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Originally Posted by LauraN
*drags out physics*
OK, in parallel circuits, the total resistance actually decreases the more branches you add. (Weird, I know.) Ohm's Law then tells us that if you decrease the resistance, you increase the current. Then, we also know that power=I2R. So the amount of power you'd be drawing would decrease because you've decreased the resitance, but then increase proportionally the the square of the corresponding increase in current.
In conclusion, I think I've totally confused myself. In terms of safety/starting fires/tripping breakers...it would depend on the specs of the circuit, unless I'm totally missing something (which I probably am, so feel free to correct me.)
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You are absoultly right. Parell branches resistivity can be calculated by taking the inverse sum of the inveresed branch resistivtys. Just keep hitting that 1/x key till you get the answer you expected, as my professor always said.
However, it's not really an issue here. The resistence of each switch when closed is probably 1 or 2 ohms at most. So there is no real power savings putting three switches in there, since the total resistence with all 3 closed wouldn't be much less then with one closed. For all intents and purposes. there is no real diffrence in the resistivity of the circuit, since far and away the major resistence there are the lightbulbs.
In terms of safety, assuming it is wired correctly, you won't have any problems with fires or such. It is a little odd though.