Circuit Measurements: HSC Physics Practice Paper with Solutions

This original tutorgum practice paper tests circuit measurements in Physics, including ammeter and voltmeter connections, predicted and measured values in fixed-resistance circuits, and the use of uncertainty to evaluate experimental results. It covers the supplied circuit-measurement scope for the Physics 11-12 Syllabus (2025). It is a focused topic paper, not a full-course mock exam or an official or NESA-endorsed paper.

01Exam instructions

Suggested reading time: 5 minutes

Suggested working time: 45 minutes

Total marks: 30

Permitted equipment: Scientific calculator, ruler, and pen or pencil. No notes are required.

These are the conditions for this practice paper, not official NESA examination rules.

Assessed topics

• Selecting and connecting ammeters and voltmeters.
• Predicting current and potential difference in fixed-resistance series and combination circuits.
• Comparing measured values with theoretical predictions.
• Estimating measurement uncertainty and assessing whether results are consistent within uncertainty.
• Evaluating experimental measurements and identifying appropriate follow-up checks.

Exclusions

This paper does not assess AC circuits, non-ohmic conductors, cell internal resistance, electrical power and energy, magnetism, or other unrelated syllabus modules.

Assumptions and conventions

1. Unless stated otherwise, resistor values and supply voltages are exact for calculations, and connecting wires have negligible resistance.
2. All stated uncertainties are absolute uncertainties.
3. When adding or subtracting measured quantities, add their absolute uncertainties.
4. For repeated readings, estimate repeatability uncertainty as half the range. Report the larger of this value and the meter’s stated uncertainty.
5. Show working, include units, and round answers to an appropriate number of significant figures.

Attempt all questions before consulting the worked solutions.

02Questions

Question 1 (4 marks)

A student wants to measure the current through a resistor and the potential difference across that same resistor in a direct-current (DC) circuit.

(a) State how the ammeter and voltmeter must each be connected relative to the resistor. (2 marks)

(b) Explain why an ideal ammeter has very low resistance, while an ideal voltmeter has very high resistance. (2 marks)

Question 2 (5 marks)

An ideal 6.00 V supply is connected to two fixed resistors, \(R_1=100\,\Omega\) and \(R_2=200\,\Omega\), in series.

(a) Calculate the total resistance of the circuit. (1 mark)

(b) Calculate the current through each resistor. (2 marks)

(c) Calculate the potential difference across each resistor. (2 marks)

Question 3 (5 marks)

An ideal 6.00 V supply feeds a combination circuit consisting of a \(100\,\Omega\) resistor, \(R_1\), in series with a parallel branch containing \(R_2=150\,\Omega\) and \(R_3=300\,\Omega\).

For this question, the equivalent resistance of the parallel branch is \(100\,\Omega\). Treat all resistor values as exact.

A student correctly connects the measuring instruments and records the following results.

(a) Calculate the predicted current in the main line and in each parallel branch. (3 marks)

(b) Use the measured values and their uncertainties to determine whether the measured total current is consistent with the sum of the measured branch currents. Show your working. (2 marks)

Question 4 (7 marks)

A student measures the current through a fixed \(150\,\Omega\) resistor connected to a 3.00 V supply. The predicted current is \(20.0\,\mathrm{mA}\).

The student takes four repeat readings using a digital ammeter with a stated uncertainty of \(\pm 0.2\,\mathrm{mA}\).

Use half the range to estimate repeatability uncertainty, and follow the uncertainty convention given in the exam instructions.

(a) Calculate the mean measured current. (2 marks)

(b) Calculate the repeatability uncertainty and state the uncertainty that should be reported with the mean. (2 marks)

(c) Determine whether the predicted current is consistent with the measured result within the reported uncertainty. Justify your answer. (2 marks)

(d) Calculate the percentage difference between the mean measured current and the predicted current, using the prediction as the reference value. (1 mark)

Question 5 (9 marks)

A circuit contains two fixed resistors, \(R_1=150\,\Omega\) and \(R_2=300\,\Omega\), connected in series.

