Physics 9702/23 — May/June 2018

Background Concept

Errors in measurements are often classified as:

• Systematic errors: shift all readings in the same direction by a similar amount (a consistent bias).
• Random errors: cause readings to scatter about a mean value in an unpredictable way.

A systematic error affects accuracy (how close you are to the true value) and cannot be reduced by repeats; it must be corrected (e.g. by calibration or technique).

Understanding the Question

You are using an analogue voltmeter to measure a constant potential difference across a resistor. The question asks for one example of a systematic error that could occur in this situation.

Approach

Choose an error mechanism that would make every reading consistently wrong in the same way when using an analogue meter.

Step-by-Step Reasoning

A common systematic error in analogue meters is zero error:

• Before any measurement, the pointer should read $0\ \text{V}$ when there is no p.d. applied.
• If the pointer rests at, say, $+0.1\ \text{V}$ (or $-0.1\ \text{V}$), then when you measure a real p.d., every reading will be too high (or too low) by about $0.1\ \text{V}$.
• This is systematic because the bias is in the same direction each time.

Other valid systematic examples for an analogue voltmeter would include a miscalibrated scale or consistently viewing the scale from a fixed wrong angle (consistent parallax).

Key Takeaways

• Systematic errors create a consistent offset or proportional bias.
• They affect accuracy and are not improved by repeating measurements.

Common Mistakes

• Giving a random-effect example (e.g. “reading fluctuates”) when asked for systematic.
• Saying only “human error” without describing a specific mechanism.

Things to Be Careful About

• Parallax can be systematic if your eye is always in the same wrong position, but often becomes random if the viewing angle changes between readings.

Background Concept

Random errors produce unpredictable variations between repeated readings. They typically arise from:

• limitations of resolution (finite scale divisions),
• judgement of the observer,
• small fluctuations/noise.

Random errors affect precision (the spread of readings) and can be reduced by repeating readings and taking a mean.

Understanding the Question

You measure a constant p.d. repeatedly with an analogue voltmeter. The question asks for an example of a random error that could occur during these readings.

Approach

Pick an effect that would make consecutive readings vary slightly even if the true p.d. is constant.

Step-by-Step Reasoning

For an analogue meter you must visually align the pointer with the scale.

• If your eye position changes slightly from one reading to the next, the apparent pointer position shifts due to parallax.
• This can make readings sometimes a little high and sometimes a little low.
• That variation is random scatter about a mean value.

Another acceptable random example is difficulty judging the pointer between two scale marks (limited resolution), especially if the needle wobbles slightly.

Key Takeaways

• Random error = scatter in repeated measurements.
• Reduced by repeats/averaging; does not shift all values the same way.

Common Mistakes

• Stating a calibration/zero error (systematic) when random is required.
• Saying “parallax” without indicating that it varies between readings (to make it random).

Things to Be Careful About

• If the viewing angle is always wrong in the same way, parallax becomes systematic; for random error, the key idea is that the viewing varies between readings.

Background Concept

Electrical power is the rate of transfer of electrical energy.

For a component with potential difference $V$, current $I$, and resistance $R$:

$$P = VI$$

Using Ohm’s law $V = IR$, we can also write:

$$P = I^2R \quad \text{and} \quad P = \frac{V^2}{R}$$

These apply for a resistor where the $V$ and $I$ relationship is ohmic (or when you are told to treat it as a resistor with given $R$).

Understanding the Question

Given:

• $V = 5.0\ \text{V}$ across the resistor,
• $R = 125\ \Omega$.

You are asked to calculate the power dissipated.

Approach

Use the power form that uses only the quantities given, $P = V^2/R$, then substitute the numbers and include the unit watt.

Step-by-Step Reasoning

Start with:

$$P = \frac{V^2}{R}$$

Substitute:

• $V^2 = (5.0\ \text{V})^2 = 25\ \text{V}^2$,
• divide by $125\ \Omega$:

$$P = \frac{25}{125} = 0.20$$

So:

$$P = 0.20\ \text{W}$$

Key Takeaways

• Choose the form of the power equation that matches the given data.
• Always give a unit for power: $\text{W}$.

Common Mistakes

• Using $P = VI$ but not finding $I$ correctly.
• Forgetting to square $V$ in $P = V^2/R$.
• Writing the unit as $\text{V}$ or $\Omega$ instead of $\text{W}$.

