The electromotive force of the power supply is e = 30V, the internal resistance is r = 1 ohm, and the coil resistance of the DC motor is r = 2 ohm. If the reading of the ideal ammeter is 2a, the voltage at both ends of the motor? The output power of the motor? (motor and er in series)

U motor = E-I * r = 30-1 * 2 = 28V
W motor = w total - W loss = 30 * 2-2 * 2 (1 + 2) = 48W

The total resistance of DC motor is composed of coil resistance and what kind of resistance
DC motor has no impedance, so what is the "what resistance",

According to Faraday's law of electromagnetic induction and the working principle of the motor, a deduction is obtained that the motor coil will produce "self induction" when it moves in the magnetic field. According to Lenz's law, the direction of self induction electromotive force is opposite to the direction of current, and the effect is equivalent to "resistance"

In the circuit shown in the figure, the power supply voltage remains unchanged. When the switch S is from open to close, the ratio of the two readings of the ammeter is 1:5. Therefore, the ratio of resistance R1 to R2 is ()
A. 5：1B. 1：5C. 4：1D. 1：4

It can be seen from the figure that when the switch is off, only R1 is connected to the circuit. When the switch is closed, the two resistors are connected in parallel. When the switch S is off, the ammeter measures the current through R1. When the switch S is closed, the ammeter measures the main circuit current. The power supply voltage remains unchanged. Before and after the switch S is closed, the current through the resistor R1 remains unchanged. The two resistors are connected in parallel, U1 = U2, i1r 1 = i2r2, r1r2 = i2i1 = 41

The electromotive force of motor power supply is constant, the internal resistance R = 2.0, the normal working voltage of motor is 40V, the current is 4a
The electromotive force of the motor power supply is constant, the internal resistance R = 2.0, the normal working voltage of the motor is 40V, the current is 4a, the output power is 120w1, do not omit other resistances, calculate: |. The input power of the motor in normal operation 2. The thermal power of the motor 3. The electromotive force of the power supply

1）P=UI
=40*4
=160w
2) P (heat) = P-P (out)
=160-120
=40w
3) U (source) = u + IR
=40+2*4
=48V

A heavy machine model is shown in the figure, the resistance of the motor coil r = 1 Ω, the mass of the heavy object m = 200g, the electromotive force of the power supply e = 6V, regardless of the internal resistance, the efficiency of the gearbox η = 60%, when the closed switch s reaches a stable state, the heavy object rises at a constant speed, and the current intensity I = 1a in the circuit, the rising speed of the heavy object is calculated

The output power of the motor: P = ie-i2r & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; the solution is p = 5W. According to the conservation of energy: P × 60% = FV & nbsp; and F = mg & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; v = 1.5m/s A: the rising speed of the weight v = 1.5m/s

In the circuit shown in the figure, a and B terminals are connected to the constant voltage power supply. When s is disconnected, the ammeter shows 2a, and when s is closed, the ammeter shows 3a. Then the ratio of resistance R1 to R2 R1: R2= , s ratio of power of R1 under open and closed conditions & nbsp; P1: P1 ′ ．

When s is off, the resistance R1 is directly connected to the power supply, then I1 = ur1 = 2A. When s is closed, three resistors are connected in parallel, and the ammeter measures the total current through resistance R1 and R2, then I2 = ur1 + UR2 = 3A

There is a DC motor with a coil resistance of 40 Ω. When it is connected to a 220 V circuit, its normal working current is 3a. If the friction between the parts during the motor's rotation is not taken into account, how much electric energy is consumed for the motor to work normally for 1 min? (2) the energy generated when the current passes through the coil? (3) what is the mechanical energy converted into?

1 W=UIT=220V*3A*60S=39600J
2q = i2rt = 3A square * 40 Ω * 60s = 21600j
3W machine = w-q = 39600-21600 = 18000j

The coil resistance of a DC motor is 0.5 ohm. It is connected to a 12V power supply, and the current is 2V. How much electric energy does the motor consume after working for 5 minutes? How much heat does the current generate through the coil? (Joule unit)

The current should be 2a, not 2V. The conversion of current into heat is: 2 * 2 * 0.5 * 5 * 60 = 600J. The electric energy consumed depends on the load of the motor. It depends on how much electric energy is converted into mechanical energy. If the resistance of the motor is large, the electric energy consumed is also large. If the motor has no load and no resistance, only the resistance of the coil consumes 600J

DC motor connected to 120V power supply, coil resistance is 20, when the working current is 2, what is the motor power
When what is the working current, what is the maximum output power of the motor

The total power of the motor is:
Ptotal = UI = 120V * 2A = 240W
If the other losses are not considered and only the heat loss is considered, the heat loss power is:
P loss = I ^ 2R = (2a) ^ 2 * 20 Ω = 80W
So the output power of the motor at this time is:
P output = P total - P loss = 240w-80w = 160W

There is a resistance of 100 ohm. If the voltage applied at both ends is 50 V, what is the current passing through it? If the voltage applied at both ends is 0 V, what is the resistance

According to Ohm's Law: I = u / r = 50 / 100 = 0.5 A, if the voltage is 0 V at both ends, the resistance is 100 ohm (resistance is an attribute of conductor and does not change with the change of voltage)

The electromotive force of the power supply is e = 30V, the internal resistance is r = 1 ohm, and the coil resistance of the DC motor is r = 2 ohm. If the reading of the ideal ammeter is 2a, the voltage at both ends of the motor? The output power of the motor? (motor and er in series)