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What is the change of total current and power factor after shunt capacitor in the improvement of fluorescent lamp circuit and power factor in electrical experiment?
After shunt capacitor, the power factor increases, the reactive current decreases, the active current remains unchanged, the total current decreases, and the current loss on the line decreases
In order to improve the power factor of fluorescent lamp circuit, parallel capacitors are needed
When a is connected in parallel with a suitable capacitor, the active component of the current decreases
When B is connected in parallel with a suitable capacitor, the apparent power decreases and the power factor increases
When C is connected in parallel with a suitable capacitor, the active power and power factor increase
The larger the D parallel capacitance, the higher the power factor
Select C. the most important power is the active power. As long as the active power is increased and the reactive power is decreased, the power factor will be increased
In the fluorescent lamp circuit, if a proper capacitor is connected in parallel, will the power factor be improved? Why?
How and why will the active power consumed by the fluorescent lamp change?
This needs to be divided into two situations. If electronic ballast is used, this kind of device basically belongs to capacitive load, and then shunt capacitor will not improve the power factor of lamps. If inductive ballast is used, shunt capacitor with appropriate capacity can advance the lagging circuit current to basic phase with voltage, and reduce reactive power loss, The purpose of improving the power factor of the circuit is achieved. No matter it is electronic or inductive, the shunt capacitor has no change to the active power of the lamp circuit
Why in the experiment of fluorescent lamp installation and power factor improvement, the current becomes larger after shunt capacitor
There is a limit to reactive power compensation. It's not necessary to use the capacitor. Generally, it's good to compensate to 0.98. If you add the capacitor on this basis, it's over compensated. That's just the opposite
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