Question

A cell of e.m.f. \(E\) and internal resistance \(r\) is connected with a variable external resistance \(R\) and a voltmeter showing potential drop \(V\) across \(R\). Obtain the relationship between \(V\), \(E\), \(R\) and \(r\).
(b) Draw the shape of the graph showing the variation of terminal voltage \(V\) of the cell as a function of current \(I\) drawn from it. How one can determine the e.m.f. of the cell and its internal resistance from this graph?

Hint:

Use \(E = V + Ir\) with \(V = IR\) for part (a). For part (b), \(V = E - Ir\) is a straight line: the V-axis intercept is \(E\) and the slope magnitude is \(r\).

Answer 1

Step 1: Understanding the Concept:
A real cell has an e.m.f. \(E\) and an internal resistance \(r\). The e.m.f. is the total energy the cell gives to each coulomb. Part of it is lost inside the cell as \(Ir\), and the rest appears as the terminal voltage \(V\) across the external resistance \(R\).

Step 2: Part (a): Apply the loop rule:
Let the current in the circuit be \(I\). The voltmeter reads the potential drop across \(R\), so \(V = IR\).

Going once round the loop (Kirchhoff's voltage law):
\[ E = IR + Ir = V + Ir \]
From \(V = IR\) we get \(I = V/R\). Put this in the loop equation:
\[ E = V + \frac{V}{R}\,r = V\left(1 + \frac{r}{R}\right) = V\,\frac{R + r}{R} \]
Solving for \(V\):
\[ V = \frac{ER}{R + r} \]
The same result also follows from \(I = E/(R + r)\) and \(V = IR\).

Step 3: Part (b): Shape of the V-I graph:
Write the relation in terms of \(I\): \(V = E - Ir\). This is a straight line, \(y = c + mx\), with \(y = V\) and \(x = I\).

So the graph is a straight line sloping downwards. It starts on the \(V\)-axis at the value \(E\) (when \(I = 0\)) and falls steadily as \(I\) increases. It meets the \(I\)-axis at \(I = E/r\), where \(V = 0\).

Diagram in words: draw \(I\) on the horizontal axis and \(V\) on the vertical axis. Draw a straight line with negative slope from a point \(E\) on the \(V\)-axis down to a point \(I_{sc} = E/r\) on the \(I\)-axis.

Step 4: Part (b): Find E and r from the graph:
Put \(I = 0\) in \(V = E - Ir\). Then \(V = E\). So the intercept of the line on the \(V\)-axis is the e.m.f. \(E\) of the cell.

The slope of the line is \(dV/dI = -r\). So the internal resistance is the magnitude of the slope:
\[ r = \left|\frac{\Delta V}{\Delta I}\right| \]
Equivalently, if the line cuts the \(I\)-axis at \(I_{sc}\), then \(r = E/I_{sc}\).

Final Answer:
The relation is given below, and the graph is a straight line of negative slope. Its \(V\)-axis intercept gives \(E\), and the magnitude of its slope gives \(r\). \[ \boxed{V = \frac{ER}{R + r}} \]