The supply voltage is \(8.98\,\mathrm{V}\). For this question, treat the supply voltage and resistor values as exact when calculating theoretical predictions.

A student obtains the following measurements.

(a) Calculate the predicted current through the circuit. (2 marks)

(b) Calculate the percentage difference between the predicted and measured currents, using the prediction as the reference value. State whether the current measurement is consistent with the prediction within the stated ammeter uncertainty. (2 marks)

(c) Calculate the predicted potential difference across each resistor. Determine whether each measured voltage is consistent with its prediction within the stated voltmeter uncertainty. (2 marks)

(d) Calculate the sum of the two measured potential differences and assess whether it agrees with the supply voltage within uncertainty. Explain why agreement in this sum does not establish that both individual voltage measurements are accurate, and suggest one specific follow-up check. (3 marks)

03Worked solutions and marking guide

Solution 1

(a) Meter connections (2 marks)

• The ammeter must be connected in series with the resistor. The same current then passes through both the resistor and the ammeter. (1 mark)
• The voltmeter must be connected in parallel across the resistor. Its terminals measure the potential difference between the two ends of the resistor. (1 mark)

(b) Meter resistance (2 marks)

An ideal ammeter has zero resistance, or as little resistance as possible in a real instrument. This prevents the ammeter from adding significant resistance to the circuit and changing the current being measured. (1 mark)

An ideal voltmeter has infinite resistance, or very high resistance in a real instrument. This ensures that it draws negligible current and has minimal effect on the circuit being measured. (1 mark)

Marking breakdown: 1 mark for each correct connection; 1 mark for each correct explanation of the resistance requirement.

Solution 2

(a) Total resistance (1 mark)

For resistors connected in series, their resistances add.

\[ \begin{aligned} R_{\text{total}} &= R_1+R_2\\ &=100+200\\ &=300\,\Omega \end{aligned} \]

Answer: \(300\,\Omega\).

(b) Current through each resistor (2 marks)

Using Ohm’s law,

\[ I=\frac{V}{R} \]

The total current is

\[ \begin{aligned} I&=\frac{6.00}{300}\\ &=0.0200\,\mathrm{A}\\ &=20.0\,\mathrm{mA} \end{aligned} \]

In a series circuit, the current is the same at every point. Therefore,

\[ I_1=I_2=20.0\,\mathrm{mA} \]

Answer: \(20.0\,\mathrm{mA}\) through each resistor.

(c) Potential difference across each resistor (2 marks)

The potential difference across each resistor is calculated using \(V=IR\).

For the \(100\,\Omega\) resistor,

\[ V_1=0.0200\times100=2.00\,\mathrm{V} \]

For the \(200\,\Omega\) resistor,

\[ V_2=0.0200\times200=4.00\,\mathrm{V} \]

Answer: \(2.00\,\mathrm{V}\) across \(R_1\) and \(4.00\,\mathrm{V}\) across \(R_2\).

The sum is \(2.00+4.00=6.00\,\mathrm{V}\), equal to the supply voltage.

Marking breakdown:

• (a) 1 mark for the correct total resistance.
• (b) 1 mark for correct substitution into Ohm’s law; 1 mark for the current and recognition that it is the same through both resistors.
• (c) 1 mark for each correct potential difference, including units.

Solution 3

(a) Predicted currents (3 marks)

The equivalent resistance of the parallel branch is given as \(100\,\Omega\). This branch is in series with \(R_1=100\,\Omega\).

Therefore, the total resistance is

\[ R_{\text{total}}=100+100=200\,\Omega \]

The predicted total current is

\[ \begin{aligned} I_{\text{total}}&=\frac{6.00}{200}\\ &=0.0300\,\mathrm{A}\\ &=30.0\,\mathrm{mA} \end{aligned} \]

The potential difference across \(R_1\) is

\[ V_1=0.0300\times100=3.00\,\mathrm{V} \]

The parallel branch therefore also has a potential difference of \(3.00\,\mathrm{V}\).