Things to Be Careful About

• Keep consistent significant figures; here $5.0$ suggests the power should be given to about 2 significant figures (matching typical exam expectations).

Background Concept

When a quantity is calculated from measured values, its uncertainty depends on how those values are combined.

For products and quotients:

• if $Q = AB$ or $Q = A/B$, then percentage uncertainties add:

$$\frac{\Delta Q}{Q} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$$

For powers:

• if $Q = A^n$, then the fractional (percentage) uncertainty multiplies by $n$:

$$\frac{\Delta Q}{Q} = |n|\frac{\Delta A}{A}$$

Understanding the Question

You calculated power using:

$$P = \frac{V^2}{R}$$

Given uncertainties:

• $V = 5.0\ \text{V} \pm 0.1\ \text{V}$,
• $R = 125\ \Omega \pm 3\%$.

You must find the percentage uncertainty in $P$.

Approach

1. Convert the voltage uncertainty into a percentage uncertainty: $\Delta V / V$.
2. Because $V$ is squared, double its percentage uncertainty.
3. Add the resistor’s percentage uncertainty (because $R$ is in the denominator, but division still adds percentage uncertainties).

Step-by-Step Reasoning

Voltage fractional uncertainty:

$$\frac{\Delta V}{V} = \frac{0.1}{5.0} = 0.020$$

Convert to percent: $0.020 \times 100\% = 2.0\%$.

Because $P \propto V^2$, the percentage uncertainty contribution from $V$ is:

$$2 \times 2.0\% = 4.0\%$$

The resistance is already given as $3\%$.

Combine for $P = V^2/R$:

$$\text{percentage uncertainty in } P = 4.0\% + 3.0\% = 7.0\%$$

Key Takeaways

• Square (or any power) multiplies the percentage uncertainty by that power.
• Multiplication/division means percentage uncertainties add.

Common Mistakes

• Adding absolute uncertainties ($0.1\ \text{V}$ and something in $\Omega$) instead of percentage uncertainties.
• Forgetting to double the percentage uncertainty because of $V^2$.
• Trying to subtract uncertainties because $R$ is in the denominator (you still add percentage uncertainties).

Things to Be Careful About

• Use the measured value for the percentage uncertainty: $\Delta V/V$.
• Keep percentage uncertainty typically to 1–2 significant figures; $7.0\%$ is acceptable here.

Background Concept

To convert a percentage uncertainty into an absolute uncertainty:

$$\Delta Q = \left(\frac{\text{percentage uncertainty}}{100}\right)Q$$

When quoting a final result with uncertainty:

• quote the uncertainty to usually 1 significant figure (sometimes 2 if the first digit is 1 or 2),
• then quote the value to the same decimal place as the uncertainty.

Understanding the Question

You have already found the power $P$ and its percentage uncertainty. You must now state the power with an absolute uncertainty, in the form:

$$P = \text{value} \pm \text{uncertainty}\ \text{W}$$

with appropriate significant figures.

Approach

1. Take the calculated power from part (i).
2. Use the percentage uncertainty from part (ii) to find $\Delta P$.
3. Round $\Delta P$ sensibly and then match the power value to that precision.

Step-by-Step Reasoning

From part (i):

$$P = 0.20\ \text{W}$$

From part (ii): percentage uncertainty in $P$ is $7.0\%$, so fractional uncertainty is $0.070$.

Absolute uncertainty:

$$\Delta P = 0.070 \times 0.20 = 0.014\ \text{W}$$

Round the uncertainty to 1 significant figure:

$$0.014\ \text{W} \approx 0.01\ \text{W}$$

Then quote the power to the same decimal place (hundredths of a watt):

$$P = 0.20 \pm 0.01\ \text{W}$$

Key Takeaways

• Absolute uncertainty = fractional uncertainty $\times$ the value.
• Final quoted value must match the precision implied by the uncertainty.

Common Mistakes

• Leaving the uncertainty as a percentage when an absolute uncertainty is required.
• Rounding the value but not the uncertainty (or vice versa) inconsistently.
• Writing $0.2 \pm 0.01\ \text{W}$ (mismatched decimal places).

Things to Be Careful About

• If you keep $\Delta P = 0.014\ \text{W}$ (2 s.f. because it starts with 1), then the value should be quoted consistently (e.g. $0.200 \pm 0.014\ \text{W}$). In most exam mark schemes, $0.20 \pm 0.01\ \text{W}$ is the expected rounded form.