The function of resistance and capacitance

The function of resistance and capacitance

Purpose of resistor: to block the passage of current: to divide voltage, reduce voltage, limit current, load, shunt, area distribution, etc. in the circuit. Capacitor is a kind of energy storage element, which can separate direct current and AC flow. It can filter, couple, bypass, tune and convert energy in the circuit. Inductance: to pass DC, resistance, etc. in the circuit

What is the function of a capacitor and a resistor in series? What is the principle?

Peak absorption circuit
Because the inductor has the function of energy storage, when the switch is cut off, there is no place to release the energy stored by the inductor. When the switch is turned on again, the electric energy stored by the inductor is added to the switch together with the power supply, resulting in high voltage on the switch and breakdown of the switch
With the peak absorption circuit, the inductor releases energy through the peak absorption circuit during the switch off period

Capacitor and resistor in series
For example, in the simplest series circuit, a known capacitor is connected in series with a known resistor, and the supply voltage is known. How to calculate the charge on the capacitor?

Since the capacitor is connected to the power supply voltage, then u is constant. According to C = q / u, q = Cu can be known. Generally, given the capacitor C, then it is easy to know the charge quantity Q

Resistor and capacitor in series
There is a circuit, which is a power braking electric cabinet (energy consumption braking). The three-phase power supply is connected to the three-phase rectifier line through the air switch - fuse - module transformer. At the same time, the wire series resistance and capacitance are led out at one end of the fuse and return to the primary side of the transformer. What is the role of the line with series resistance and capacitance?

High order harmonic elimination by RC filter

The rated voltage is 380V, the rated current is 45A, and the rated power is 22KW. Is the power factor of this motor calculated according to the formula P = u * I * 1.732 * power factor
So, some p = u * I * 1.732 * power factor * efficiency, and some p = u * I * 1.732 * power factor is not multiplied by efficiency, which formula is more reliable?
According to the rated voltage of 380V, rated current of 45A and rated power of 22KW, is the power factor of this motor calculated according to the formula P = u * I * 1.732 * power factor, and the power factor obtained is the rated power factor, right? Is there so many words about the actual power factor?
What is the actual power consumption of my motor per hour?

P = u * I * 1.732 * power factor * efficiency, so according to the data you know now, you can't calculate the power factor
According to your parameters, the power factor of 22KW motor is 0.9 for 2-pole motor, 0.86 for 4-pole motor, 0.83 for 6-pole motor and 0.78 for 8-pole motor

The rated power P = 1.5KW; rated voltage U = 380V; rated power factor cos = 0.88; rated efficiency 88%; maintenance
After measuring the no-load current is 9a, try to judge whether its no-load current is qualified
Write the calculation steps

Y160m2-2 motor's rated power P = 15kw, current I ≈ 30a, no-load current is 20 ~ 30% of the rated current (data of electrician's manual), measured no-load current is 9a, qualified!

It is known that the rated power of three-phase asynchronous motor is 55kW, the rated voltage is 380V, the rated power factor is 0.89, and the rated efficiency is 91.5%
Calculate the rated current of the motor

Rated current of three-phase motor = rated power (W) / (1.732 × rated voltage × power factor × efficiency)
About 103 a

The rated power, voltage and current of three-phase asynchronous motor are 200kW, 1140V and 90a respectively?

The rated power of motor is p = 1.732 * u * I * cos φ * η, so cos φ * η = P / 1.732/u/i = 200 / 1.732/1.14/90 = 1.13
From the calculation results, cos φ * η > 1 is just impossible, which means that the output power is greater than the input power. Therefore, the problem is wrong. According to the given rated voltage and rated current, even if the power factor is 1 and the efficiency is 1, the input power is only 177.7kw

Why the actual power of electrical appliances is less than the rated power
The physics teacher said that the rated power on the appliance is useless, and the actual power is always less than the rated power. Why

Because there is no work

What if the actual power of the consumer is higher than the rated power?

It will burn out the electrical appliances. If you are at home, it will blow out the fuse or trip