Current electricity: charge flow, energy and a.c.
Key idea: Connect microscopic charge transport to circuit energy and power, then distinguish instantaneous, peak and r.m.s. a.c. quantities before analysing resistive power and a single diode.
Before you start: Quantities & Measurement objective chainOscillations objective chain
By the end, you can
- Relate current to charge flow and derive the charge-carrier drift equation.
- Use energy-per-charge definitions and electrical power relationships while distinguishing e.m.f. from p.d.
- Describe sinusoidal a.c. with period, frequency, peak, angular frequency and r.m.s. values.
- Deduce mean resistive power and explain single-diode half-wave rectification.
Starting-point self-check
1. Check your starting point
Attempt all four groups without notes and mark the first charge, energy, waveform or power decision you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Mean a.c. power and half-wave rectification 15(h), 15(j)
Question 1
A sinusoidal current of peak 4.0 A passes through 10 Ω. Find maximum and mean power, then describe the output after one ideal series diode.
Check the model response
Pmax = I₀²R = 160 W and mean power = ½Pmax = 80 W. A single diode passes one polarity of half-cycle and blocks the other, producing a pulsating half-wave output.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Mean a.c. power and half-wave rectification 15(h), 15(j)
Check this idea
Misconception: Mean power is found by averaging current before squaring it.
Repair: Instantaneous resistive power is i²R; average sin² over the cycle, giving one half.
Check this idea
Misconception: One diode converts a.c. into steady d.c.
Repair: One diode produces a pulsating half-wave output; smoothing would require additional components.
worked example
3. Follow four worked models
Follow how each solution fixes carrier direction, the energy-transfer boundary, waveform assumptions or the averaging interval before calculating.
Mean a.c. power and half-wave rectification 15(h), 15(j)
Model 1
Show why a sinusoidal current produces mean power equal to half its peak power in a resistor.
Check the model response
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.
guided practice
4. Guided practice
Use each hint only to select the correct definition, waveform relation or power average.
Mean a.c. power and half-wave rectification 15(h), 15(j)
Question 1
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 the model response
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.
independent practice
5. Independent practice
Solve without repair notes and state the current direction, energy boundary, sinusoidal assumption and load type.
Mean a.c. power and half-wave rectification 15(h), 15(j)
Question 1
Sketch verbally the input, diode state and load output throughout one cycle of half-wave rectification.
Check the model response
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.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Mean a.c. power and half-wave rectification 15(h), 15(j)
Question 1
A resistor has peak sinusoidal power 200 W. Find mean power and explain how one diode changes a full sinusoid.
Check the model response
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.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Mean a.c. power and half-wave rectification 15(h), 15(j)
Question 1
State one difference between half-wave rectified output and steady d.c., and give its pulse frequency for a 60 Hz input.
Check the model response
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.