The rated power of an electric fan is 60W, the internal resistance is 2 Ω, and the rated voltage is 220. If it is connected to the power supply, it can't rotate for some reason. Calculate the current and actual electric power

The rated power of an electric fan is 60W, the internal resistance is 2 Ω, and the rated voltage is 220. If it is connected to the power supply, it can't rotate for some reason. Calculate the current and actual electric power

When the motor can not run normally, its work is pure resistance heating, P = I * I * r i = u / r = 110A, P = 24.2KW

The resistance of the motor coil is 10 Ω, which is connected to the home circuit. The current passing through it is 5 A. the efficiency of this motor is high

It is impossible to know the efficiency of a motor only by measuring the DC resistance. The loss of a motor includes: 1. Mechanical loss, bearing friction and air friction of rotor and impeller. 2. Coil resistance loss. It is caused by coil resistance (copper loss). 3. Eddy current loss and hysteresis loss. Eddy current caused by electromagnetic induction on the surface of iron core, and

The power supply of electric vehicle is 36V lead-acid battery. The resistance of motor coil is 1 ohm. The current through motor is 5A
So the electric power consumed by the motor is w. the heat produced in 1 minute is J
Which one of you is right? 2v2 draws

P=UI=36V*5A=180W
Q=I^2Rt=25*1*60=1500J

The voltage of the motorcycle is 36V, the coil resistance of the motor is 1 ohm, and the current of the motor in normal operation is 5A
The voltage of the motorcycle is 36V, the coil resistance of the motor is 1 ohm, the current of the motor in normal operation is 5a, and the power of the motor is 0____ w. The efficiency of motorcycles is_____
Is to find the power of the motor, the efficiency of the motorcycle

Motorcycle motor is to convert electrical energy into mechanical energy, so when calculating its power and work, it should be noted that not all electrical energy is converted into mechanical energy, and a small part is converted into coil heat (coil heating). In the topic, only the extra work of coil heat is considered, so the extra power of current on coil should be calculated first, and then the total power should be subtracted, To sum up, P = P total - P amount = ui-i ^ 2R = 36V × 5A - (5a) ^ 2 × 1 Ω = 155W, efficiency = w real / W total = P / P total = 155W / 180W = 86.1%