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Topic: Is this circuit correct? (Read 498 times) previous topic - next topic


Just need to know if this circuit is correct. Making a 4 bulb lamp and needed to wire everything in parallel.


what's up with lamp3 and 4 having 0 ohms of resistance ?
Electrical Engineer


The zero-ohms makes no sense, but this is how lights are wired in parallel.

I can't tell from your drawing what is the power source.  120VDC ??
If the lamps are truly zero ohms, expect lots of smoke.
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Lamps are loads and each load has some kind of Impedance= Resistive + Inductive. In your case, the lamps are resistive loads (with a really small value of resistance) that dissipate energy (this energy converted into heat and light). The zero ohm resistance indicates a short circuit between the positive and negative side and this will leed into a smoky situation.  8)

So if you want to simulate your circuit in terms of power dissipation, current consumption, voltage drop, etc you need to take into account this resistance.

I am not sure what the PWL and CSV stand for but as SteveMan pointed, 120VDC it's a kind of strange. May you need to replace your power source with an AC one?

The parallelization is correct


OPs circuit;

If they are all 120V bulbs then this config will be okay.
Some incandescent bulbs measure near zero ohms when measured with a multi-meter and COLD.
Their resistance increases as they heat up and glow, hence you usually have  HIGH instantaneous switch on current.

Tom... :)
Everything runs on smoke, let the smoke out, it stops running....


what's up with lamp3 and 4 having 0 ohms of resistance ?
Sorry that was just a mistake in the drawing. Disregard the 0 ohms. Also the power source is 120v AC


Disregard the 0 ohms. Also the power source is 120v AC
So disregard "Negative" and "Positive" as well.


The only law for me; Ohms Law: U=R*I       P=U*I
Note to self: "Damn! Why don't you just fix it!!!"


I don't see any reason why that wouldn't work

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