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How much wire and switch should be used when the total power of 380V power equipment is 60kW?
60kW load selection conductor and switch rated current I = P / 1.732ucos Φ = 60 / 1.732/0.38/0.8 = 60 / 0.53 = 113a, select 150A three-phase main switch, if it is close (within tens of meters), copper wire 25 square mm, aluminum wire 35 square mm
1 Q: add the voltage of 220 V to calculate the power consumed by the electric fan, the power converted into mechanical energy and the heating power
The current is I = P / u = 66 / 220 = 0.3A
(1) The power consumption is 66W
(2) I = 0.8W * 3.8W
(3) Mechanical power 66w-1.8w = 64.2w
1. 66W 2. 1.8w 3. 64.2w
AC: 1.66w 2.1.8w 3.64.2w
If the current through a constant resistance R increases from 0.3A to 0.4A and the consumed electric power increases by 1.4W, then the current through resistance R increases______ A. The voltage at both ends of the conductor increases______ V．
The current increases from 0.3A to 0.4A, and the current through the resistor increases by 0.4a-0.3a = 0.1A; when the current is 0.3A, the power of the resistor is p = I2R = (0.3A) 2R & nbsp; ①; when the current is 0.4A, the power of the resistor is p ′ = I ′ 2R = (0.4A) 2R & nbsp; ②
The output of 1 with the specification of "220v1000w" is 400m, and the wire resistance is 0.0165 ohm per meter. When 1 works in series with resistance R, the actual power is
1 is iodine tungsten lamp
From P = u ^ 2 / R
When the lamp works, the resistance R = u ^ 2 / P = 220 * 220 / 1000 = 48.4 Ω
Wire resistance R = 400 * 0.0165 = 6.6 Ω
Actual current I = u / (R + R) = 220 / (48.4 + 6.6) = 4A
Actual power of lamp p '= UI = I ^ 2 * r = 4 * 4 * 48.4 = 774.4w
A motor is a device that converts electrical energy into mechanical energy. When a motor works normally, the voltage is 220 V and the current through the coil is 10 A. if the coil resistance is 1
0 ohm, what's its electric power?
If the coil resistance is 2 Ω, not 0 Ω, wrong number
P=UI=220v*10a=2200W
The motor is not pure resistance and Ohm's law cannot be used
When 3V is applied at both ends of a resistor, the current is 1a. If the current reaches 3a, the voltage at both ends of the resistor is several v
I = u divided by R, r = 3V divided by 1A = 3 Ω, problem, 3A times 3 Ω = 9V
9V
If the voltage is 2V, r = 6 / 0.3 = 20 ohm, v = 0.1 * 20 = 2V, are you from the queven experiment
9V
3V divided by 1A equals 3 ohm, then 3 ohm multiplied by 3a to get 9V.
I don't know how old people like to ask such questions...
How old are you?
Nine
Safety lamp, specification: 220V 1000W iodine tungsten lamp, constant output voltage of electric control room is 220V, 400m wire is required, the selected wire
The resistance is the negative second power of 1.65 * 10 per meter. The actual circuit is equivalent to the series connection of the iodine tungsten lamp and the resistance. When the iodine tungsten lamp works, its actual power is? Figure: --- lamp L --- resistance R----
--------. 220 V --- [both sides are connected]
1000W
How to calculate the power voltage of 220 V resistance 4.3
A formula or result for calculating power
P=U^2/R=220X220/4.3=11.256KW
This is calculated with r = 4.3 Ω
P=U.U/R!
P=UI I=U/R
P=(U^2)/R
The resistance is written in units
After the voltage at both ends of the resistor changes from 6V to 9V, the current changes by 0.3A. How much does the resistance value and electric power change respectively?
U1-u2 = 9-6 = 3 R = 3 / 0.3 = 10 ohm
P1-P2=9*9/10-6*6/10= 4.5 W
So the resistance is 10 ohm and the electrical power changes by 4.5 W
The resistance value remains unchanged.
Power increased by 0.3 * 3 = 0.9W
Constant resistance, electric power 0.9
In the circuit as shown in the figure, two transmission lines with 0.8 Ω resistance are used to supply power to a "220 V, 1000 W" electric oven. For 10 hours, the power consumption of the two transmission lines is ----- kW &; h, the power consumption of the electric oven is ----- kW &; h, and the heat generated by the wire and the electric oven is ----- J
I can't see the picture, but you've described enough
Electric oven resistance R1 = (u ^ 2) / P = 48.4 Ω, total resistance R = 48.4 Ω + 0.8 Ω * 2 = 50 Ω
I=U/R=220V/50Ω=4.4A
Energy consumption of transmission line q = (2R) Ti ^ 2 = 1.6 Ω * 10h * 19.36a ^ 2 = 0.30976kwh
Energy consumption of electric oven q = r1ti ^ 2 = 48.4 Ω * 10h * 19.36a ^ 2 = 9.37024kwh
Total energy consumption q = 0.30976kwh + 9.37024kwh = 9.68kwh = 34848000j
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