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.
Sinusoidal a.c., peak and r.m.s. values 15(f)–(g), 15(i)
Question 1
For v = 170 sin(100πt) V, state the peak voltage, angular frequency, frequency, period and r.m.s. voltage.
Check the model response
V₀ = 170 V, ω = 100π rad s⁻¹, f = ω/(2π) = 50 Hz, T = 1/f = 0.020 s and Vᵣₘₛ = V₀/√2 = 120 V.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Sinusoidal a.c., peak and r.m.s. values 15(f)–(g), 15(i)
Check this idea
Misconception: The r.m.s. value is the arithmetic mean of a sinusoid.
Repair: A full sinusoid has zero arithmetic mean; r.m.s. is the d.c.-equivalent value for mean resistive power.
Check this idea
Misconception: Peak/√2 applies to every alternating waveform.
Repair: The stated peak/√2 relation is for a sinusoidal current or voltage.
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.
Sinusoidal a.c., peak and r.m.s. values 15(f)–(g), 15(i)
Model 1
A 60 Hz sinusoidal current has Iᵣₘₛ = 3.0 A and is zero, increasing positively, at t = 0. Write i(t).
Check the model response
I₀ = √2 Iᵣₘₛ = 4.24 A and ω = 2πf = 120π rad s⁻¹, so i = 4.24 sin(120πt) A.
guided practice
4. Guided practice
Use each hint only to select the correct definition, waveform relation or power average.
Sinusoidal a.c., peak and r.m.s. values 15(f)–(g), 15(i)
Question 1
A sinusoidal voltage has period 8.0 ms and peak 20 V. Find f, ω and Vᵣₘₛ.
Hint: Convert milliseconds before taking the reciprocal.
Check the model response
f = 1/T = 125 Hz, ω = 2πf = 785 rad s⁻¹ and Vᵣₘₛ = 14.1 V.
independent practice
5. Independent practice
Solve without repair notes and state the current direction, energy boundary, sinusoidal assumption and load type.
Sinusoidal a.c., peak and r.m.s. values 15(f)–(g), 15(i)
Question 1
Define period, frequency, peak and r.m.s. value, and explain the physical meaning of r.m.s.
Check the model response
T is time per cycle, f = 1/T is cycles per second, and peak is maximum magnitude. The r.m.s. current or voltage is the steady d.c. value producing the same mean power in a resistor; for a sinusoid it is peak/√2.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Sinusoidal a.c., peak and r.m.s. values 15(f)–(g), 15(i)
Question 1
A 50 Hz supply has Vᵣₘₛ = 230 V. Find V₀, T and ω, then write v(t) for zero phase.
Check the model response
V₀ = √2(230) = 325 V, T = 0.020 s, ω = 100π rad s⁻¹ and v = 325 sin(100πt) V.
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.
Sinusoidal a.c., peak and r.m.s. values 15(f)–(g), 15(i)
Question 1
For i = 5.0 sin(400πt) A, find f and Iᵣₘₛ.
Check the model response
ω = 400π rad s⁻¹, so f = 200 Hz. Iᵣₘₛ = 5.0/√2 = 3.54 A.