Eletricidade e eletromagnetismo · GCSE Physics Topic 4
Simulador da lei de Joule e aquecimento elétrico
Simulador da lei de Joule e efeito Joule: Q = I²·R·t e calor gerado por corrente
- No navegador
- Sem cadastro
- Dedução ao vivo
- Modo projetor
- Folha de exercícios A4
Janela de simulação 3D● 60 FPS
Variáveis de controle do experimento
A
Ω
Ω
s
Deduções e métricas em tempo real
Calor da resistência Q
--
Calor no fio Q
--
Potência da resistência P = I²R
--
Estado do fusível
--
Dedução de fórmulas em tempo real
Fórmula fundamental:
--Substituição de valores:
--Estado físico: --
Exemplos clássicos e armadilhas de exame
Worked Problem
Problema: A 50Ω electric heater is plugged into 220V for 10 minutes. How much heat is generated?
Derivação passo a passo:
- For pure resistor: Q = (U²/R)*t = (48400/50)*600 = 580,800 J ≈ 581 kJ.
Worked Problem
Problema: If a broken electric blanket wire is twisted back together, why does the joint scorch and catch fire?
Derivação passo a passo:
- Twisted joints introduce high localized contact resistance. Equal series current concentrates intense heat via Q=I²Rt, scorching insulation.
Folha de investigação em aula e exercícios
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.
Perguntas frequentes e erros comuns
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.