Hello WorldwideHIDCollectorUSA,
basically funkybulb gave you the answer however he took one thing out of consideration. 120x15A = 1800VA, VA being volt amps. You can have a "load" of 1800VA and drawing a current of 15A and still have not a single Watt of power dissipation. Possible is this for example with an (ideal) inductor or a large capacitor. If you ad inductance or capacitance in an AC Circuit you creat reactive power and blind current. reactive current is an current. This current is flowing thruw you conductors between it sources and the effecting component, it is flowing with all the consequence, and inducing loses and wear on the system, however it cannot be used to do work. Your ballast is also producing reactive current, this is why you put capacitors in parallel to the input (line), to compensate these reactive currents. So the question is how much reactive current does your ballast produce? Well thats what the power factor says to you, that information is missing.
However i don't think that the current flowing to the line to your lamp (The geometrical addition of reactive and real power) will reach 15A. But I can't tell.
The second thing, a circuit breaker will not break the circuit immediately if it is slightly overloaded. Every breaker has a time current chart wich shows how long an overload must be present until a trips. If you would draw a few amps over it creating it can still take 30min to several hours according to its exact rating and timecurve. Is it good to overload a cable even for short time? Certainly not. So think carefully what your doing, your responsible for it.
Best regards Alex
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