For \(R_2=150\,\Omega\),

\[ I_2=\frac{3.00}{150} =0.0200\,\mathrm{A} =20.0\,\mathrm{mA} \]

For \(R_3=300\,\Omega\),

\[ I_3=\frac{3.00}{300} =0.0100\,\mathrm{A} =10.0\,\mathrm{mA} \]

The predicted currents are:

• Main line: \(30.0\,\mathrm{mA}\).
• Through \(R_2\): \(20.0\,\mathrm{mA}\).
• Through \(R_3\): \(10.0\,\mathrm{mA}\).

Marking breakdown:

• 1 mark for the correct total resistance and total current.
• 1 mark for the correct potential difference across the parallel branch.
• 1 mark for both correct branch currents.

(b) Comparing measured currents (2 marks)

First, add the measured branch currents:

\[ \begin{aligned} I_2+I_3&=19.9+10.1\\ &=30.0\,\mathrm{mA} \end{aligned} \]

The absolute difference between this sum and the measured total current is

\[ |30.0-29.7|=0.3\,\mathrm{mA} \]

The uncertainty in the sum of the branch currents is

\[ 0.2+0.2=0.4\,\mathrm{mA} \]

The uncertainty in the difference between the branch-current sum and the measured total current is

\[ 0.4+0.2=0.6\,\mathrm{mA} \]

The difference of \(0.3\,\mathrm{mA}\) is smaller than the combined uncertainty of \(0.6\,\mathrm{mA}\).

Therefore, the measured total current is consistent with the sum of the measured branch currents within uncertainty.

Marking breakdown:

• 1 mark for calculating the branch-current sum and its difference from the measured total.
• 1 mark for correctly combining uncertainties and concluding that the measurements are consistent.

Solution 4

(a) Mean measured current (2 marks)

The mean is the sum of the readings divided by the number of readings.

\[ \begin{aligned} \bar I&=\frac{19.6+20.2+20.0+19.8}{4}\\ &=\frac{79.6}{4}\\ &=19.9\,\mathrm{mA} \end{aligned} \]

Answer: \(19.9\,\mathrm{mA}\).

Marking breakdown: 1 mark for the correct averaging method; 1 mark for the correct mean with units.

(b) Repeatability uncertainty (2 marks)

The range is the highest reading minus the lowest reading.

\[ \text{Range}=20.2-19.6=0.6\,\mathrm{mA} \]

The repeatability uncertainty is half the range:

\[ \Delta I_{\text{repeat}}=\frac{0.6}{2} =0.3\,\mathrm{mA} \]

The meter’s stated uncertainty is \(\pm0.2\,\mathrm{mA}\). The repeatability uncertainty is larger, so the reported uncertainty is \(\pm0.3\,\mathrm{mA}\).

The result should be reported as

\[ I=(19.9\pm0.3)\,\mathrm{mA} \]

Marking breakdown: 1 mark for the range and half-range calculation; 1 mark for selecting the larger uncertainty and reporting the result correctly.

(c) Agreement within uncertainty (2 marks)

The measured interval is

\[ 19.9-0.3=19.6\,\mathrm{mA} \]

to

\[ 19.9+0.3=20.2\,\mathrm{mA} \]

The predicted current is \(20.0\,\mathrm{mA}\), which lies within this interval.

Therefore, the predicted and measured currents agree within the reported uncertainty.

This does not prove that the measurement is perfectly accurate. It means the observed difference is small enough to be consistent with the estimated uncertainty.

Marking breakdown:

• 1 mark for identifying the interval from \(19.6\) to \(20.2\,\mathrm{mA}\).
• 1 mark for correctly comparing the prediction with the interval and stating the conclusion.

(d) Percentage difference (1 mark)

Using the predicted current as the reference,

\[ \begin{aligned} \text{Percentage difference} &=\frac{|19.9-20.0|}{20.0}\times100\%\\ &=0.5\% \end{aligned} \]

Answer: \(0.5\%\).

Marking breakdown: 1 mark for the correct calculation and percentage.

