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. The current drawn from the battery becomes $10I$ . The value of $'n'$ is?
A) $20$
B) $10$
C) $9$
D) $11$

Answer 1

Here we have to apply the formula to calculate the effective resistance for series and parallel combination. When the current drawn is $I$ , all the resistors are connected in series including the internal resistance of the battery. But when the current drawn is $10I$ , $'n'$ resistors are in parallel but the internal resistance is in series. Calculate accordingly.

Complete step-by-step solution: In the given problem, we are given with two cases of series combination and parallel combination. In the first case, we have $'n'$ equal resistors having resistance $'R'$ each are connected in series. Also, the internal resistance of the battery will be in series. Therefore, the total resistance in this series combination ${R_S}$ will be:
${R_s} = nR + R$
$\Rightarrow {R_s} = R\left( {n + 1} \right)$
Thus, using ohm’s law, let the emf voltage in the circuit be $E$ . Therefore, the current $I$ can be written as:
$I = \dfrac{E}{{R\left( {n + 1} \right)}}$ --equation $1$
In the second case, we have $'n'$ equal resistors having resistance $'R'$ each are connected in parallel. Here, the internal resistance of the battery will be in series. Therefore, the total resistance in this series combination ${R_p}$ will be:
${R_p} = \dfrac{R}{n} + R$
Here, $\dfrac{R}{n}$ is the effective resistance for $'n'$ equal resistors having resistance $'R'$ each are connected in parallel. This is obtained using formula for effective resistance in parallel combination as follows:
$\dfrac{1}{{{R_p}}} = \dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} + ... + \dfrac{1}{R}$
$\Rightarrow \dfrac{1}{{{R_p}}} = \dfrac{1}{R}(n)$
$\therefore {R_p} = \dfrac{R}{n}$
And the internal resistance is in series with this parallel combination thus the total resistance will be ${R_p} = \dfrac{R}{n} + R$
The current using ohm’s law will be:
$10I = \dfrac{V}{{\dfrac{R}{n} + R}}$
$\Rightarrow 10I = \dfrac{{nE}}{{R\left( {1 + n} \right)}}$
Taking ratio of this value with equation $1$ we get
$\dfrac{{10I}}{I} = \dfrac{{\dfrac{{nE}}{{R\left( {1 + n} \right)}}}}{{\dfrac{E}{{R\left( {n + 1} \right)}}}}$
$\Rightarrow n = 10$
Therefore, the value of $n = 10$ .
Thus, B is the correct option.

Note: Be careful in calculating the effective resistance. Remember the formula for calculating the effective resistance for $'n'$ equal resistors, of value $'R'$ each. Don’t forget that the internal resistance of the battery will be in series for both the cases.