When the resistance of a motor coil is r, the voltage at both ends is u and the current is I, the maximum mechanical power output of the motor is () A. UIB. I2RC. UI+I2RD. UI-I2R

When the resistance of a motor coil is r, the voltage at both ends is u and the current is I, the maximum mechanical power output of the motor is () A. UIB. I2RC. UI+I2RD. UI-I2R

The maximum power of the motor is p = UI, the power consumed by the coil is p = I2R, so the maximum mechanical power output is p output = P-P coil = ui-i2r

A light bulb L, marked with "6V, 12W", a DC motor D, its coil resistance is 2 Ω. When the motor works normally, the light bulb will also light normally. When the motor works normally, the light bulb's actual power is 34% of the rated power. Then, when the motor works normally, the power is converted into mechanical energy (assuming that the bulb resistance remains unchanged) （　　）
A. 4.39wB. 275wC. 4wD. 3w

When the motor is connected in parallel with the bulb L, both of them work normally, which indicates that when the rated voltage of the motor is 6V in series, the actual power of the bulb is 34% of the rated power, P = 9W. When the filament resistance of the bulb RL = u, 2p = 3612 = 3 Ω in series, the current I = PRL = 93 = 3A & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; at this time, the motor

In the circuit shown in the figure, the input voltage U is always 12V, the words "6V, 12W" are marked on the bulb L, and the resistance of the motor coil RM = 0.50 Ω. If the bulb can light normally, the correct statement in the following is ()
A. The input power of the motor is 24 & nbsp; WB. The output power of the motor is 12 & nbsp; WC. The thermal power of the motor is 2.0 & nbsp; WD. The electric power consumed by the whole circuit is 22 & nbsp; W

A. The voltage at both ends of the motor U1 = u-ul = 12-6v = 6V, the current in the whole circuit I = plul = 126a = 2A, so the input power of the motor P = u1i = 6 × 2W = 12W. So a is wrong

The small bulb with the specification of "8V 4W" is connected in parallel with the small DC motor (the internal resistance of the coil is r = 0.4 Ω) and connected to the power supply with the electromotive force of 10V and the internal resistance of R = 0.5 Ω. The small bulb just lights up normally
1. Internal voltage of power supply u
2. The total current in the circuit and the current passing through the motor
3. Input power, heating power and output power of motor

1. U inside = E-U outside = 10-8 = 2V
2. I total = u inside / r = 2 / 0.5 = 4A
I lamp = P lamp / u outer = 4 / 8 = 0.5A
I electricity = I total - I lamp = 4-0.5 = 3.5A
3. P in = u out * I power = 8 * 3.5 = 28W
P heat = I electricity ^ 2 * r electricity = 3.5 ^ 2 * 0.4 = 0.49w
P machine = P inlet - P heat = 28-0.49 = 27.51w