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.

Spec mapping: AQA 7408 A-Level Physics, Section 3.7.5 (Alternating currents). This lesson develops the sinusoidal model V = V₀ sin(ωt) for AC supplies, the derivation and use of root-mean-square values V_rms = V₀/√2 and I_rms = I₀/√2, mean power in a resistive AC circuit P = V_rms I_rms, the use of the oscilloscope to measure peak voltage, period and frequency, and the role of high transmission voltage in minimising I²R losses on long-distance cables. Refer to the official AQA 7408 specification document for the authoritative wording.

Synoptic links: (i) Capacitor charge / discharge (3.7.4) — the sinusoidal model used here also describes the steady-state response of a capacitor or inductor to AC excitation, and is the foundation of impedance treatments encountered at university. (ii) Electromagnetic induction (3.7.5) — the rotating-coil generator produces ε = ε₀ sin(ωt), and the AC supply at every wall socket has its sinusoidal form because of the geometry of a rotating coil cutting field lines. (iii) Energy and power in DC circuits (3.5.1) — the derivation P_mean = V_rms × I_rms = V₀I₀/2 sits on top of P = VI for DC, with the time-averaging step ⟨sin²(ωt)⟩ = ½ being the new physics.

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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.

From the oscilloscope trace:

• Peak voltage V₀ = (number of divisions from centre to peak) × (voltage sensitivity in V/div)
• Period T = (number of divisions per cycle) × (time base in s/div)
• Frequency f = 1/T
• V_rms = V₀/√2

Worked Example 3 — Reading an Oscilloscope

Question: An oscilloscope displays a sinusoidal wave. The voltage sensitivity is set to 2.0 V per division and the time base is 5.0 ms per division. The wave has a peak-to-peak height of 6.0 divisions and one complete cycle spans 4.0 divisions. Determine: (a) the peak voltage; (b) the rms voltage; (c) the frequency.

Solution:

(a) Peak-to-peak voltage = 6.0 × 2.0 = 12.0 V V₀ = 12.0/2 = 6.0 V

(b) V_rms = V₀/√2 = 6.0/1.414 = 4.24 V

(c) T = 4.0 × 5.0 × 10⁻³ = 0.020 s = 20 ms f = 1/T = 1/0.020 = 50 Hz

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Transformers and Power Transmission

Why High Voltages Are Used for Transmission

Electrical power is transmitted at very high voltages (up to 400 kV in the UK National Grid) to minimise power losses in the transmission cables.

The power lost as heat in a cable of resistance R carrying current I is:

P_loss = I²R

For a given power P transmitted at voltage V: I = P/V

Therefore: P_loss = (P/V)² × R = P²R/V²

P_loss = P²R/V²

By transmitting at a higher voltage V, the current I is reduced, and the power loss (proportional to I²) decreases dramatically.

Worked Example 4 — Power Transmission Losses

Question: A power station generates 100 MW of power. The electricity is transmitted through cables with a total resistance of 4.0 Ω. Compare the power lost when transmitting at (a) 25 kV and (b) 400 kV.

Solution:

(a) At 25 kV: I = P/V = 100 × 10⁶ / (25 × 10³) = 4000 A P_loss = I²R = (4000)² × 4.0 = 16 × 10⁶ × 4.0 = 64 MW (64% of generated power is lost!)

(b) At 400 kV: I = P/V = 100 × 10⁶ / (400 × 10³) = 250 A P_loss = I²R = (250)² × 4.0 = 62 500 × 4.0 = 250 kW = 0.25 MW (only 0.25% lost)

Increasing the transmission voltage by a factor of 16 reduces the power loss by a factor of 16² = 256.

The National Grid

The UK National Grid uses step-up transformers at power stations to increase the voltage to 275 kV or 400 kV for long-distance transmission, then step-down transformers at substations to reduce the voltage to 33 kV, 11 kV, and finally 230 V for domestic use.

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AC Power Calculations

For a purely resistive load:

Quantity	Formula
Mean power	P = V_rms I_rms
Mean power	P = ½V₀I₀
Mean power	P = V_rms²/R = I_rms²R

The instantaneous power in a resistive AC circuit is always positive (or zero) — it fluctuates between 0 and V₀I₀ with a mean of ½V₀I₀.

Instantaneous and Mean Power — Worked Detail

Consider an AC supply V = V₀ sin(ωt) connected across a resistor R. The instantaneous current is I(t) = (V₀/R) sin(ωt) = I₀ sin(ωt), and the instantaneous power dissipated is:

P(t) = V(t) × I(t) = V₀ I₀ sin²(ωt)

Using the identity sin²(ωt) = ½(1 − cos(2ωt)):

P(t) = ½ V₀ I₀ (1 − cos(2ωt)) = ½ V₀ I₀ − ½ V₀ I₀ cos(2ωt)

The instantaneous power has two components:

• A constant term ½ V₀ I₀ — this is the mean power.
• An oscillating term −½ V₀ I₀ cos(2ωt) — this oscillates at twice the supply frequency, with mean zero over any complete cycle.

So the time-averaged power is exactly ½ V₀ I₀ = V_rms I_rms, as derived. Notice that the instantaneous power never goes negative for a resistive load — it ranges from zero (when V and I are both zero, twice per cycle) to a peak of V₀ I₀ (when V and I are both at their maximum, twice per cycle). The mean lies halfway between, at ½ V₀ I₀.

Worked Example 5 — Instantaneous Power on the UK Mains

Question: A 100 W tungsten-filament light bulb is connected to the UK 230 V (rms) mains supply. Calculate (a) the rms current; (b) the peak voltage and peak current; (c) the peak instantaneous power; (d) the energy dissipated in 1 hour of continuous operation.

Solution:

(a) I_rms = P / V_rms = 100 / 230 = 0.435 A (rms)

(b) V₀ = V_rms × √2 = 230 × 1.414 = 325 V I₀ = I_rms × √2 = 0.435 × 1.414 = 0.615 A

(c) P_peak = V₀ × I₀ = 325 × 0.615 = 200 W

The peak instantaneous power is exactly twice the mean power. This is a general result for any purely resistive AC load: P_peak = 2 P_mean.

(d) Energy = P_mean × t = 100 × 3600 = 3.6 × 10⁵ J = 360 kJ

Equivalent to 0.1 kWh — the unit the consumer is billed in, currently ~28 p per kWh in the UK as of mid-2026.

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Capacitive and Inductive Loads (Brief Context)

While the AQA A-Level specification focuses on resistive loads, real AC circuits often contain capacitors or inductors. These react to AC in a way that introduces a phase shift between current and voltage:

• Pure capacitor: current leads voltage by π/2 (90°). The instantaneous power oscillates symmetrically about zero, with zero mean — a pure capacitor stores energy half the cycle and returns it the other half, with no net dissipation.
• Pure inductor: voltage leads current by π/2 (90°). Same result — zero mean power.
• Real load (resistive + reactive): the current lags or leads the voltage by some phase angle φ. The mean power is V_rms × I_rms × cos φ, where cos φ is the power factor. A power factor of 1 corresponds to a pure resistor; lower power factors mean some energy is stored and returned each cycle without doing useful work, but the cable I²R losses still depend on the full rms current.

National Grid engineers spend significant effort improving the power factor of large industrial loads (motors, fluorescent lighting), because a low power factor means higher I²R losses for the same useful power delivered. This is beyond the AQA A-Level spec but useful synoptic context for the physics behind grid engineering.

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Three-Phase AC (Brief Context)