2024 A/L Physics Past Paper · Question 24 (English Medium)

Answer: (3) Graph (3)

What the question asks

This question is about how much heat a metal wire gives off when a current flows through it.

The wire is kept at a constant temperature, so its resistance $R$ does not change. The current can flow either way, so $I$ can be positive or negative.

What we must find: which graph shows how the power $P$ wasted as heat changes with the current $I$.

How we find it: write $P$ using $I$ and the fixed $R$. Then see what the graph of that looks like for positive and negative $I$.

What each letter means:

$I$ ($\text{A}$): current in the wire.

$V$ ($\text{V}$): potential difference across the wire.

$R$ ($\Omega$): resistance of the wire, fixed.

$P$ ($\text{W}$): power turned into heat in the wire.

Relevant theory

Heat is mainly liberated when a current passes through a resistor and this is known as "joule heating".

Passive resistors are those resistors in which the electrical energy flowing through them gets completely converted into heat.

What is happening: as the current flows, the wire gets warm. All the electrical energy becomes heat.

The power is $P=VI$. But a bigger current also needs a bigger voltage, $V=IR$. So both go up together, and $P=I^2R$.

The direction of the current does not matter: $(-I)^2=I^2$, so the wire heats the same either way.

For example, an electric toaster. Turn the current up a little and the wires glow much brighter, because the heat grows with the square of the current.

In words: double the current, and the heat is four times as much.

Required equations

$P=VI$

$V=IR$

$P=I^2R$

Reasoning and calculation

Two regions: $I>0$ and $I<0$, the current one way and the other way.

In both, $P=I^2R$ is positive. A current of $-I$ gives the same $P$ as $+I$, so the graph is the same on both sides of the $P$ axis.

$0$

$P(0)=0^2R$

$P_1$

$P(I)=I^2R$

$4P_1$

$P(2I)=(2I)^2R$

$P(-2I)=(-2I)^2R$

The shape: $P=I^2R$ is a parabola. It touches the origin, is flat there, and curves up more and more steeply on both sides.

The graph we expect: a U-shaped curve with its lowest point at the origin, the same on both sides.

$x*x$

Does it make sense? A wire with no current stays cold, so $P=0$ at $I=0$. Heat is never negative, so the curve never goes below the $I$ axis. Both fit the parabola.

Reading the options

Now match it with the five graphs.

Graph (3) is a U-shaped curve through the origin. It matches.

Graph (1) is two straight lines, (2) bends the wrong way, (4) is highest at zero current and (5) is flat. None of these match.

Why the other options fail

Graph (1) is straight lines. $P$ grows as $I^2$, not as $I$.

Graph (2) bends the wrong way. $I^2$ rises slowly at first.

Graph (4) is highest at $I=0$, where there is no heat at all.

Graph (5) is flat. $P$ changes with $I$.

Conclusion

$P=I^2R$

With $R$ fixed, $P=I^2R$: a parabola through the origin. That is graph (3).

The answer is (3).

At constant temperature $R$ is fixed, so the heat goes as the square of the current.

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