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.
Charge flow, current and drift velocity 15(a)–(b)
Question 1
A 2.4 A current flows for 35 s. Find the charge transferred. Then find the drift speed in a wire with n = 8.5 × 10²⁸ m⁻³, A = 1.2 × 10⁻⁶ m² and carrier-charge magnitude 1.60 × 10⁻¹⁹ C.
Check the model response
Q = It = 84 C. From I = nAvq, v = 2.4/[(8.5 × 10²⁸)(1.2 × 10⁻⁶)(1.60 × 10⁻¹⁹)] = 1.47 × 10⁻⁴ m s⁻¹.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Charge flow, current and drift velocity 15(a)–(b)
Check this idea
Misconception: Current is the amount of charge in a wire.
Repair: Current is the rate of charge flow: I = ΔQ/Δt.
Check this idea
Misconception: Electron drift and conventional current point in the same direction.
Repair: Because electrons are negative, their drift direction is opposite conventional current.
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.
Charge flow, current and drift velocity 15(a)–(b)
Model 1
Derive I = nAvq for carriers crossing a wire section in time Δt.
Check the model response
In Δt, carriers within length vΔt cross area A. Their number is nAvΔt and charge magnitude is ΔQ = nAvqΔt. Therefore I = ΔQ/Δt = nAvq. For electrons, conventional current is opposite their drift direction.
guided practice
4. Guided practice
Use each hint only to select the correct definition, waveform relation or power average.
Charge flow, current and drift velocity 15(a)–(b)
Question 1
The carrier number density doubles at fixed A, q and current. State the drift-speed factor.
Hint: Hold every named quantity except n and v fixed.
Check the model response
From I = nAvq, v is inversely proportional to n, so it halves.
independent practice
5. Independent practice
Solve without repair notes and state the current direction, energy boundary, sinusoidal assumption and load type.
Charge flow, current and drift velocity 15(a)–(b)
Question 1
Explain the meanings and directions in I = nAvq, including why slow drift does not imply a slow circuit response.
Check the model response
n is carrier number per unit volume, A is perpendicular area, v is mean drift speed and q is carrier-charge magnitude. Electron drift is opposite conventional current. Drift is slow, while the electric-field change that establishes current propagates through the circuit much faster.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Charge flow, current and drift velocity 15(a)–(b)
Question 1
A wire carries 0.80 A for 2.0 min. Find Q. With n = 6.0 × 10²⁸ m⁻³, A = 0.50 mm² and q = 1.60 × 10⁻¹⁹ C, find v.
Check the model response
Q = It = 96 C. A = 5.0 × 10⁻⁷ m² and v = I/(nAq) = 1.67 × 10⁻⁴ m s⁻¹.
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.
Charge flow, current and drift velocity 15(a)–(b)
Question 1
At fixed n and q, the wire area triples and drift speed halves. State the current factor.
Check the model response
I = nAvq, so the factor is 3 × ½ = 1.5.