HID Ballast Circuit - Kirchoffs Law
HID Ballast Circuit - Kirchoffs Law
(OP)
Hi All,
its understood that HId lamps operate at 110V while its running, eventhough the i/p supply voltage is 230V...........the remaining voltage being dropped across the ballast.
Now what confuses , is the current drawn by the lamp and the input current.
for example.. if the lamp wattage is 1000W, input current to the circuit will be 1000/230V (neglecting losses,p.f etc).the current drwn by the lamp should be 1000/110v (if im not mistaken)..........
but it makes no sense as one end of the lamp remains connected to the neutral,........
Going back to basics what Kirchoffs law says is current entering in the circuit should be equal to current leaving the circuit.........
its understood that HId lamps operate at 110V while its running, eventhough the i/p supply voltage is 230V...........the remaining voltage being dropped across the ballast.
Now what confuses , is the current drawn by the lamp and the input current.
for example.. if the lamp wattage is 1000W, input current to the circuit will be 1000/230V (neglecting losses,p.f etc).the current drwn by the lamp should be 1000/110v (if im not mistaken)..........
but it makes no sense as one end of the lamp remains connected to the neutral,........
Going back to basics what Kirchoffs law says is current entering in the circuit should be equal to current leaving the circuit.........





RE: HID Ballast Circuit - Kirchoffs Law
RE: HID Ballast Circuit - Kirchoffs Law
Kirchoffs law doesn't apply because the transformer makes two separate isolated circuits, thus the current can be different.
Steve
RE: HID Ballast Circuit - Kirchoffs Law
RE: HID Ballast Circuit - Kirchoffs Law
Yes, Cbarn the ballast is just a choke in this case, the ordinary reactor type not the Autotransformer type.
If PF is considered as Steve suggests
For a PF of 0.85, Lamp wattage = 1000W, Ballast loss = 50W
Input Current, Ip = (1000 + 50)/(230*0.85) = 5 .37 A
As HID lamps operate at 110V whole they are working,
Lamp Current, Ic = 1000/ 110 = 9 A
Which is contradicting as the circuit is a series one……
RE: HID Ballast Circuit - Kirchoffs Law
RE: HID Ballast Circuit - Kirchoffs Law
pls. note these arent measured values...im going by the theory....
RE: HID Ballast Circuit - Kirchoffs Law
RE: HID Ballast Circuit - Kirchoffs Law
"if the lamp wattage is 1000W, input current to the circuit will be 1000/230V "
This is incorrect. Nothing in the ballast or lamp operates at 1000W and 230V. You are taking a wattage from the lamp and the voltage from the input to the ballast. You can't use the voltage at one part of a circuit and the power at another part to find current.
Steve
RE: HID Ballast Circuit - Kirchoffs Law
RE: HID Ballast Circuit - Kirchoffs Law
There are three power factors: the Displacement Power Factor, the Distortion Power Factor and the Total Power Factor.
The former is defined to be cos(phi1) where phi1 is the angular displacement between voltage fundamental and current fundamental.
The total power factor is defined as P/S where P is power and S is appearent power. The symbol for it is often the greek letter Lamda (looks like an inversed "y", which can be seen on ballasts.
If power factor is being used for "conversion efficiency", then that particular manufacturer does not know what he is doing.
RE: HID Ballast Circuit - Kirchoffs Law
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