When cables of different specifications are used in parallel, which one heats up first?
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As the title says. Although Clause 3.7.11 of the national standard (GB50217-2007 Code for Design of Cables in Electrical Engineering) does not permit the parallel use of cables of different specifications, there are still many examples of such applications in actual production. I searched Baidu for some related discussions, and generally people believe there will be negative effects. However, most people say that cables with smaller specifications generate a lot of heat, which seems unbelievable to me. Assumption: The supply voltage is 240V; a 240V cable and a 120V cable are used in parallel. If the resistance of the 240V cable is 1 ohm, then the resistance of the 120V cable is 2 ohms. Simple calculation: The current actually flowing through a 240 cable is twice that of a 120 cable. In other words, the current flowing is inversely proportional to the cross-sectional area of the cable. Rated current-carrying capacity of the cable: The thicker the cable cross-section, the lower the current-carrying coefficient. It follows that it will be the cables with thicker cross-sections that first exceed their rated current-carrying capacity, and they will also generate more heat. Why is it said that cables with smaller specifications generate a lot of heat? What do the sea friends think?
How do you learn physics? ! :lol
I really haven’t thought about that; let me think. . . A bit square. . .
This post was last edited by RongshuOfDream on 2017-3-30 at 14:15. The heat generated is given by Q = I2RT (the “2” denotes squaring; superscripts cannot be typed here). Heat Q equals the square of the current multiplied by the resistance, then multiplied by time. I is the current flowing through the wire, R is the resistance of the wire, and T is the duration of time
I suggest you take a close look at what the cable current-carrying capacity actually means
What I mean is that the person whose wire has a higher resistance is more likely to overheat!
How can we use this formula to determine which component has a higher resistance and is therefore more likely to overheat?
How am I supposed to answer that? Then I can only ask you to take a closer look; if you still can’t tell, there’s nothing more I can do! Q is directly proportional to R; can you understand that?
Uh. . . . Thank you for your patience, but I think there might be a misunderstanding on your part; you can calculate it by using the actual values of current and resistance.
Simple calculation: The current actually flowing through a 240 cable is twice that of a 120 cable. In other words, the current flowing is inversely proportional to the cross-sectional area of the cable. Rated current-carrying capacity of the cable: The thicker the cable cross-section, the lower the current-carrying coefficient. It follows that it will be the cables with thicker cross-sections that first exceed their rated current-carrying capacity, and they will also generate more heat. Why is it said that cables with smaller specifications generate a lot of heat? What do the sea friends think? By 240 and 120, you mean the cross-sectional area of the cable, right? The larger the cross-sectional area, the lower the resistance will be for the same length and voltage. Where do you think I misunderstood it?