The electromotive force of the power supply is e, the internal resistance is r, and the power supply supplies power to the variable resistance R. is the following statement about the terminal voltage correct? Why Because u = IR, when R increases, does the terminal voltage also increase

The electromotive force of the power supply is e, the internal resistance is r, and the power supply supplies power to the variable resistance R. is the following statement about the terminal voltage correct? Why Because u = IR, when R increases, does the terminal voltage also increase

This is not strict, because when R increases, I will decrease, strictly speaking, it does not directly indicate that u increases
According to u = E-IR, when R increases, I decreases, IR decreases and e remains unchanged, so u increases

The electromotive force of the power supply e = 10, the internal resistance R = 0.2, the bulb L marked with 8V, 16W just lights up, the coil resistance RO of the motor is 0.15, the motor and the lamp are connected in parallel to calculate the output power of the power supply

The bulb just glows, indicating that the parallel voltage is 8V
Then I = (10-8) / 0.2 = 10A
Bulb current I1 = P / u = 16 / 8 = 2A
Motor current I2 = i-i1 = 10-2 = 8A
Motor power = u * I = 8 * 8 = 64W
Total output power = 64 + 16 = 80W

Given the power supply voltage us = 24 V, the internal resistance ro = 4 Ω, and the load resistance 20 Ω, the current I in the circuit=___ A. Power supply internal resistance voltage drop uo=___ V
Load voltage U = - V

Given the power supply voltage us = 24 V, power supply internal resistance ro = 4 Ω, load resistance 20 Ω, then the current in the circuit I = 1a, power supply internal resistance voltage drop uo = 4 V, load voltage U = 20 v

When a motor with coil resistance of 2 Ω is connected to a 110V circuit and the current passing through the motor is 10a, what is the electric power consumed by the motor? What is the heating power? What is the conversion to mechanical power?

The internal resistance of the motor is 2 Ω, which is connected in the 110V circuit, and the current passing through the motor is 10a, so: Electric Power: P = UI = 110 × 10 = 1100W thermal power: P thermal = I2R = 102 × 2 = 200W mechanical power: P output = P-P thermal = 1100-200 = 900W A: the electric power consumed by the motor is 1100W, the heating power is 200W, and the mechanical power converted into 900W