Solution 5

(a) Predicted current (2 marks)

The total resistance of the series circuit is

\[ \begin{aligned} R_{\text{total}}&=150+300\\ &=450\,\Omega \end{aligned} \]

Using the supply voltage of \(8.98\,\mathrm{V}\),

\[ \begin{aligned} I_{\text{predicted}}&=\frac{8.98}{450}\\ &=0.01996\,\mathrm{A}\\ &=19.96\,\mathrm{mA} \end{aligned} \]

Answer: \(19.96\,\mathrm{mA}\), or \(20.0\,\mathrm{mA}\) to three significant figures.

Marking breakdown:

• 1 mark for the correct total resistance.
• 1 mark for the correct predicted current with units.

(b) Percentage difference and consistency (2 marks)

The absolute difference between the predicted and measured currents is

\[ |19.96-19.4|=0.56\,\mathrm{mA} \]

The percentage difference, using the prediction as the reference, is

\[ \begin{aligned} \text{Percentage difference} &=\frac{|19.96-19.4|}{19.96}\times100\%\\ &\approx2.8\% \end{aligned} \]

The current difference of \(0.56\,\mathrm{mA}\) exceeds the stated ammeter uncertainty of \(0.2\,\mathrm{mA}\). The measured current is therefore not consistent with the predicted current within the stated uncertainty.

Marking breakdown:

• 1 mark for the correct percentage difference.
• 1 mark for correctly comparing the absolute difference with the stated uncertainty and drawing the conclusion.

(c) Predicted and measured potential differences (2 marks)

The predicted current is approximately \(0.01996\,\mathrm{A}\).

Across the \(150\,\Omega\) resistor,

\[ \begin{aligned} V_1&=IR_1\\ &=0.01996\times150\\ &\approx2.99\,\mathrm{V} \end{aligned} \]

Across the \(300\,\Omega\) resistor,

\[ \begin{aligned} V_2&=IR_2\\ &=0.01996\times300\\ &\approx5.99\,\mathrm{V} \end{aligned} \]

Compare these predictions with the measurements.

For \(R_1\), the difference is approximately

\[ |2.99-2.92|=0.07\,\mathrm{V} \]

For \(R_2\), the difference is approximately

\[ |6.06-5.99|=0.07\,\mathrm{V} \]

Both differences exceed the stated voltmeter uncertainty of \(0.02\,\mathrm{V}\). Neither measured voltage is consistent with its predicted value within the stated uncertainty.

Marking breakdown:

• 1 mark for both correct predicted voltage drops.
• 1 mark for correctly identifying that both measured readings differ from their predictions by more than the stated uncertainty.

(d) Sum of measured voltages and follow-up check (3 marks)

First, calculate the sum of the measured potential differences:

\[ \begin{aligned} V_1+V_2&=2.92+6.06\\ &=8.98\,\mathrm{V} \end{aligned} \]

The uncertainty in this sum is

\[ \Delta V_{\text{sum}}=0.02+0.02 =0.04\,\mathrm{V} \]

Therefore, the measured sum is

\[ V_{\text{sum}}=(8.98\pm0.04)\,\mathrm{V} \]

This sum is consistent with the supply voltage of \(8.98\,\mathrm{V}\).

However, this agreement does not establish that each individual voltage reading is accurate. The first reading is approximately \(0.07\,\mathrm{V}\) below its prediction, while the second is approximately \(0.07\,\mathrm{V}\) above its prediction. These opposing discrepancies cancel when the readings are added.

One suitable follow-up check is to repeat both voltage measurements using a calibrated second voltmeter. The student should also verify the meter range and probe contact points. If the individual readings change significantly, the original measurements may have been affected by an instrument or connection problem.

Marking breakdown:

• 1 mark for calculating the voltage sum, its uncertainty, and correctly assessing agreement with the supply.
• 1 mark for explaining that opposing discrepancies can cancel, so agreement in the sum does not establish individual accuracy.
• 1 mark for proposing a specific and appropriate follow-up check.

End of paper – 30 marks

After marking your responses, focus on any lost marks involving meter connections, the treatment of uncertainty, or the distinction between repeatability and agreement with a theoretical prediction. Those are separate skills, and a result can be repeatable without necessarily agreeing with the model.