How many kilowatts can 6 square copper wire and 10 square copper wire pass through with 220 V voltage

How many kilowatts can 6 square copper wire and 10 square copper wire pass through with 220 V voltage

6 square meters can pass 36a, 36 * 220 = 7.9kw. 10 square meters can pass 60A, 220 * 60 = 13kw
What is the role of resistance in the circuit?
Resistance in the circuit, can play the role of voltage and shunt. Adjust the voltage, the role of circuit parameters. For example, resistance step-down starting, star delta starting motor
There are a lot of them, such as common partial pressure and current limiting. Just read more e-books.
It can also generate heat and protect the circuit
When a "220 V, 100 W" bulb works, the resistance is 484 Ω. When a bulb is not working, the resistance should be ()
A. Equal to 484 Ω B. less than 484 Ω C. more than 484 Ω D. cannot be determined
When the 220 V, 100 W "bulb works, the voltage is u = 220 V, the power is p = 100 W, and the resistance is r = U2 P = 220 2100 Ω = 484 Ω. When the bulb doesn't work, the temperature decreases, the resistance decreases, and the resistance will be less than 484 Ω
How many kilowatts can four square copper wires bear
Answer after
What role does resistance play in the circuit
Resistance is a physical property
It doesn't exist to say what role it plays. It plays different roles in different places
Such as limiting the current, or releasing heat, or emitting light and so on
The bulb of "pz220-25" is connected in the circuit of 220 v. what is the resistance through the bulb? (2) what is the resistance of the bulb? (3) what is the current and resistance of the bulb of "pz220 100"? (4) what is the difference between the filaments of the two bulbs? (5) what is the actual power consumption if the first bulb is connected in the circuit of 210 v? (6) what is the actual power consumption if it is connected in the circuit of 230 V?
1. P = U2 / R, so r = U2 / P, then r = 220 (2) / 25 = 19363. I = P / u, so I = 100 / 220 = 5 / 11, r = 220 (2) / 100 = 4844
The first question is about the current‘‘‘
？？
(1)P=UI I=P/U=220/25=8.8A
(2)R=U/I=220/8.8=25Ω
(3)I=P/U=220/100=2.2A
R=U/I=220/2.2=100Ω
(4) The 25W filament is thicker than the 100W filament because the larger the resistance, the smaller the wire radius
Who knows the maximum load of 4 square copper wire is kW
If it is close (within tens of meters), three-phase 10kW, single-phase 3KW
If it is a long distance (several hundred meters), three-phase 3KW or so, single-phase 1kW or so
If it is between short distance and long distance, three-phase 5kW or so, single-phase 2kW or so
What is the role of resistance in the circuit?
In the form of active power loss in the circuit, the resistor is usually connected in series to limit current and divide voltage
What is the filament resistance of a bulb with a rated voltage of 220 V and a rated power of 40 W? If this bulb is connected to a 110 V circuit, the bulb is
What is the filament resistance of a bulb with rated voltage of 220 V and rated power of 40 W? If the bulb is connected to a 110 V circuit, what is the actual power of the bulb?
According to P = u ^ 2 / R, r = u ^ 2 / P = 220 * 220 / 40 = 1210 ohm
P real = u real ^ 2 / r = 110 * 110 / 1210 = 10W
It is 2 times of rated power
1210 ohm, 10W.
R = square of rated voltage / rated power = 1210 ohm; actual power = square of rated voltage / resistance = 10W
1210Ω，10W
Excuse me: 2 kW induction cooker at least with how many square copper wire?
Conservative point 2.5 square