Question

A cell of emf E and internal resistance 'r' is connected across a variable resistor 'R'. Plot a graph showing variation of terminal voltage 'V' of the cell versus the current T. Using the plot, show how the emf of the cell and its internal resistance can be determined.

Answer 1

Hint: The internal resistance of a cell can be seen as just another resistor connected in series with the load. The emf of the cell is a constant voltage dependent on the composition of the battery.

Complete step by step answer:
The emf of a cell is the voltage generated by the chemical reactions happening inside the cell. This emf is a constant value independent of the load connected to the cell.

But along with emf, each cell also has some internal resistance. This resistance is due to the resistance of the connectors and the dilution of the chemicals inside and hence, it cannot be removed. So this means whenever we connect a cell, we are actually connecting a cell and a resistor combination.

The voltage across this combination is what we call the terminal potential and denoted by the letter $V$. When we connect some source to a cell, a current starts flowing through it. Due to the presence of internal resistance, some voltage drops across it and hence, the terminal potential is always less than the emf.

The question asks us to find the relation between $V$ and the current $I$ in the circuit. We can see from figure 1 that the terminal potential is the voltage across AB, which is actually the voltage across the load. Also, we see that there are two resistors- one internal resistance and one load resistance that we have to consider while finding the current.
In the circuit, the emf E drops across the load as the terminal potential and also across the internal resistance.

So we see that $E = V + Ir$.Thus $V$ and $I$ are related as :
$V = E - Ir$

Plotting a graph between V and I, we see that the graph would be a straight line which intersects the V axis at E and I axis at $\dfrac{E}{r}$.
We see from the graph that the slope of the graph –
$\dfrac{{\Delta V}}{{\Delta i}} = - r$
So the magnitude of slope of the graph gives us the internal resistance of the cell.

Note:
The internal resistance of a cell is generally a small value. So for small currents, the output voltage would be approximately the emf of the cell. But as current becomes significantly large, The drop across internal resistance becomes significantly large and Terminal voltage deviates significantly from the emf.