2023 A/L Physics Past Paper · Question 43 (English Medium)

Answer: (1) Graph (1)

What the question asks

We want: the graph of the voltmeter reading $V$ against $R$.

What each letter means:

$V$ ($\text{V}$): the reading of the voltmeter, with its $+$ terminal on $B$.

$V_A$, $V_B$ ($\text{V}$): the potentials of $A$ and $B$, measured from the earth at the bottom of the circuit.

$R$ ($\Omega$): the resistance of the part of the $100\,\Omega$ resistor between the earth and the sliding contact $B$.

Relevant theory

When a certain fraction only of an input voltage is required to be used for another electrical

Also, by using a potential divider consisting of a single variable resistor with a sliding key,

The first quote says that a potential divider gives a fraction of the input voltage. Two resistors $R_1$ and $R_2$ in series divide it in the ratio $R_1 : R_2$, so the output across $R_2$ is $V_0 = \dfrac{R_2}{R_1 + R_2}V_1$. The second quote says that one resistor with a sliding key gives any output voltage from zero up to the input voltage.

For example, the volume knob of a radio is a sliding key on a resistor. Turn it up and a bigger part of the battery voltage reaches the speaker.

Required equations

$V_A = \dfrac{20}{60 + 20} \times 12 = 3$

$V_B = \dfrac{R}{100} \times 12 = 0.12R$

$V = V_B - V_A = 0.12R - 3$

Reasoning and calculation

The potential of $A$:

The $60\,\Omega$ and the $20\,\Omega$ resistors are in series across the $12\,\text{V}$ battery. The point $A$ is between them.

The voltage divides in the ratio of the resistances, so the $20\,\Omega$ resistor takes $\dfrac{20}{80}$ of $12\,\text{V}$, which is $3\,\text{V}$. It does not change when $R$ changes, because the ideal voltmeter takes no current.

$3$

$V_A = \dfrac{20}{60 + 20} \times 12$

The potential of $B$:

The whole $100\,\Omega$ resistor is across the $12\,\text{V}$ battery. The part $R$ at the bottom takes $\dfrac{R}{100}$ of the $12\,\text{V}$.

So $V_B = \dfrac{12R}{100} = 0.12R$ in volts. It grows in a straight line from $0$ at $R = 0$ to $12\,\text{V}$ at $R = 100\,\Omega$.

$0.12R$

$V_B = \dfrac{R}{100} \times 12$

$0.12R - 3$

$V = V_B - V_A = 0.12R - 3$

$-3$

$V = 0 - 3$

$25$

$R = \dfrac{3}{0.12}$

$V = 0.12 \times 50 - 3$

$9$

$V = 0.12 \times 100 - 3$

$12 * r / 100 - 12 * 20 / (60 + 20)$

Reading the options

Now match it with the five graphs.

Graph (1) is a straight line from $-3\,\text{V}$, through zero at $25\,\Omega$, up to $+9\,\text{V}$. It matches.

Graph (2) starts at zero. Graph (3) is a curve. Graph (4) falls. Graph (5) crosses zero at $50\,\Omega$. None of these match.

Why the other options fail

Graph (2) starts at zero, but the point $A$ is at $3\,\text{V}$, so the reading at $R = 0$ is $-3\,\text{V}$.

Graph (3) is a curve, but $V_B - V_A$ is a straight line in $R$.

Graph (4) falls from $+3\,\text{V}$, but the $+$ terminal is on $B$, so the reading rises from $-3\,\text{V}$.

Graph (5) crosses zero at $50\,\Omega$, but the whole $12\,\text{V}$ is across the $100\,\Omega$ resistor, so it crosses at $25\,\Omega$.

Conclusion

$V = 0.12R - 3$

The reading is $V = 0.12R - 3$, a straight line from $-3\,\text{V}$ to $+9\,\text{V}$. That is graph (1).

The answer is (1).

The quotes give the potential divider and the sliding key on one resistor. That the voltmeter reads $V_B - V_A$ with its $+$ terminal on $B$, and that $R$ is the part of the $100\,\Omega$ resistor below the slider, are our reading of the figure.

Teacher's note

Find the potential of each point from the earth, then subtract: the $+$ terminal minus the $-$ terminal. Find two or three points, such as $R = 0$, the zero and $R = 100\,\Omega$, and join them.

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