Question

The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10$\Omega $ is
$
  {\text{A}}{\text{. 0}}{\text{.2}}\Omega \\
  {\text{B}}{\text{. 0}}{\text{.5}}\Omega \\
  {\text{C}}{\text{. 0}}{\text{.8}}\Omega \\
  {\text{D}}{\text{. 1}}{\text{.0}}\Omega \\
 $

Answer 1

Hint: The current produced by a cell is equal to the voltage of the cell divided by the total resistance of the circuit. We are given the values of the voltage of the cell, current produced by the cell and the external resistance connected to this cell. Using these values in the above expression, we can get the required answer.

Formula used: The electric current produced by a cell is given in terms of its internal resistance and the voltage by the following expression.
$I = \dfrac{E}{{R + r}}$

Complete step by step answer:
We are given an electric cell whose voltage is given as
$E = 2.1V$
The amount of current produced by this cell is given as
$I = 0.2A$
The external resistance connected to this cell is given as
$R = 10\Omega $
Let r be the internal resistance of the cell. We know that the current produced by a cell is given in terms of its internal resistance and the voltage by the following expression.
$I = \dfrac{E}{{R + r}}$
We need to find out the internal resistance r of this cell, so we can modify the above expression in the following way.
$r = \dfrac{E}{I} - R$
We are given the values of the voltage of the cell, current produced by the cell and the external resistance connected to this cell. So, we can now insert these known values in the above expression for the internal resistance of the cell. Doing so, we get
 $r = \dfrac{{2.1}}{{0.2}} - 10 = 10.5 - 10 = 0.5\Omega $
This is the required value of internal resistance.

So, the correct answer is “Option B”.

Note: It should be noted that the value of the internal resistance of a battery is very small. This resistance is caused by the electrolyte and electrodes present within the cell. They cause a small potential drop inside the battery.