Электричество и магнетизм · GCSE Physics Topic 4
Симулятор закона Джоуля-Ленца и теплового действия тока
Симулятор закона Джоуля-Ленца и тепловыделения тока: Q = I²·R·t
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Окно 3D симуляции● 60 FPS
Параметры эксперимента
A
Ω
Ω
s
Расчёты и показатели в реальном времени
Тепло нагревателя Q
--
Тепло в проводе Q
--
Мощность нагревателя P = I²R
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Состояние предохранителя
--
Вывод формул в реальном времени
Основная формула:
--Подстановка значений:
--Физическое состояние: --
Разбор типовых задач и анализ экзаменационных ошибок
Worked Problem
Условие задачи: A 50Ω electric heater is plugged into 220V for 10 minutes. How much heat is generated?
Пошаговый вывод формул:
- For pure resistor: Q = (U²/R)*t = (48400/50)*600 = 580,800 J ≈ 581 kJ.
Worked Problem
Условие задачи: If a broken electric blanket wire is twisted back together, why does the joint scorch and catch fire?
Пошаговый вывод формул:
- Twisted joints introduce high localized contact resistance. Equal series current concentrates intense heat via Q=I²Rt, scorching insulation.
Рабочий лист для урока и домашние задания
Q1. Variant 1: Why connect two different resistors in series when investigating heat vs resistance?
Answer Key: Control of variables! Series connection guarantees identical current and duration, isolating resistance as the sole variable.
Q2. Variant 2: Why are fuses made of high-resistivity low-melting-point lead-antimony? Why never copper wire?
Answer Key: Lead alloy melts rapidly at ~200°C under overload. Copper melts above 1085°C and will not blow, causing fires.
Q3. Variant 3: How do pure resistors differ from non-pure resistors (motors) in work vs heat formulas?
Answer Key: Pure resistor: all electrical energy turns to heat (W = Q = UIt = I²Rt). Motor: energy turns mostly to mechanical work (W = UIt > Q = I²Rt); heating must use Q = I²Rt.
Часто задаваемые вопросы и типичные заблуждения
When are Q = I²Rt, Q = UIt, and Q = (U²/R)t interchangeable?
Only for purely resistive loads where all electricity becomes heat. For motors or battery charging, only Q = I²Rt calculates heat correctly.
Why does the grid use ultra-high voltage (hundreds of kV) for transmission?
Power loss is P_loss = I²*R. Boosting voltage 10x drops current to 1/10x, slashing heat dissipation on power lines by 100 times (1/I²).
If a heater element is made of zero-resistance superconductor, does it generate infinite heat?
Zero heat! By Q = I²Rt, R = 0 means heat generation is strictly zero. Superconductors cannot heat, but make perfect loss-less transmission cables.
Why do computer CPUs require thermal paste and fan/water cooling heatsinks?
Billions of switching transistors generate immense Joule heat density. Without heatsinks, silicon temps exceed 100°C, causing thermal throttling or silicon destruction.