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How many kilowatts of electrical appliances can 4 square millimeter copper wire use
Basic appliances are OK
How many kilowatt water heaters can 4 square lines be used in
Copper wire
8000w air switch with leakage protection over 40A
How many kilowatts of household appliances can a 1.5 square millimeter copper core drive at most?
If calculated by "pure resistance load", the current per 1kW should be about 4.5a, so if it is a single-phase 220 V household line, 1.5 square millimeter copper core wire, it can pass the rated current of about 1-1.5kw power. However, for safety, line resistance loss and other problems, it is recommended that the power should not exceed 1kW
There is a copper wire with a cross-sectional area of S, and the current flowing through it is I. suppose that there are n free electrons in each unit volume of the wire, and the charge quantity of the electrons is Q, then the directional moving speed of the electrons is v. in △ t time, the number of free electrons passing through the cross-section of the wire can be expressed as ()
A. nvS△tB. nv△tC. I△tqD. I△tSq
In the time of △ T, the length of the electron moving at the velocity V in the copper wire is v △ T. because the cross-sectional area of the copper wire is s, the volume of the wire is v △ ts in the time of △ T. and because there are n free electrons in the unit volume of the wire, the number of free electrons passing through the cross-section of the wire in the time of △ t
Why is the size of the resistance related to the thickness of the material
The current is originally the directional movement of many electrons, so many electrons pass through a resistance. If the resistance is small, the space for electrons to pass through is small, the amount of electrons passing through per unit time is less, and of course, the resistance is large. If I haven't learned the formula, I think it's relatively simple
According to r = resistivity * (L / s), I don't know if you have learned that there are no different materials with different resistivities (the symbol is similar to the density of the mobile phone). The resistance is not only related to the size s of the cross-sectional area, but also related to the length l of the resistance. It is directly proportional to the length of the resistivity and inversely proportional to the cross-sectional area
How many kilowatt hours is one joule
1 kwh = 3 600 000 joules
Urgent: how to connect the 380V electricity to the 220V AC contactor?
380V three live lines, you connect a zero line, any one of the live line and the voltage between the zero line is 220V
220 V contactor, then contacts A1 and A2 of contactor are connected to 220 v,
There is a copper wire with a cross-sectional area of S, and the current flowing through it is I. suppose that there are n free electrons in each unit volume of the wire, and the charge quantity of the electrons is Q, then the directional moving speed of the electrons is v. in △ t time, the number of free electrons passing through the cross-section of the wire can be expressed as ()
A. nvS△tB. nv△tC. I△tqD. I△tSq
In the time of △ T, the length of the electron moving at the speed of V in the copper wire is v △ T. because the cross-sectional area of the copper wire is s, the volume of the wire is v △ ts in the time of △ T. because there are n free electrons per unit volume of the wire, the number of free electrons passing through the cross-section of the wire can be expressed as NVS △ t in the time of △ T, so a is correct If the current is I, the amount of charge flowing through the wire is I △ t in △ t time, and the amount of charge of the electron is Q, then the number of free electrons passing through the cross section of the wire in △ t time can be expressed as I △ TQ, so C is also correct
When the two wires are connected in series, the total resistance is 2 Ω and the total power is 6 v?
P=U^2/R=6*6/(3+2)=36/5=7.2W
According to P = u ^ 2 / r = 6 * 6 / 5 = 7.2W
7.2 watts
Square of 6 divided by 3 + 2 = 7.2 w
In series, the total resistance is 2 + 3 = 5 Ω
The total power is 6 * 6 / 5 = 7.2W
How many joules is a kilowatt hour
1kW · h is what we call one degree of electricity, 1kW · H = 3.6 * 10 to the sixth Joule
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