2021 A/L Physics Past Paper · Question 43 (English Medium)

Answer: (1) Graph (1)

What the question asks

We want: the graph that shows how the ammeter reading $I$ changes as the angle $\theta$ grows.

The cell pushes current into the ring at one of the points $A$ and $B$. It leaves the ring at the other, and goes through the ammeter. The cell and the ammeter are ideal, so they have no resistance of their own. The only resistance in the circuit is the ring.

The plan: see how the ring looks to the cell. Then find the resistance for a few angles and see what happens to the current.

Relevant theory

(1) When the length of a conductor increases its resistance increases.

The current flowing in a conductor is directly proportional to the potential difference between its ends provided temperature and its other physical factors remain constant

the total charge entering a junction of a circuit should be equal to the total charge leaving the junction.

Applying it to this question

Think of a round pond with two places on its shore, $A$ and $B$. There are two paths round the pond between them: a short way round and a long way round. Cars can use both paths at the same time.

The ring is like the pond. The current can go from $B$ to $A$ by the short arc and by the long arc at once. A long arc is harder to get through, because the resistance grows with length.

What each letter means:

$\theta$: the angle at the centre between $A$ and $B$.

$R_0$ ($\Omega$): the resistance of the whole ring.

$R_1$ and $R_2$ ($\Omega$): the resistances of the two arcs between $A$ and $B$.

$R$ ($\Omega$): the resistance of the two arcs together.

$E$ ($\text{V}$): the e.m.f. of the cell.

$I$ ($\text{A}$): the ammeter reading.

Read the figure. The short arc is $\theta$ out of the full circle of $2\pi$, and the long arc is the rest. So the short arc is the fraction $\dfrac{\theta}{2\pi}$ of the ring, and the long arc is the fraction $\dfrac{2\pi - \theta}{2\pi}$.

The two arcs start and end at the same two points, $A$ and $B$. So they have the same voltage across them. They are side by side, in parallel.

Reasoning and calculation

Let us try numbers. Say the whole ring is $R_0 = 12\ \Omega$ and the cell is $E = 12\ \text{V}$. A full turn is $360^\circ$. We try three angles.

For two arcs in parallel with the same voltage $E$, the currents add, so $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$, which gives $R = \dfrac{R_1R_2}{R_1 + R_2}$. Here $R_1 + R_2 = R_0 = 12$.

$1$

$R_1 = 12 \times \dfrac{30}{360}$

$11$

$R_2 = 12 - 1$

$0.92$

$R = \dfrac{1 \times 11}{12}$

$13.1$

$I = \dfrac{12}{0.92}$

$3$

$R_1 = 12 \times \dfrac{90}{360}$

$9$

$R_2 = 12 - 3$

$2.25$

$R = \dfrac{3 \times 9}{12}$

$5.3$

$I = \dfrac{12}{2.25}$

$6$

$R_1 = R_2 = 12 \times \dfrac{180}{360}$

$R = \dfrac{6 \times 6}{12}$

$4$

$I = \dfrac{12}{3}$

So the current is $13.1\ \text{A}$ at $30^\circ$, $5.3\ \text{A}$ at $90^\circ$ and only $4\ \text{A}$ at $180^\circ$. It comes down as $\theta$ grows towards the middle.

After the middle, the two arcs just swap places. At $270^\circ$ the arcs are $9\ \Omega$ and $3\ \Omega$, the same as at $90^\circ$. So the current climbs back up on the other side by the same steps.

As $\theta$ goes to $0$ or to $360^\circ$, one arc shrinks to nothing and $R$ goes to nothing, so the current grows very large. At the middle it is smallest, and it is never zero.

Reading the options

The graph must have: a large current at the left end, a smallest value in the middle that is above zero, and a large current again at the right end. The two sides must look the same.

Only graph (1) does this. It is option (1).

Why the other options fail

The current never reaches zero, because the two arcs are always joined to the cell.

The current is largest when one arc is very short and smallest in the middle. Graph (3) is the wrong way round.

The two sides of the middle are mirror images. A graph that keeps rising cannot be the same on both sides.

The ring is not a plain wire. The two arcs change their length as $B$ moves, so the current changes.

Checks

Does it make sense?

The smallest current, $4\ \text{A}$, is when the resistance is largest, $3\ \Omega$. This is a quarter of the ring resistance: $\dfrac{R_0}{4}$.

The ring is two equal arcs only at $\theta = \pi$. This is the most symmetric place, and the current is least there.

$e * r / (a * b)$

Conclusion

The two arcs of the ring are in parallel, and their combined resistance is $R = \dfrac{R_1R_2}{R_0}$.

It is small when $\theta$ is near $0$ or $2\pi$, and largest, $\dfrac{R_0}{4}$, in the middle. So $I$ is large at both ends, smallest in the middle, and never zero. This is graph (1). The answer is (1).

When a contact moves round a ring, the ring is two arcs in parallel. Try three angles: near the start, the middle, and past the middle.

Ask: where is the resistance largest? There the current is least.

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