac{E}{{(\dfrac{R}{n} + R)}}$
$10.1 = \dfrac{{nE}}{{R + nR}}$ (Say equation 4)
Here, 10.1 is the current drawn in the parallel combination.
Now, we are going to substitute the value of equation (2) in the equation (4)
So, we get,
$10. \times \dfrac{E}{{nR + R}} = \dfrac{{nE}}{{R + nR}}$
Then,
$n = 10$

Hence, the correct value of ‘$n$ ’ is 10.

Note: While solving such kind of problems always remember the concept of internal resistance of cells and combination in series and parallel, change potential ‘$V$ ’ to $'E'$ as an emf in the Ohm’s law formula.