The amp draw was most likely below 48 amps (80% rule) but the wire is probably only a 12 awg (which would normally go on a 20 amp breaker). Therefore the amperage draw was not enough to trip the breaker, but too much amperage for the wire causing it to overheat.
@@rachidqir8770Скорее всего автомат исправен, а ток через него недостаточен для отключения (низкое напряжение, сопротивление подводящей цепи велико...). Не зря введена величина "сопротивление цепи «фаза-ноль»". Её измеряют (по крайней мере должны!) перед началом проектных работ. И обязательно - при приёмке объекта, для наиболее удалённых точек. С занесением в протокол.
@@marktupapi5609 why no way? I wire houses and that combination of breakers and wire is kind of common for large electric heaters etc. Obviously the breaker is inadequate to protect the thin wire in the video
@@marktupapi5609 In the United States, the reference for ampacity sizing is contained in the National Fire Protection Association's Publication 70E, or the National Electric Code. 4 AWG copper or 3 AWG aluminum is minimum. 6 stranded has about the same ampacity than 4 solid, but since my job with the government requires me to make repairs that follow code, 4 it is.
"Entertainment" "Now I need to call the electrician" "We don't have one..." "They will charge us thousands!" "Don't call them!" "Hello? My brother has caused a smokey BBQ with our circuit and we need a repair" "Okay that will be €200" *After he sees and fixes it* "Nevermind this was more than I expected so it's going to be €5000" "RUN!" *And they both run away to an orphanage*
Это не короткое замыкание, а лабораторный эксперимент, который наглядно показал, что постоянный ток 100-120 Ампер разогреет до красна медный провод 4 квадратных миллиметра раньше, чем отключится автоматический выключатель.
@@deerhunter1670 Причем здесь сеть? На прибор подано напряжение 1.5-2 вольта с мощного лабораторного источника питания (или может со сварочного инвертора, он подешевле будет)
He’s not putting 220V directly across the wire. He either has one leg of it in series with a load upstream of the breaker) or he’s using a much lower voltage. It’s probably passing 100 something amperes or maybe a few hundred at most judging by how long it’s taking to trip. And the voltage drop across the wire is probably well under 10V
That is a short circuit because theres no resistive load. If no fuse is available the cable will just keep heating up until eventually it breaks. Assuming the rest of the surrounding wnvironment hasnt caught fire by then
@@dogwalker666 the circuit breaker was rated at 60 amps. That coil wasn't drawing that much, so it didn't trip. This illustrates why it's important to size the fuse or circuit breaker correctly, because if it's oversized a fault condition won't blow/trip it.
@@dlg9309 it should have tripped, That cable was dead short for way longer than the tripping curve of even a D rated breaker, A "B" should have gone instantly the fault current was hundreds of amps.
@@berkyt19 если правильно подобраннное по току срабатіваеия по кз ,то не будет никакого горения ,а так как никто не читает а только снимает имеем такой итог .
That's not a short circuit. A short circuit will have disarmed that circuit breaker of 60A. It's because a short circuit has currents in thousands-level (Amperes) in low voltange and hundred-level (A) in medium voltage depending on total Apparent Power (P+jQ). What is happening there is just someone that had put a cable that does not support that level of current (less than 60A for sure). And this cable is burning. How he did that? He has a load in series with that circuit breaker, so the current that goes inside of this load (before circuit breaker) goes inside the cable.
В смысле положенную нагрузку? Тебя не смущает что он не выбивает при кз, при перегреве провода? Этот автомат мусор... А твои познания в электрике удручают😎
Eso paso por qué el diámetro del conductor no estaba calculado para el amperaje del break. Por ejemplo pones un térmico de 32A con un conductor #14 AWG el 14 solo aguanta 15amperios entonces cuando haya una sobrecarga el disyuntor no cumplirá su función por qué el amperaje del disyuntor excede a la del conductor. Y pasará lo del vídeo.
Un cortocircuito activa cualquier interruptor térmico a no ser que esté mismo falle, por otra parte la cantidad de vueltas que tiene el cable lo convierte en una bobina lo cual ofrece una impedancia y sin especificar cómo suministra voltaje es lógico que no explote o no se active, puede estar usando un transformador el cual limita su corriente a la saturación de su núcleo
