Go Down

Topic: Is this circuit correct? (Read 488 times) previous topic - next topic

V4D3R

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

KASSIMSAMJI

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

SteveMann

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.
Fritzing pictures are NOT schematics. I don't speak Fritzing.

Please do not ask for help by PM. I will not respond. If you need help, post a question on the appropriate forum.

Click on Add Karma if I helped you.

Nikosant03

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

TomGeorge

Hi,
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....

V4D3R

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

GaryP

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

Cheers,
Kari

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

awesome_tornado

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

Go Up