Monitoring and measuring alternating current
Back to Higher hubWhat alternating current is
Key idea
a.c. is a current that changes direction and instantaneous value with time; d.c. flows one way only.
- Alternating current (a.c.) continually changes direction, and its instantaneous value changes with time (the mains supply, signal generators). Direct current (d.c.) flows in one direction only.
- On an oscilloscope, a.c. shows as a repeating wave (sinusoidal for the mains); steady d.c. is a flat horizontal line.
- The peak value is the maximum the wave reaches from zero; the period T is the time for one complete cycle; the frequency f is the number of cycles per second (f = 1 ÷ T).
Model answer — State the difference
State one difference between alternating current and direct current. 1 mark
Model answer
a.c. changes direction (and instantaneous value) with time, whereas d.c. flows in one direction only.
Peak & rms values
On the relationship sheet — given to you in the exam
Derived — just the same equation rearranged
On the relationship sheet
rms voltage
Vrms =
Vpeak√2
rms = peak ÷ √2
Derived
Peak voltage
Vpeak = √2 Vrms
rearranged: peak = √2 × rms (≈ 1.41 × rms)
On the relationship sheet
rms current
Irms =
Ipeak√2
and rearranged: Ipeak = √2 Irms
- The rms (root mean square) value of an a.c. is the d.c. value that would deliver the same power to a resistor. It is what a.c. meters read and what "230 V mains" means (230 V is the rms, not the peak).
- The relationship sheet gives Vrms = Vpeak ÷ √2 (and Irms = Ipeak ÷ √2). Rearranged, Vpeak = √2 × Vrms ≈ 1.41 × Vrms.
- Always check whether a value is peak or rms before using it — mixing them is the classic error. Meters and quoted mains values are rms; an oscilloscope shows you the peak.
Substitute & solve
The mains supply has an rms voltage of 230 V. Calculate the peak voltage of the supply. 3 marks
Vpeak=√2 × Vrmson the sheet
Vpeak=1.41 × 230√2 ≈ 1.41
Vpeak=325 V (3.25 × 10² V)
Rearrange — numbers in first
An oscilloscope shows that an a.c. signal has a peak voltage of 12 V. Calculate the rms voltage. 3 marks
Vpeak=√2 × Vrms
12=1.41 × Vrmsnumbers in first
Vrms=12 ÷ 1.41now rearrange
Vrms=8.5 V
Peak or rms? A value from a meter or quoted for the mains is rms; a value read off an oscilloscope is peak. Decide which you have before you substitute.
Practice 1
An a.c. supply has a peak current of 0.50 A. Calculate the rms current. 3 marks
Answer
I_rms = I_peak ÷ √2 = 0.50 ÷ 1.41 = 0.35 A
Reading the oscilloscope
On the relationship sheet
Period
T =
1f
period = 1 ÷ frequency
Derived
Frequency
f =
1T
rearranged: frequency = 1 ÷ period
- An oscilloscope has two key control settings:
- volts/div (Y-gain): volts per vertical division → peak voltage = (divisions from the centre line to the peak) × (volts/div);
- time-base (time/div): seconds per horizontal division → period T = (divisions for one complete wave) × (time/div).
- From the period, f = 1 ÷ T. From the peak voltage, rms = peak ÷ √2.
- Method: (1) count divisions for one full cycle → × time/div → T → f = 1 ÷ T; (2) count divisions from the centre to the peak → × volts/div → Vpeak → Vrms.
Substitute & solve — frequency from the time-base
On an oscilloscope, one complete wave occupies 4.0 divisions. The time-base is set to 5.0 ms per division. Calculate the frequency of the a.c. signal. 4 marks
Convert first
5.0 ms = 5.0 × 10⁻³ s ms → s
T=divisions × time/div
T=4.0 × 5.0 × 10⁻³
T=0.020 s
f=1 ÷ T = 1 ÷ 0.020
f=50 Hz
Several steps — peak then rms
The same trace reaches 3.0 divisions above the centre line. The Y-gain is 2.0 V per division. Calculate (a) the peak voltage and (b) the rms voltage. 4 marks
(a) Vpeak=divisions × volts/div
Vpeak=3.0 × 2.0 = 6.0 V
(b) Vrms=Vpeak ÷ √2peak → rms
Vrms=6.0 ÷ 1.41 = 4.2 V
Practice 2
An a.c. trace shows a peak that is 2.5 divisions high with the Y-gain at 4.0 V/div, and one full cycle spanning 5.0 divisions with the time-base at 2.0 ms/div. Determine (a) the peak voltage, (b) the rms voltage and (c) the frequency. 5 marks
Answer
(a) V_peak = 2.5 × 4.0 = 10 V
(b) V_rms = 10 ÷ 1.41 = 7.1 V
(c) T = 5.0 × 2.0 × 10⁻³ = 0.010 s (ms → s)
f = 1 ÷ 0.010 = 100 Hz