2020 A/L Physics Past Paper · Question 38 (English Medium)

Answer: (5) $X$: $400\ \text{mA}$, $-12\ \text{V}$; $Y$: $400\ \text{mA}$, $+8\ \text{V}$

What the question asks

We must find: the readings of the ammeter and the centre-zero voltmeter when $K$ is at $X$ and at $Y$.

Relevant theory

Also, by using a potential divider consisting of a single variable resistor with a sliding key, any output voltage varying from zero to V can be obtained.

V =IR, where R is a constant.

Also, if the internal resistance (r) of the source is zero V=E.

"The algebraic sum of the currents flowing across a junction of a circuit is zero".

For this purpose, a resistor of resistance R is connected to in series with the milli-voltmeter.

Applying it to this question

Objects in the figure: a $20\ \text{V}$ battery with no internal resistance; an ideal ammeter on the top rail; two branches between the top rail and the earthed bottom rail — $40\ \Omega$ and $60\ \Omega$ in series (junction $M$), and the whole length of the rheostat $X$–$Y$; a sliding key $K$; an ideal voltmeter with its positive terminal at $K$ and its negative terminal at $M$.

Interactions: the voltmeter's negative-terminal wire bypasses the bottom rail (it loops around) — it is joined only to $M$. No current flows through an ideal voltmeter, so moving $K$ does not change the current in either branch — $K$ is only a point where a potential is "tapped".

Reasoning and calculation

Variables from the figure: top rail $V_T = 20\ \text{V}$ (ideal battery, $V = E$), bottom rail $V_G = 0$ (earth); $X$ is on the bottom rail and $Y$ on the top rail; the rheostat in full is $R = 100\ \Omega$ (its largest value).

Rules: $V = IR$ in each branch; Kirchhoff's first law at the top-rail junction — the ammeter current equals the sum of the two branch currents; the potential at $M$ comes from the divider formula.

$I_1 = \frac{20}{40 + 60} = 0.2\ \text{A}$

$I_2 = \frac{20}{100} = 0.2\ \text{A}$

$I = I_1 + I_2 = 0.4\ \text{A} = 400\ \text{mA}$

$V_M = \left(\frac{60}{40 + 60}\right) \times 20 = 12\ \text{V}$

$V = V_X - V_M = 0 - 12 = -12\ \text{V}$

$V = V_Y - V_M = 20 - 12 = +8\ \text{V}$

Reading the options

Comparing with the options: $400\ \text{mA}$ in both positions; $-12\ \text{V}$ at $X$, $+8\ \text{V}$ at $Y$ — option (5).

Why the other options fail

Takes only one branch's current, and treats the voltmeter as joined to earth.

Takes the voltmeter's negative terminal as earth.

Leaves the rheostat branch's current out of the ammeter.

Reverses the voltmeter's terminal sign.

Checks

Physical plausibility: moving $K$ from $X$ to $Y$, the voltmeter reading changes continuously from $-12\ \text{V}$ to $+8\ \text{V}$; somewhere between (where $K$'s potential is $12\ \text{V}$, $60\%$ of the way from $X$) it is zero — that is exactly why a centre-zero meter is needed ✓. The difference between the two readings, $8 - (-12) = 20\ \text{V}$, equals the battery ✓.

Conclusion

Ammeter $0.2 + 0.2 = 400\ \text{mA}$ (unaffected by $K$); $V_M = 12\ \text{V}$: with $K$ at $X$, $-12\ \text{V}$; at $Y$, $+8\ \text{V}$ — the answer is (5).

Assuming that moving the sliding key changes the resistance of the circuit.

An ideal voltmeter draws no current: the rheostat's whole $100\ \Omega$ is always in the circuit; $K$ is only a point that taps a potential.