Alternating Currents

Alternating current (AC) is central to electrical power generation, transmission, and use. AQA specification 3.7.5 requires you to understand sinusoidal AC waveforms, the relationship between peak and rms values, and the role of transformers in power transmission.

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Sinusoidal AC Waveforms

An AC voltage (or current) varies sinusoidally with time:

V = V₀ sin(ωt) or V = V₀ sin(2πft)

where V₀ is the peak voltage (the maximum value), ω = 2πf is the angular frequency (rad s⁻¹), f is the frequency (Hz), and t is time (s).

Similarly for current:

I = I₀ sin(ωt)

Described diagram — AC voltage waveform: A sinusoidal curve oscillating between +V₀ and −V₀ about the horizontal time axis. One complete cycle takes a period T = 1/f. The peak voltage V₀ is the maximum displacement from zero. The voltage is positive for the first half-cycle and negative for the second half-cycle, representing the alternating direction of current flow.

For the UK mains supply: V₀ ≈ 325 V, f = 50 Hz, T = 1/50 = 0.02 s.

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Peak and RMS Values

The peak value (V₀ or I₀) is the maximum instantaneous value of the AC quantity.

The root mean square (rms) value is the effective value of an AC quantity — it is equivalent to the DC value that would produce the same power dissipation in a resistor.

Derivation of V_rms

Power dissipated in a resistor: P = V²/R

For an AC voltage V = V₀ sin(ωt), the instantaneous power is P = V₀² sin²(ωt) / R.

The mean power over a full cycle is: P_mean = V₀² × ⟨sin²(ωt)⟩ / R

The mean value of sin²(ωt) over a complete cycle is ½.

Therefore: P_mean = V₀²/(2R)

The rms voltage is defined as the DC voltage that gives the same power: P_mean = V_rms²/R = V₀²/(2R)

V_rms = V₀/√2 ≈ 0.707 V₀

Similarly:

I_rms = I₀/√2 ≈ 0.707 I₀

Important Values

Quantity	Relationship
V_rms	V₀/√2
I_rms	I₀/√2
V₀ from V_rms	V₀ = V_rms × √2
Mean power	P = V_rms × I_rms = V₀I₀/2

For the UK mains:

• V_rms = 230 V
• V₀ = 230 × √2 = 325 V

Worked Example 1 — RMS and Peak Values

Question: A sinusoidal AC supply has a peak voltage of 12 V and a frequency of 50 Hz. Calculate: (a) the rms voltage; (b) the rms current through a 100 Ω resistor; (c) the mean power dissipated.

Solution:

(a) V_rms = V₀/√2 = 12/√2 = 12/1.414 = 8.49 V

(b) I_rms = V_rms/R = 8.49/100 = 0.0849 A = 84.9 mA

(c) P_mean = V_rms × I_rms = 8.49 × 0.0849 = 0.721 W

Check: P = V₀²/(2R) = 144/200 = 0.720 W ✓

Worked Example 2 — UK Mains

Question: Calculate the peak current drawn by a 2.0 kW electric heater connected to the 230 V mains supply.

Solution:

I_rms = P/V_rms = 2000/230 = 8.70 A

I₀ = I_rms × √2 = 8.70 × 1.414 = 12.3 A

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Using an Oscilloscope to Measure AC

An oscilloscope displays voltage on the vertical axis and time on the horizontal axis.

Described diagram — Oscilloscope trace of AC: A sinusoidal wave displayed on a grid. The vertical axis has a sensitivity setting (e.g., 5 V per division) and the horizontal axis has a time base setting (e.g., 2 ms per division). The peak-to-peak amplitude spans from the top to the bottom of the wave.