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.

  • H2 Physics 9478 · 2027
  • Internally reviewed by MiniEducation Team
  • Recorded selected-response study loop available

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.

Potential difference, power and e.m.f. 15(c)–(e)

Question 1

A source supplies 24 J to 4.0 C; a resistor transfers 15 J from the same charge in 5.0 s. Find the e.m.f., resistor p.d. and resistor power.

Check the model response

E.m.f. = 24/4.0 = 6.0 V. P.d. = 15/4.0 = 3.75 V. P = W/t = 3.0 W, also VI because I = Q/t = 0.80 A.

repair

2. Repair the common breaks

Use only the correction matching an error, then retry the corresponding diagnostic.

Potential difference, power and e.m.f. 15(c)–(e)

Check this idea

Misconception: E.m.f. and p.d. are interchangeable names for voltage.

Repair: Both have units J C⁻¹, but e.m.f. describes energy supplied per charge and p.d. describes energy transferred from electrical form per charge.

Check this idea

Misconception: Current is used up when a component transfers energy.

Repair: Charge flow is conserved at a steady junction; the component transfers energy, not 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.

Potential difference, power and e.m.f. 15(c)–(e)

Model 1

A 12 V heater takes 2.5 A. Find its power, resistance and energy transferred in 3.0 min.

Check the model response

P = VI = 30 W. R = V/I = 4.8 Ω, so checks give I²R = V²/R = 30 W. W = Pt = 30(180) = 5.4 kJ.

guided practice

4. Guided practice

Use each hint only to select the correct definition, waveform relation or power average.

Potential difference, power and e.m.f. 15(c)–(e)

Question 1

A 6.0 Ω resistor carries 2.0 A. Find p.d. and power using two power forms.

Hint: Use Ohm's law only to connect the supplied operating values.

Check the model response

V = IR = 12 V. P = VI = 24 W and I²R = 24 W.

independent practice

5. Independent practice

Solve without repair notes and state the current direction, energy boundary, sinusoidal assumption and load type.

Potential difference, power and e.m.f. 15(c)–(e)

Question 1

Distinguish e.m.f. from p.d. using energy, then derive all three resistor-power forms.

Check the model response

E.m.f. is energy supplied by a source per unit charge; p.d. is energy transferred from electrical form per unit charge in a component. P = W/t = (W/Q)(Q/t) = VI. With V = IR, P = I²R = V²/R.

Practice exit check

6. Practice assessment

Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.

Potential difference, power and e.m.f. 15(c)–(e)

Question 1

A cell gives 9.0 J to 1.5 C. A lamp transfers 6.0 J from that charge in 3.0 s. Find e.m.f., p.d., current and power, and state the energy distinction.

Check the model response

E.m.f. = 6.0 V, lamp p.d. = 4.0 V, I = 0.50 A and P = VI = 2.0 W. The source supplies energy per charge; the lamp transfers it from electrical form per charge.

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.

Potential difference, power and e.m.f. 15(c)–(e)

Question 1

A 24 Ω resistor dissipates 6.0 W. Find its current and p.d.

Check the model response

I = √(P/R) = √(6/24) = 0.50 A and V = IR = 12 V.

Continue with established practice

Use the established six-question structured set after the delayed re-test. It directly samples nine outcomes; the mean-power deduction in 15(h) remains assessed only inside this review chain.

Open Current Electricity structured practice