Question

A cell of emf $E$ and internal resistance $r$ is connected across a resistor of variable resistance $R$. Draw a plot showing the variation of (i) the current ($I$) in the circuit with $R$, and (ii) the terminal potential difference of the cell with $R$.

Hint:

Use \(I = E/(R+r)\) and \(V = IR\). Find the values at R = 0, R = r and very large R, then join them.

Answer 1

Step 1: Understanding the Concept:
A cell of emf \(E\) and internal resistance \(r\) drives a current through an external resistance \(R\). The total resistance is \(R + r\).

Step 2: Key Formulas:
\[ I = \frac{E}{R + r} \]
The terminal potential difference is the voltage across \(R\):
\[ V = IR = \frac{E R}{R + r} \]

Step 3: Plot (i): current against R:
Draw \(R\) on the horizontal axis and \(I\) on the vertical axis.
At \(R = 0\), \(I = E/r\). This is the highest value, and the curve starts on the \(I\) axis at this point.
At \(R = r\), \(I = E/2r\), half the maximum.
As \(R\) grows, \(I\) falls steadily. For very large \(R\), \(I\) approaches zero but never reaches it.
The graph is a smooth falling curve (a rectangular hyperbola shifted to the left by \(r\)). The \(R\) axis is its asymptote.

Step 4: Plot (ii): terminal potential difference against R:
Draw \(R\) on the horizontal axis and \(V\) on the vertical axis.
At \(R = 0\), \(V = 0\), so the curve starts at the origin.
At \(R = r\), \(V = E/2\).
As \(R\) grows, \(V\) rises quickly at first, then flattens. For very large \(R\), \(V\) approaches \(E\) but stays below it.
The graph is a rising curve that bends over and approaches a horizontal line \(V = E\) drawn as a dashed asymptote.

Final Answer:
(i) \(I\) falls from \(E/r\) at \(R = 0\) towards zero as \(R\) increases. (ii) \(V\) rises from zero at \(R = 0\) towards \(E\) as \(R\) increases. \[ \boxed{I = \frac{E}{R + r}, \quad V = \frac{E R}{R + r}} \]