Question

When a resistor of $11\Omega $is connected in series with an electric cell, the current flowing in it is $0.5A$. Instead when a resistor of $5\Omega $is connected to the same electric cell in series, the current increases by $0.4A$. The internal resistance of the cell is.

Answer 1

Hint: The internal resistance of the cell can be found out by using the formula to find the current flowing through a cell and equating it for both cases. We can use this method because in both cases the resistor is connected in series to the cell.

Formula Used:
The current flowing through a cell is given by the following mathematical expression:
$i = \dfrac{E}{{R + r}}$
In this mathematical expression:
E is the electromotive force or EMF of the cell, R is the load resistance and r is the internal resistance of the cell.

Complete step by step answer:
In the first case, when the resistor of resistance $11\Omega $ is connected in series with the electric cell, the current flowing through it is given equal to $0.5A$. Now substituting these values in the mathematical equation for finding the electric current of a cell we have:
$0.5 = \dfrac{E}{{11 + r}}$
Now, if we make E the subject of the formula, we have:
$E = 0.5 \times \left( {11 + r} \right)$
We now take into consideration the second case. In this case, the resistor of resistance $5\Omega$ is connected in series with the electric cell. The current flowing through the cell is increased by $0.4A$. Thus, we can say that the total current flowing through the cell is equal to $0.5 + 0.4 = 0.9A$.
Now, substituting these values in the mathematical formula to find current flowing through the cell, we have:
$0.9 = \dfrac{E}{{5 + r}}$
Again, making E as the subject of the formula, we have:
$E = 0.9 \times (5 + r)$
Now, since we know that in the series combination, the electromotive force of the cell remains the same; we can equate both of the equations of E. Thus, we have:
$5 \times (11 + r) = 9 \times (5 + r)$
Solving this equality to find the unknown internal resistance r, we have:
$
4r = 10 \\
\Rightarrow r = 2.5\Omega \\
 $
Thus, the internal resistance of the electric cell is equal to $2.5\Omega$.

Note: This method can only be used if the load resistance is connected in series in both the cases. It cannot be used in case the load resistance is connected in parallel.