Mean a.c. power and half-wave rectification
Key idea: H2 Physics lessons on drift velocity, electrical energy, sinusoidal supplies and half-wave rectification.
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The core idea
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Big question: How can an alternating supply deliver steady heating or one-way current?
For a resistive load, mean power is VrmsIrms = I²rmsR. A diode in series produces half-wave rectification by blocking one half-cycle, so current becomes unidirectional but pulsating. Distinguish the input a.c. waveform from the load output and account for diode orientation.
Calculate mean power with r.m.s. values
For a pure resistor, instantaneous power is p = vi = i²R. Averaging over a sine-wave cycle gives Pmean = I_rms²R = V_rms²/R = V_rms I_rms. Peak values may be used only after the correct factor of one half is included.
Voltage and current are in phase for a resistor, so both reverse together and their product stays positive. This lesson does not assume the same expression without care for reactive components.
Check your understanding: A 6.0 V r.m.s. supply is connected to 12 Ω. Find mean power.
P = V_rms²/R = 36/12 = 3.0 W.
Follow the conducting path through a diode
A diode conducts mainly in one direction. In a half-wave rectifier it passes one half-cycle and blocks the other, producing a pulsating but unidirectional load current. Reversing the diode passes the opposite half-cycles.
Draw input and output on aligned time axes. The output is not a smaller sine wave: the blocked intervals sit at zero in the ideal-diode model. Rectification changes direction pattern; smoothing would require additional components and is a separate idea.
Check your understanding: Does half-wave rectification produce direct current?
It produces unidirectional, pulsating current. It is not constant unless further smoothing is used.
Key ideas to keep
- Mean voltage and r.m.s. voltage are different quantities.
- Half-wave rectification uses only alternate half-cycles.
- Reversing the diode reverses which half-cycle appears across the load.
See the reasoning
Worked example
Derive mean resistor power before rectifying the waveform
Question: Show why a sinusoidal current produces mean power equal to half its peak power in a resistor.
Step 1: Write instantaneous power
Why: Resistor power depends on current squared.
Working: With i = I₀sinωt, p = i²R = I₀²R sin²ωt.
Step 2: Average over one cycle
Why: The mean of sin² over a complete cycle is one half.
Working: Pmean = ½I₀²R = I_rms²R.
Step 3: Compare with the peak
Why: Peak power occurs when sin²ωt = 1.
Working: Pmax = I₀²R, so Pmean = Pmax/2.
Answer: With i = I₀ sin ωt, p = i²R = I₀²R sin²ωt. The mean of sin² over a complete cycle is ½, so ⟨P⟩ = ½I₀²R = Iᵣₘₛ²R and Pmax = I₀²R.
Check: Power never becomes negative even though current reverses.
Use a hint if needed
Practise with support
Try this
Peak power in a sinusoidally driven resistor is 72 W. State mean power and explain why a single diode does not produce steady d.c.
Hint: Separate direction from constancy.
Check your answer
Mean power is 36 W. The diode blocks alternate half-cycles, so the output is unidirectional but falls to zero for half of every cycle.
Now work without the hint
Practise independently
Your turn
Sketch verbally the input, diode state and load output throughout one cycle of half-wave rectification.
Check your answer
During the forward-biased half-cycle the diode conducts and the load follows that half-sinusoid. During the reverse-biased half-cycle it blocks and load current is zero. The result is one same-polarity pulse per input cycle, not smooth d.c.
Avoid these traps
Common mistakes
Common mistake
Mean power is found by averaging current before squaring it.
What is wrong with this reasoning?
Show better thinking
Instantaneous resistive power is i²R; average sin² over the cycle, giving one half.
Common mistake
One diode converts a.c. into steady d.c.
What is wrong with this reasoning?
Show better thinking
One diode produces a pulsating half-wave output; smoothing would require additional components.
Write for the examiner
Exam guidance
Sketch input and output on aligned time axes and label the conducting half-cycle.
Exam-style practice [5 marks]
A resistor has peak sinusoidal power 200 W. Find mean power and explain how one diode changes a full sinusoid.
Plan before you answer
- Use the sine-wave mean-power relation.
- Trace the diode during both half-cycles.
- Describe output direction and variation separately.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Mean power is 100 W because ⟨sin²⟩ = ½. A single diode conducts for one half-cycle and blocks the other, giving a pulsating half-wave output with one pulse per cycle.
Come back in three days
Check what stayed with you
Recall question
State one difference between half-wave rectified output and steady d.c., and give its pulse frequency for a 60 Hz input.
Check the answer
Half-wave output varies from zero to a peak rather than remaining constant. It has one pulse per input cycle, so its pulse frequency is 60 Hz.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 15 states no explicit exclusions. q in I = nAvq is treated as carrier-charge magnitude when calculating current magnitude; electron drift is opposite conventional current. E.m.f. is energy supplied per unit charge, while p.d. is energy transferred from electrical form per unit charge. The peak/√2 and half-maximum-power results are restricted to sinusoidal waveforms and a resistive load; a single ideal diode gives unsmoothed half-wave rectification.
- GCE A-Level H2 PhysicsTopic 15(h) / Topic 15(j) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027