If the transformer capacity PS is known and the power factor is 0.8, the reactive power is () COSφ=P/S P=S×COSφ=0.8S Q=√(S²-P²)=√[S²-(0.8S)²]=0.6S Reactive power is (0.6s) How is 0.6 calculated?

S²-(0.8S)²
=(S-0.8S)(S+0.8S)
=0.2S×1.8S
=0.36S²
If you square it again, you get 0.6s

Given the active power and reactive power, calculate the power factor

Apparent power s = (square of active power P + square of reactive power q) re square
And the power factor cos φ = active power P / apparent power s

Transformer rated capacity PS, power factor 0.8 to calculate the rated power

If the rated capacity of the transformer is Sn (usually apparent capacity in kVA) and the power factor is p = 0.8, then the rated power (in kW) PS = pxsn = 0.8xsn

How to calculate the power factor when the power meter is connected to the circuit with inductance and resistance in series?

The current coil of the power meter is connected in series with the inductance and resistance circuit, and the voltage coil is connected at both ends of the circuit. The apparent power of the circuit is equal to the voltage multiplied by the current, that is, s = u × I, the reading of the power meter is the active power P, and the power factor cos φ = P △ s

In sinusoidal alternating current, what are the power factors of pure inductance, pure resistance and pure capacitance? Why?

The power factor of pure resistor load is 1. The power factor of pure inductor is equal to 0. The power factor of pure capacitor is also 0. The power factor is the ratio of active power to apparent power. When pure inductor passes through alternating current, there is only reactive power, and the active power is zero. Therefore, the power factor of pure inductor is equal to 0

What's the power factor of pure resistance circuit? What's the power factor of pure inductance circuit? What's the power factor of pure capacitance circuit? Help~

Resistance element circuit, or inductance and capacitance elements, but their impact on the circuit can be ignored. Voltage and current are in the same frequency and phase. Resistance will get all the energy from the power source into internal energy, this kind of circuit is called pure resistance circuit. For example: electric lamp, electric iron, iron, and so on, they are just heating

A series circuit with inductive load, when providing 220 V AC power supply voltage, the current is 10 A and the power factor is 0.8, what is the apparent power of the circuit
Please give the formula and law

The apparent power of the circuit is s = UI, and 220 * 10 = 2200va. Because the current in the series circuit is equal everywhere, and the total voltage is equal to the sum of the voltages on each load, the total apparent power is equal to the sum of the apparent power on each load

How much capacitance is needed to increase the power factor of u = 220 V, P = 40 W, cos φ = 0.5 to 0.9

P = 40W, cos φ = 0.5 by P = s × cos φ, so s = P / cos φ = 40 / 0.5 = 80 (VA), q = √ (S & # 178; - P & # 178;) = √ (80 & # 178; - 40 & # 178;) = 40 √ 3 (VaR)

The power factor of the circuit can be improved by shunt capacitor of inductive load. Why not use series capacitor to improve the power factor of inductive load?

Parallel circuit can put capacitor into operation without disconnecting the original circuit. Series circuit can only put capacitor into operation with disconnecting the original circuit. Do you think the power supply circuit can disconnect the circuit at any time

If inductive load adopts series capacitor method to improve power factor, does the reactive power provided by power supply change? Why?

First of all, a practical case method: inductive load commonly uses shunt capacitor to improve power factor, rarely series capacitor
When the reactive power of the load is compensated by capacitor, the power factor is increased. The change of reactive power depends on the type of power supply
If the power source is a generator, as long as the excitation current does not change, the output reactive power of the generator is constant. Reason: the output reactive power of the generator is determined by the excitation
If the power source is a transformer, the output reactive power of the transformer will be reduced if the power factor of the load is increased, and vice versa. Reason: the total capacity of the transformer is determined by the active power and reactive power. If the reactive power output is less, the active power output can be more

If the transformer capacity PS is known and the power factor is 0.8, the reactive power is () COSφ=P/S P=S×COSφ=0.8S Q=√(S²-P²)=√[S²-(0.8S)²]=0.6S Reactive power is (0.6s) How is 0.6 calculated?