2021 A/L Physics Past Paper · Question 23 (English Medium)

Answer: (1) $+5\ \text{V}$

What the question asks

We want: what the voltmeter reads.

The voltmeter is ideal. That means it takes no current, so it does not change the circuit. It only tells us how much higher the voltage is at its $+$ end than at its $-$ end.

The plan: the cell pushes current down two separate paths. Find the voltage at the left end of the voltmeter and at the right end. Then take the difference.

Given information

We know:

The cell gives $12\ \text{V}$ and its own resistance can be ignored.

Left path: $50\ \Omega$ and $70\ \Omega$ in a line. Right path: $100\ \Omega$ and $20\ \Omega$ in a line. The top wire goes to the $+$ side of the cell.

What each letter means:

$V$ ($\text{V}$): voltage (potential difference).

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

$R$ ($\Omega$): resistance.

$V_L$ ($\text{V}$): voltage of the left end of the voltmeter, measured up from the bottom wire.

$V_R$ ($\text{V}$): voltage of the right end of the voltmeter, measured up from the bottom wire.

Relevant theory

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

are connected in series with the input voltage V By this the voltage V is divided in the ratio

the resistance of the voltmeter lowers the potential difference between the terminals of the cell

Reasoning and calculation

Which points have the same voltage?

Plain wire has no resistance, so the whole top wire is at the same voltage, and so is the whole bottom wire. Call the bottom wire $0\ \text{V}$. The top wire is then $12\ \text{V}$.

Both paths run from the top wire to the bottom wire, so each path has the full $12\ \text{V}$ across it.

Think of a slide that is $12$ m tall, with two steps on the way down. The $50\ \Omega$ and $70\ \Omega$ are the two steps. The bigger the resistance, the longer the step.

The ideal voltmeter takes no current, so the same current goes through both resistors of a path, like a queue of children going through two gates one after the other.

First the current in each path, then the drop on the lower resistor.

$120$

$R_1 = 50 + 70$

$0.1$

$I_1 = \dfrac{12}{120}$

$7$

$V_L = 0.1 \times 70$

$R_2 = 100 + 20$

$I_2 = \dfrac{12}{120}$

$2$

$V_R = 0.1 \times 20$

The $+$ end of the voltmeter is on the left. So it reads the left voltage minus the right voltage.

$5$

$V_L - V_R = 7 - 2$

Required equations

$V = IR$

$R = R_a + R_b$

$V_L = I_1 \times 70, \quad V_R = I_2 \times 20$

$V_L - V_R$

Reading the options

The reading is $+5\ \text{V}$. This is option (1).

Options (2) and (4) measure the two heights from different wires. Option (3) forgets that the two paths split the voltage differently. Option (5) has the right size but the wrong sign.

Checks

Does it make sense?

Both middle points must lie between $0\ \text{V}$ and $12\ \text{V}$. The left one, $7\ \text{V}$, and the right one, $2\ \text{V}$, do.

The left one is higher, and the $+$ end is on the left, so the reading is positive. If we swapped the two ends of the voltmeter, it would read $-5\ \text{V}$.

$v * a / (a + b) - v * c / (c + d)$

Conclusion

$V = V_L - V_R = 7 - 2 = +5\ \text{V}$

Each path has $120\ \Omega$ across $12\ \text{V}$, so each carries $0.1\ \text{A}$.

The left end of the voltmeter is at $7\ \text{V}$ and the right end is at $2\ \text{V}$. The reading is $7 - 2 = +5\ \text{V}$. The answer is (1).

For a voltmeter between two points, find the voltage of each point from the same wire. Then subtract.

Mark which end is $+$ before you decide the sign.

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