One ampere means one coulomb per second. The same 4.0 C passes another point of that steady loop in the same 2.0 s. If the same charge takes 4.0 s instead:
Teacher-written: a charger supplies a constant 0.90 A for 25 minutes. Find charge in coulombs, then predict the effect of doubling the time at the same current.
In the second part of this practical, connect two equal resistors in series, then in parallel. With the ammeter measuring total current and the voltmeter across the whole combination, measure total potential difference and current and calculate total resistance. Compare with one resistor: series has greater total resistance; parallel has smaller total resistance. For two identical 10 Ω resistors, ideal totals are 20 Ω in series and 5 Ω in parallel. The parallel result can be explained from the doubled total current at the same potential difference, without needing a reciprocal-resistance formula.
探索
One operating point: voltage, current and resistance
$I=\frac{V}{R};\quad P=VI$
At the worked operating point, 6 V across 20 ohms produces 0.30 A and 1.8 W.
This ideal model explains an unlit lamp with a dangerous downstream terminal. Both lamp terminals are at approximately the same potential, so the lamp pd is near zero; either can still have a large pd relative to earth. It is inconsistent to assign 230 V both across this unlit ideal lamp and from each of its terminals to earth in the stated single-phase model.
Worked example — AQA June 2025 8463/1H Q09.2. The question gives pump-motor power 4.86 W and resistance 6.0 Ω and asks for charge flow in 30 minutes. Use the question's prescribed $P=I^2R$ model; this is not a general statement that all electrical input to a real running motor is resistance heating. Power and resistance give current; current and time give charge.
Compare power ratings — teacher-written. Two devices transfer the same 120 kJ of input energy at constant powers 1.0 kW and 2.0 kW. Convert 120 kJ to 120 000 J and kW to W before using $t=E/P$.
At the same run time, the 2.0 kW device transfers twice the energy. For the stated equal input energy, it takes half the time. A greater rating alone does not prove a greater total energy use or total cost for a job: duration and, for useful output, efficiency matter. For the same material and amount of water, a larger useful heating power raises temperature faster when losses are comparable.
2.7
国家送電網
シラバス
高圧送電網（AQA 8463 記述 4.2.4.3）
発電所と消費者をケーブルや変圧器で結ぶシステムとして高圧送電網を説明せよ。
昇圧変圧器は送電電圧を上昇させ、降圧変圧器は家庭用電圧を下げることを述べよ。
P = VI およびケーブルの電力損失 P = I^2 R を用いて、高圧送電網がエネルギー移送において効率的な方法である理由を説明せよ。
出典: Cambridge International シラバス
National Grid は、電力発電所から送電線および変圧器を通じて電気エネルギーを消費者へ伝達する。
Twice the sending pd gives half the current and one quarter of the cable loss. Input power is unchanged; output power increases because less is lost. A current-and-resistance calculation gives the loss, but cannot by itself give efficiency: total input power or energy is also needed.
Sheet 2.7 comparison. At 2000 A through 40 Ω, $P_{\text{loss}}=I^2R=(2000\ \text{A})^2\times40\ \Omega=1.6\times10^8\ \text{W}$. At 500 A through the same resistance, $P_{\text{loss}}=I^2R=(500\ \text{A})^2\times40\ \Omega=1.0\times10^7\ \text{W}$. Current is one quarter, so loss is one sixteenth. Without a stated input, do not claim these losses are a small percentage of the total.
実際の効率計算 — AQA June 2023 8463/1H Q01.5. 入力は34.2 GJ、効率は0.992である。$\eta=E_{\text{useful}}/E_{\text{in}}$ を用いて式を変形してから数値を代入する。ここでは両方のエネルギー単位がGJであるため、Jへの変換は不要である。
Neutral-object extension for sheet 2.8. A charged rod can attract neutral paper because it slightly separates positive and negative charge within the paper. The nearer opposite charges feel stronger attraction than the repulsion of the further like charges. In an insulating wall, bound charges shift slightly; do not assume electrons flow freely through it. Attraction alone does not prove opposite net charges. In the sheet's rod question, all rods are stated to be charged, so the unlike-charge rule applies.
