Alternating currents (A-level only) — AQA A-Level Physics
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Alternating current (a.c.) is a current that continuously reverses direction, typically described by a sinusoidal variation with time. For AQA A-Level Physics, you must understand root mean square (rms) values, the relationship between peak and rms quantities, and how a.c. behaves in resistive, capacitive, and inductive circuits.
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Alternating currents (A-level only) explained
Alternating currents (A-level only) covers 3 specification points, including Sinusoidal voltages and currents only; root… and Use of an oscilloscope as a dc and ac….
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Open a point below for its explanation and revision guidance.
Sinusoidal voltages and currents only; root mean square, peak and peak-to-peak values for sinusoidal waveforms only.
For a sinusoidal alternating voltage or current, the value changes continuously, so a single representative value is needed for power calculations. The root mean square value is the steady direct value that would dissipate the same average power in a given resistor. For a sinusoid, V_rms = V₀/√2 and I_rms = I₀/√2, where V₀ and I₀ are peak values. The peak-to-peak value is the difference between the maximum positive and maximum negative values, so V_pp = 2V₀. Average power is P = I_rms²R = V_rms²/R = I_rms V_rms for a resistor. These relationships apply only to sinusoidal waveforms; other waveforms need their own rms calculation.
Use of an oscilloscope as a dc and ac voltmeter, to measure time intervals and frequencies, and to display ac waveforms.
A cathode-ray oscilloscope displays how a potential difference varies with time. The electron beam sweeps horizontally at a known timebase setting, so horizontal distance represents time; the beam is deflected vertically by the input voltage, so vertical distance represents potential difference. To use it as a dc voltmeter, set the timebase to a slow sweep or off and read the steady vertical deflection, multiplying by the volts-per-division setting. For ac, a sinusoidal trace appears; its amplitude gives peak voltage and its period is the time for one complete cycle, measured in divisions and multiplied by the time-per-division setting. Frequency is the reciprocal of period, f = 1/T. Time intervals between events are found from horizontal separation. Always check the zero line and account for any vertical offset before measuring.
No details of the structure of the instrument are required but familiarity with the operation of the controls is expected.
You are not expected to explain how an oscilloscope is constructed inside, such as the electron gun, deflection plates or phosphor screen. Instead, you must know what the main controls do and how to operate them to obtain a usable trace. The timebase control sets the horizontal sweep speed in time per division, so it controls the time scale. The vertical gain or volts-per-division control sets the vertical scale, so it controls the voltage scale. The vertical position control shifts the trace up or down to set a convenient zero line. The brightness and focus controls adjust the visible spot. To make a measurement, adjust the timebase and gain so the trace fills the screen, centre the trace, then read divisions and multiply by the settings.
Your focus
- Define the root mean square value of an alternating current or voltage in terms of average power.
- Convert between peak, peak-to-peak and rms values for sinusoidal waveforms.
- Calculate average power dissipated in a resistor using rms current and voltage values.
Show all 9 objectives
- Read a stated vertical scale and determine a dc or peak ac potential difference from an oscilloscope trace.
- Read a stated timebase and determine a time interval or period from an oscilloscope trace.
- Calculate the frequency of an ac waveform from its measured period using f = 1/T.
- Describe the function of the timebase, gain, and vertical position controls on an oscilloscope.
- Explain how adjusting these controls affects the appearance and scale of a displayed waveform.
- Select appropriate control settings to display a waveform clearly for measurement.
Which command word is this?
You must use the correct formula, substitute the given values, and show your working. The final answer should be given to an appropriate number of significant figures with correct units. Marks are awarded for correct substitution and accurate arithmetic.
Alternating currents (A-level only) exam tips
Topic Overview
Alternating currents (a.c.) are a fundamental topic in A-Level Physics that describes how current and voltage vary sinusoidally with time. Unlike direct current (d.c.), which flows in one direction, a.c. continuously reverses direction, typically at frequencies of 50 Hz in the UK mains supply. This topic introduces key concepts such as peak and root mean square (rms) values, which allow us to compare the heating effect of a.c. to that of d.c.
Understanding a.c. is crucial for analysing circuits containing resistors, capacitors, and inductors, and for explaining phenomena like phase differences and reactance. It builds on earlier work on electricity and extends into more advanced areas such as electromagnetic induction and power transmission. Mastery of this topic is essential for tackling exam questions on circuit behaviour and for appreciating the practical applications of a.c. in everyday technology.
Key Concepts
- →Peak and rms values: For a sinusoidal a.c., the rms (root mean square) value is the equivalent steady d.c. value that would produce the same power dissipation in a resistor. I_rms = I_0 / sqrt(2) and V_rms = V_0 / sqrt(2).
- →Power in a.c. circuits: The average power dissipated in a resistor is P = I_rms^2 R = V_rms^2 / R = I_rms V_rms. This is half the product of peak voltage and peak current for a sinusoidal supply.
- →Phase relationships: In a pure resistor, current and voltage are in phase. In a pure capacitor, current leads voltage by 90 degrees (pi/2 rad). In a pure inductor, voltage leads current by 90 degrees.
