Pfc For Flood Light

[AB2005]

Dear all

 

I have some confusion for calculating the active and reactive power. I have a flood light with ballast 400W, 220V, Cosф 0.45. I calculated a capacitor 44uf for PF improvement upto 0.96 but available capacitor was 50uf. first I checked the current of light before PFC was 3.4A. I connected the capacitor then checked and found the following;

Current in main line (from switch to light) = 2A

Current in out going cable from capacitor = 2.3A

Current at light terminals = 3.4A

I couldn’t understand that why the current difference? 2 + 2.3 = 4.3A but we have 3.4A at light terminals. how can we calculate these figures. I think I am missing some thing.

 

[marke]

hello AB2005

 

You can not add the currents together.

The current into the capacitor is at phase quadrature to a resistive current. It is a capacitive current.

 

The current into the lamp is a vector made up of resistive current, and inductive current.

 

The current in the supply is a combination of the resistive component of the lamp current and the inductive current of the ballast and the capacitive current of the capacitor.

 

If you break the lamp current down to ists resistive component and it's inductive component, then you can subtract the inductive component and the capacitive component to end up with a resultant reactive current. The you square the reactive current and square the resistive current, add these together and take the square root to get the supply current.

 

Best regards,

[AB2005]

Thanks Mark for your valuable comments.

 

Now we have the values of current Im=2A at main cable, Ic=2.4A at capacitor out put and Il=3.4A at lamp cable. How can we calculate the both currents Im and Ic while capacitor is 50uf?