2019 A/L Physics Past Paper · Question 46 (English Medium)

A battery with a variable source: the I against V graph

Consider a battery of electromotive force (emf) $E_0$ and internal resistance $r$ . As shown in the figure, it is connected in series with a resistor $R$ and a variable dc voltage source which can be reversible. When the voltage of the variable source $V_{VR}$ is varied, the graph of $I$ vs $V$ is best represented by

(1) Graph (1)
(2) Graph (2)
(3) Graph (3)
(4) Graph (4)
(5) Graph (5)

Answer and explanation

Answer: (4) Graph (4)

What the question asks

We want:

How the current $I$ changes with the voltage $V$ across the battery.

The variable source is changed.

Then we choose the right graph.

Relevant theory

The voltage across a cell with internal resistance $r$ :

When the current enters at the $+$ end, the cell is charging.

$V = E_0 + Ir$

When the current comes out, the cell is discharging.

$V = E_0 - Ir$

If we give $I$ its sign, both are the same equation, $V = E_0 + Ir$ .

That is, when no current is flowing along an external circuit, the terminal potential difference of a source is equal to its electromotive force.

Required equations

$V = E_0 + Ir$

$I = \frac{V - E_0}{r} = \frac{1}{r}V - \frac{E_0}{r}$

Reasoning and calculation

The axes:

Across: the voltage $V$ across the battery.

Up: the current $I$ .

The direction of the arrow in the figure is positive. It is into the $+$ end of the battery.

The start:

When the source is at $V_{VR} = E_0$ there is no current.

Then $V = E_0$ .

So the graph cuts the $V$ axis at $V = E_0$ , on the positive side.

The two parts:

If $V > E_0$ , the battery is charging and $I > 0$ .

If $V < E_0$ , the battery is discharging and $I < 0$ .

The source can be reversed, so both are possible.

The same equation holds in both.

In both parts:

$I = \dfrac{V - E_0}{r}$

It is one straight line.

There is no break where the two parts meet.

The two ends:

At $I = 0$ , $V = E_0$ .

At $V = 0$ , $I = -\dfrac{E_0}{r}$ . That is a negative $I$ .

The slope is $\dfrac{1}{r}$ . It is positive and it does not change.

So the line rises, with no bend.

It cuts the $I$ axis below zero and the $V$ axis on the positive side.

Reading the options

We need a straight line with a positive slope that cuts the $V$ axis on the positive side.

That is graph (4).

Why the other options fail

It goes through the origin.

It has a negative slope.

It cuts at a negative $V$ .

It shows only the size of $I$ .

Conclusion

$I = \dfrac{V - E_0}{r}$

The slope $\dfrac{1}{r}$ is positive.

$I$ is zero at $V = E_0$ .

This is graph (4).

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