- →Reactance: Capacitive reactance X_C = 1/(2 pi f C) decreases with increasing frequency. Inductive reactance X_L = 2 pi f L increases with increasing frequency. Reactance limits current in a.c. circuits similarly to resistance.
- →Phasor diagrams: These are used to represent the phase relationship between voltage and current in a.c. circuits, with rotating vectors (phasors) whose projections give instantaneous values.
Marking Points
- Defines the root mean square value of an alternating current or voltage as the steady direct value that produces the same average power in a resistor.
- States and applies the sinusoidal relationships V_rms = V₀/√2 and I_rms = I₀/√2, where V₀ and I₀ are peak values.
- Calculates peak-to-peak values from peak values using V_pp = 2V₀, and converts between peak, peak-to-peak and rms values for sinusoidal waveforms.
- Uses rms values to calculate average power in a resistor with P = I_rms²R, P = V_rms²/R or P = I_rms V_rms.
- Explains that the √2 factor arises from averaging the square of a sinusoidal function over a complete cycle, and that it applies only to sinusoidal waveforms.
- Identify the vertical scale in volts per division and multiply the measured vertical deflection in divisions by this value to obtain the potential difference.
- Identify the horizontal timebase in seconds (or milliseconds) per division and multiply the measured horizontal distance in divisions by this value to obtain a time interval.
- For a dc input, read the steady vertical displacement from the zero line; for an ac input, measure the amplitude from the centre line to a peak, not peak-to-peak, when asked for peak voltage.
- Measure the period as the horizontal distance for one complete cycle, then calculate frequency using f = 1/T, converting units consistently.
- Recognise that the trace is a graph of potential difference against time, so gradients and areas are not usually required at this level; the display directly encodes V and t.
- Check and state the zero position and any vertical shift control setting, because an offset changes the apparent deflection without changing the true input voltage.
- State the function of the timebase control: it sets the horizontal sweep rate in time per division and therefore the time scale of the display.
- State the function of the vertical gain or volts-per-division control: it sets the vertical scale in volts per division and therefore the voltage scale.
- Explain that the vertical position control shifts the trace without changing the measured potential difference, so it is used to set a reference or zero line.
- Describe a sensible operating sequence: select ac or dc coupling as appropriate, adjust timebase and gain to obtain a clear trace, centre the trace, then measure.
- Recognise that brightness and focus controls affect visibility only and do not change the voltage or time measurements.
- Explain that changing the timebase changes the horizontal width of a cycle but not the amplitude, while changing the gain changes the vertical height but not the period.
Examiner Tips
- 💡Write down whether a quoted value is peak, peak-to-peak or rms before substituting it, since mixing them is the most common source of error.
- 💡Show the conversion steps explicitly, for example peak-to-peak to peak by halving, then peak to rms by dividing by √2.
- 💡When calculating power, choose the form of the equation that matches the quantity given, such as P = V_rms²/R when the rms voltage is known.
- 💡Write the scale settings as a conversion factor before substituting numbers, for example 2.0 V div⁻¹ and 5.0 ms div⁻¹, so units cancel correctly.
- 💡When finding frequency, convert milliseconds to seconds before taking the reciprocal, because f = 1/T requires T in seconds.
- 💡Sketch the trace you are interpreting and label the amplitude and period on the sketch; this reduces confusion between vertical and horizontal measurements.
- 💡Use the phrase 'time per division' for the timebase and 'volts per division' for the gain; these precise terms show understanding of the controls.
- 💡When describing a measurement, state the control settings you would choose and why, such as increasing the timebase to spread out a high-frequency waveform.
- 💡Link each control to its effect on the trace: timebase changes horizontal scale, gain changes vertical scale, position shifts the trace.
- 💡Always show your working clearly, especially when converting between peak and rms values. Marks are often awarded for correct substitution and rearrangement, even if the final answer is wrong.
- 💡When explaining phase relationships, use the correct terminology such as 'current leads voltage by pi/2 radians' and avoid vague statements like 'they are out of sync'. Diagrams can help but must be labelled correctly.
- 💡For calculations involving reactance, ensure you use the correct formula and units. Remember that frequency must be in hertz and capacitance in farads, inductance in henries.
Common Mistakes
- Using peak values directly in power equations such as P = I₀²R; the correction is to use rms values, so P = I_rms²R = I₀²R/2 for a sinusoid.
- Confusing peak-to-peak with peak and dividing by √2 twice; the correction is to halve the peak-to-peak value first to obtain the peak, then divide by √2 for the rms value.
- Applying the √2 relationships to non-sinusoidal waveforms such as square or triangular waves; the correction is to use the defining average-power method or the specific rms value for that waveform.
- Reading peak-to-peak distance as the amplitude: the correction is to measure from the centre line to one peak for peak voltage, or to halve the peak-to-peak value.
- Forgetting to multiply the number of divisions by the scale setting: the correction is to convert every measured division into volts or seconds using the relevant control value.
