Alternating Current

Understand peak and rms values, power in AC circuits, and how rectification and smoothing work.

AQA A-Level Physics · Section 7 · AC Circuits

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Define peak and rms voltage and current

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Use V_rms = V₀/√2 and I_rms = I₀/√2

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Calculate power in AC circuits

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Explain half-wave and full-wave rectification

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Describe smoothing using a capacitor

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Sketch and interpret AC waveforms

AC Waveforms & Peak Values

An alternating current (AC) is one that periodically reverses direction. The voltage and current vary sinusoidally with time:

V = V₀ sin(2πft) I = I₀ sin(2πft)

Where:

V₀ = peak voltage (V)
I₀ = peak current (A)
f = frequency (Hz)
t = time (s)

The peak value is the maximum value reached during one cycle. The UK mains supply has a peak voltage of approximately 325 V and frequency of 50 Hz.

RMS Values

The root mean square (rms) value is the equivalent DC value that would deliver the same power to a resistor.

V_rms = V₀ / √2 I_rms = I₀ / √2

These can be rearranged:

V₀ = V_rms × √2 I₀ = I_rms × √2

UK mains: V_rms = 230 V, so V₀ = 230 × √2 ≈ 325 V

Voltmeters and ammeters in AC circuits display rms values. The factor 1/√2 arises from averaging sin²(θ) over a full cycle, which equals 1/2.

Power in AC Circuits

For a purely resistive AC circuit, the mean (average) power is:

P = V_rms × I_rms = I_rms² × R = V_rms² / R

The instantaneous power varies between 0 and a maximum of V₀I₀:

P_max = V₀ × I₀ = 2 × P_mean

Quantity	DC equivalent	AC rms
Voltage	V	V_rms = V₀/√2
Current	I	I_rms = I₀/√2
Power	P = VI	P = V_rms × I_rms

Power formulae using peak values give the wrong answer for mean power. Always convert to rms first, or use P = ½ V₀I₀.

Rectification

Rectification converts AC to DC by allowing current to flow in one direction only, using diodes.

Half-wave rectification: A single diode blocks the negative half-cycles. Only positive half-cycles pass, giving a pulsed DC output.

Full-wave rectification: A bridge rectifier (four diodes) inverts the negative half-cycles, so both halves of the AC waveform contribute to the output. The output frequency is doubled — 100 Hz for 50 Hz mains.

A bridge rectifier consists of four diodes arranged so that current always flows through the load in the same direction, regardless of the polarity of the AC input.

Smoothing with a Capacitor

The pulsed DC output from a rectifier is smoothed by connecting a large capacitor in parallel with the load resistor.

The capacitor charges when the rectified voltage rises
The capacitor discharges slowly through R when the voltage falls
This reduces the ripple (fluctuation) in the output

Ripple ∝ 1 / (f × C × R)

A larger capacitance C or larger resistance R gives less ripple and a smoother output. Higher frequency also reduces ripple — another advantage of full-wave over half-wave rectification.

If C is too small or R too small (heavy load), the capacitor discharges too quickly between cycles and ripple becomes significant.

Example 1: Finding rms voltage from peak voltage

The peak voltage of an AC supply is 340 V. Calculate the rms voltage and the mean power delivered to a 50 Ω resistor.

1 Find V_rms: V_rms = V₀ / √2 = 340 / 1.414 = 240 V

2 Find mean power: P = V_rms² / R = (240)² / 50 = 57 600 / 50

P = 1152 W ≈ 1150 W

Example 2: Peak current from rms power

A resistor of 200 Ω dissipates a mean power of 4.0 W when connected to an AC supply. Find the peak current.

1 Find I_rms: P = I_rms² × R → I_rms = √(P/R) = √(4.0/200) = √0.02 = 0.141 A

2 Find I₀: I₀ = I_rms × √2 = 0.141 × 1.414

I₀ = 0.200 A

Example 3: Comparing AC and DC heating

An AC supply with V_rms = 12 V is connected to a 6.0 Ω resistor. What DC voltage would deliver the same mean power?

1 Mean power from AC: P = V_rms² / R = (12)² / 6.0 = 144 / 6.0 = 24 W

2 For DC to give same power: V_DC² / R = 24 → V_DC² = 24 × 6.0 = 144

V_DC = 12 V — confirming that rms voltage IS the DC equivalent

Example 4: Smoothing capacitor ripple

A full-wave rectifier outputs 100 Hz pulsed DC. A 470 μF capacitor is connected in parallel with a 1.0 kΩ load. Estimate whether smoothing will be effective.

1 Time constant τ = RC = 1000 × 470 × 10⁻⁶ = 0.47 s

2 Period of ripple: T = 1/f = 1/100 = 0.010 s

3 Compare: τ = 0.47 s >> T = 0.010 s

τ >> T, so the capacitor discharges very little between cycles — smoothing is highly effective.

Q1. What is the relationship between peak voltage V₀ and rms voltage V_rms for a sinusoidal AC supply?

Q2. The UK mains supply has V_rms = 230 V. What is the peak voltage?

Q3. A sinusoidal AC supply with peak voltage 20 V is connected to a 100 Ω resistor. What is the mean power dissipated?

Q4. What does a diode bridge rectifier achieve that a single diode cannot?

Q5. To reduce the ripple on a rectified supply, you should increase which component?

Challenge 1. An AC supply has a frequency of 50 Hz and delivers a mean power of 500 W to a 25 Ω resistor. Calculate (a) the rms current, (b) the peak current, and (c) the peak voltage.

Challenge 2. Explain why the mean power in a purely resistive AC circuit equals ½ V₀I₀. Your answer should reference the mathematical averaging of the power waveform.

Challenge 3. A full-wave rectifier supplies a 2.2 kΩ load. A student uses a 100 μF capacitor to smooth the output at 100 Hz ripple frequency. Calculate the time constant and comment on the quality of smoothing. What capacitance would reduce the ripple voltage to less than 1% of the peak?

Challenge 4. A transformer steps down 230 V rms mains to 12 V rms. After full-wave rectification and smoothing, the output is approximately 12 × √2 V. Explain why the smoothed DC output is close to the peak AC voltage rather than the rms voltage, and calculate this output voltage.