Alternating currentsAQA A-Level Physics: Revision notes
Section 1
Sinusoidal voltages and currents
An alternating current (ac) changes direction periodically. For a sinusoidal supply:
V = V₀ sin 2πft and I = I₀ sin 2πft
where V₀ and I₀ are the peak values, f is the frequency and the period T = 1/f. The peak-to-peak value is twice the peak value, 2V₀. A sinusoidal waveform has a mean value of zero over a cycle.
Section 2
Root mean square values
The root mean square (rms) value of an alternating current is the value of the direct current that dissipates the same mean power in a resistor. For sinusoidal waveforms:
I_rms = I₀/√2 and V_rms = V₀/√2
The power is P = I²R, which varies between 0 and I₀²R; the mean power is ½I₀²R = I_rms²R = V_rms²/R. Rms values are used because the mean of the current is zero but the heating is not.
Do not use the rms value for peak-to-peak or vice versa: V₀ = √2 × V_rms, and V_pp = 2V₀.
Section 3
Mains electricity
UK mains is quoted as 230 V rms at 50 Hz.
- Peak voltage = 230 × √2 = 325 V.
- Peak-to-peak voltage = 2 × 325 = 651 V.
- Period = 1/50 = 20 ms.
A 46 Ω heater therefore has a mean power of V_rms²/R = 230²/46 = 1150 W, and a maximum instantaneous power of 325²/46 = 2300 W.
Quote the peak voltage to 3 s.f. (325 V) and keep the unrounded value in later calculations.
Section 4
Using an oscilloscope
An oscilloscope displays voltage against time.
- Y-gain (V per division): voltage = vertical divisions × Y-gain. Use as a dc voltmeter (the trace shifts up or down) or ac voltmeter (peak or peak-to-peak height).
- Time-base (time per division): time = horizontal divisions × time-base. The period of an ac signal is the length of one cycle, and f = 1/T.
- Adjust the controls so the trace fills the screen; set the input to dc coupling to see both a dc level and an ac signal.
Worked example: Y-gain 2.0 V/div, time-base 5.0 ms/div, trace 5.2 divisions peak-to-peak and 4.0 divisions per cycle: V_pp = 10.4 V, V₀ = 5.2 V, V_rms = 3.7 V, T = 20 ms, f = 50 Hz.
Use the peak, not the peak-to-peak, height to find the rms voltage.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Alternating currents
- A heating element of resistance 46 Ω is connected to the UK mains supply, which has an rms voltage of 230 V and a frequency of 50 Hz. The supply voltage is sinusoidal.Explain why the mean power dissipated in the element is half of its maximum instantaneous power.2 marks
- An oscilloscope is used to display a sinusoidal alternating voltage. The Y-gain is set to 2.0 V per division and the time-base to 5.0 ms per division. The trace has a peak-to-peak height of 5.2 divisions, and one complete cycle of the trace occupies 4.0 divisions horizontally.Calculate the rms voltage of the signal.2 marks
- A signal generator supplies a sinusoidal alternating current of peak value 1.5 A and frequency 400 Hz to a resistor of resistance 12 Ω.Calculate the rms current and the mean power dissipated in the resistor.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).