ECE major here that slept through all his power classes. Say I have a basic single-phase generator like the kind you get at Home Depot. If I have long string of irons and light bulbs (essentially purely resistive loads) and draw more than the rated wattage, what happens? My intuition says that the engine must work harder, but what is the specific mechanism that causes this.
In: Engineering
Think of it through conservation of energy. If you want to output more power, then you have to work harder to produce the power.
If you’re running a generator and try to draw more then it is rated for, then it could get hot from the resistance and burn out / catch fire. Another possibility is that the voltage will drop to alow for the increase in current (same power output). Another possibility is that it cuts out because it has sensors that detect that the voltage is dropping.
The reson that the voltage would drop is a very simple proof thankfully. First, the setup:
* For your purely resistive load, V=IR
* By putting lots of lights in parralell, R is made very small, but has a set value independant of your power source
* P=VI, and P has a max (it’s rated wattage), so let’s assume that its running at max.
So we have 2 equations with 2 unknowns (V and I). Solving gives:
I=P/V=V/R => V=sqrt(PR)
V=P/I=IR => I=sqrt(P/R)
You can then see that as the resistance decreases, then the current increases and the voltage decreases.
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