- Using the timebase setting to convert vertical distance: the correction is to use volts per division vertically and time per division horizontally.
- Counting a half-cycle as a full period: the correction is to measure between equivalent points on successive cycles, such as peak to next peak.
- Confusing the timebase and gain controls: the correction is to associate timebase with horizontal time per division and gain with vertical volts per division.
- Believing the vertical position control changes the voltage reading: the correction is to treat it as a display offset that must be allowed for when reading from the zero line.
- Attempting to describe internal electron optics when asked only about controls: the correction is to focus on the function and effect of each control on the trace.
- Students often think that rms voltage is the same as average voltage. In fact, the average of a sinusoidal voltage over a full cycle is zero; the rms value is the root of the mean of the squares, which is non-zero and represents the effective d.c. equivalent.
- Many students believe that current and voltage are always in phase in any a.c. circuit. They must remember that phase differences occur in capacitors and inductors, and that only in a pure resistor are they in phase.
- Some students confuse the terms 'peak' and 'peak-to-peak'. Peak-to-peak voltage is twice the peak voltage, and it is not used in rms calculations directly.
Revision Plan
- 1Week 1, Day 1-2: Review the definitions of alternating current and voltage, and learn how to calculate peak, peak-to-peak, and rms values. Practice converting between these quantities.
- 2Week 1, Day 3-4: Study the power equations for a.c. circuits and solve problems involving average power in resistive circuits. Ensure you understand why rms values are used.
- 3Week 1, Day 5-7: Learn about phase relationships in capacitors and inductors, and how to draw phasor diagrams. Memorise the reactance formulas and practice calculations.
- 4Week 2, Day 1-3: Work through past exam questions on a.c. circuits, focusing on multi-step calculations and explanations. Check your answers against mark schemes.
- 5Week 2, Day 4-5: Create a summary sheet of key formulas and concepts, and test yourself with active recall questions. Review any weak areas.
Exam Question Types
- 📋Calculation questions: These often ask you to calculate rms values, average power, or reactance. Show all steps and pay attention to units. Typically worth 2-4 marks.
- 📋Explanation questions: You may be asked to explain why rms values are used or to describe the phase relationship between current and voltage in a capacitor or inductor. Use precise language and refer to the mnemonic 'CIVIL' if helpful. Typically worth 3-4 marks.
- 📋Graph and phasor diagram questions: You might need to sketch or interpret graphs of voltage and current against time, or draw phasor diagrams. Label axes and key points clearly. Typically worth 2-3 marks.
- 📋Practical application questions: These could involve calculating the number of turns in a transformer or analysing a circuit with both resistance and reactance. Break down the problem into steps. Typically worth 4-6 marks.
Command Word Expectations (AQA)
You must use the correct formula, substitute the given values, and show your working. The final answer should be given to an appropriate number of significant figures with correct units. Marks are awarded for correct substitution and accurate arithmetic.
You must provide a clear, logical reasoning that links cause and effect. Use correct physics terminology and refer to relevant principles. For example, explaining why current leads voltage in a capacitor should mention the charging and discharging process and the rate of change of voltage.
You should give a detailed account of the phenomenon or process, including key features. For instance, describing how an alternating current varies with time should mention sinusoidal shape, frequency, and peak value. No explanation of why is required unless specified.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: An alternating voltage has a peak value of 340 V and a frequency of 50 Hz. It is applied across a 100 ohm resistor. Calculate: (a) the rms voltage, (b) the rms current, (c) the average power dissipated in the resistor.
- 1.Step 1: Identify given values: peak voltage V_0 = 340 V, resistance R = 100 ohms, frequency f = 50 Hz (not needed for this calculation).
- 2.Step 2: Calculate rms voltage using V_rms = V_0 / sqrt(2) = 340 / 1.414 = 240.4 V (to 3 s.f.).
- 3.Step 3: Calculate rms current using Ohm's law: I_rms = V_rms / R = 240.4 / 100 = 2.404 A.
- 4.Step 4: Calculate average power using P = I_rms^2 R = (2.404)^2 * 100 = 578 W (to 3 s.f.). Alternatively, P = V_rms^2 / R = (240.4)^2 / 100 = 578 W.
Question: A sinusoidal alternating current of frequency 60 Hz flows through a pure capacitor of capacitance 20 microF. The peak voltage across the capacitor is 12 V. Calculate the peak current and the rms current.
- 1.Step 1: Identify given values: C = 20e-6 F, f = 60 Hz, V_0 = 12 V.
- 2.Step 2: Calculate capacitive reactance: X_C = 1/(2 pi f C) = 1/(2 * pi * 60 * 20e-6) = 1/(0.00754) = 132.6 ohms.
- 3.Step 3: Use Ohm's law for peak values: I_0 = V_0 / X_C = 12 / 132.6 = 0.0905 A.
- 4.Step 4: Calculate rms current: I_rms = I_0 / sqrt(2) = 0.0905 / 1.414 = 0.0640 A.