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In order to adjust the speed of DC motor, a variable resistor is often connected in series. In the circuit diagram, M is a small DC motor, which is marked with "12V, 24W", and the power supply voltage is 20V. When the motor works normally, (1) what is the resistance value of the variable resistor? (2) What is the total power supplied by the power supply and the power consumed by the variable resistor?
It can be seen from the circuit diagram that the motor is connected in series with the sliding rheostat, and the voltmeter measures the voltage at both ends of the sliding rheostat. (1) the voltage of the motor in normal operation is 12V, and the power is 24W, ∵ the current in the series circuit is equal, ∵ according to P = UI, the current in the circuit: I = pmum = 24w12v = 2A, ∵ the total voltage in the series circuit is equal to the sum of the partial voltages, ∵ the current at both ends of the rheostat Voltage: ur = u-um = 20v-12v = 8V, according to Ohm's law, the resistance value of variable resistor: r = URI = 8v2a = 4 Ω; (2) the total power of power supply: P = UI = 20V × 2A = 40W, the power consumed on variable resistor: PR = URI = 8V × 2A = 16W
For a DC motor, the resistance of its rotor coil is r, which is connected to the power supply with the voltage set as U. when the motor is in normal operation, the current through the coil is I, and the mechanical power output of the motor is ()
A. UIB. I2RC. UI+I2RD. UI-I2R
The input power of the motor is p = UI & nbsp; & nbsp; the internal heating power is I2R & nbsp; & nbsp; & nbsp; P output = P-P heat = ui-i2r & nbsp; a & nbsp; UI is the total power. So a error & nbsp; & nbsp; B & nbsp; & nbsp; I2R is the internal heating power. So B error & nbsp; C is the total power plus heating power
As shown in the figure, the electromotive force of the power supply is 12V, the internal resistance is 1 Ω. R1 = 1 Ω. R2 = 6 Ω, and the coil resistance of the motor is 0.5 Ω. If the switch is closed, the current passing through the power supply is 3a. Calculate: (1) the electric power consumed on R1; (2) the mechanical power output by the motor; (3) the efficiency of the power supply
(1) The electric power consumed by R1 is: P1 = i2r1 = 32 × 1W = 9W (2) voltage at both ends of R1: U1 = IR1 = 3 × 1V = 3V & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; voltage at both ends of R2: U2 = e-uine-u1 = 12-3 × 1-3v = 6V & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; current through R2: I2 = u2r2 = 66A = 1A & nbsp; current through motor: Im = i-i2 = (3-1) a = 2A, um = U2 = 6V & nbsp; & nbsp; & nbsp; The electric power consumed by the motor is: PM = imum = 2 × 6W = 12W, output power P = pm-im2r = 12-22 × 0.5 = 10W (3) the efficiency of the motor is η = P PM × 100% = 1012 × 100% = 83%). Answer: (1) the electric power consumed by R1 is 9W; (2) the electric power consumed by the motor is 10W; (3) the efficiency of the motor is 83%
In the circuit as shown in the figure, the electromotive force of the power supply e = 12V, the internal resistance R = 1 Ω, R1 = 1 Ω, R2 = 6 Ω, and the coil resistance of the motor RM = 0.5 Ω. If the current passing through the power supply after closing the switch is 3A, calculate the output power of the motor. There is no figure. I describe that R1 is on the trunk road. R2 is in parallel with the motor, and then in series with R1!
Total power P = EI = 36W;
The power consumption of internal resistance is 9W
The power of R1 is p = I * I * r = 9W;
Parallel part voltage U = e-u1-u2 = 12-3-3v = 6V
So the current of R2 is I = 1a and the power is 6W
Therefore, the total power consumed by the motor is 18W
In this 18W, the thermal power is 2W
So the output power of the motor is 16W
RELATED INFORMATIONS
- 1. The electromotive force of the power supply is 12 V, the internal resistance is 1 ohm, R1 = 1, R2 = 6, R1 and R2 are in series, the motor and R2 are in parallel, and the motor resistance is 0.5 ohm The current in the closed circuit is 3a. Calculate the electric power consumed by the motor
- 2. The physical heat is generated. A motor can work normally. The voltage at both ends of the motor is 380V, the coil resistance is 2 Ω, and the current through the coil is 10A. The electric energy consumed by the motor in normal operation for 1s is__ J. What is the heat generated by the coil resistance__ J Don't be too detailed, just a simple formula, Voltage at both ends of coil when a motor works normally
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- 11. 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
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