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### Topic: Is this circuit correct? (Read 449 times)previous topic - next topic

#### V4D3R

##### Aug 17, 2019, 04:51 am
Just need to know if this circuit is correct. Making a 4 bulb lamp and needed to wire everything in parallel.

#### KASSIMSAMJI

#1
##### Aug 17, 2019, 05:14 am
what's up with lamp3 and 4 having 0 ohms of resistance ?
Electrical Engineer

#### SteveMann

#2
##### Aug 17, 2019, 05:49 am
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.

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#### Nikosant03

#3
##### Aug 17, 2019, 02:25 pm
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.

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

#4
##### Aug 17, 2019, 03:57 pm
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

#5
##### Aug 17, 2019, 05:14 pm
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

#6
##### Aug 18, 2019, 10:12 am
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!!!"