[Tiszy]

Im a part time elecitrician, and as the other says, if you take a piece of cobber(which has low resistance) it will connect "L" and "N" which we call it here in Norway.
A battery has two poles, + and -

If you make connection through + and - with a direct type of resistance kinda, like cobber(without any form of other resistances) it will shut down. So yeah, I'd not recommend this setup

[roger_smith]

Im a part time elecitrician, and as the other says, if you take a piece of cobber(which has low resistance) it will connect "L" and "N" which we call it here in Norway.
A battery has two poles, + and -

If you make connection through + and - with a direct type of resistance kinda, like cobber(without any form of other resistances) it will shut down. So yeah, I'd not recommend this setup

But bulbs 1 and 2 would use up the voltage, and have an amperage draw to them, so it wouldn't be like a direct ground fault.

[Tiszy]

But bulbs 1 and 2 would use up the voltage, and have an amperage draw to them, so it wouldn't be like a direct ground fault.

Let say the battery is 12V

Bulb 1 and 2 would have 4V each, and the other bubs would either have 2V or 4V(2V for 3,4 and 4,5, and 4V for 7)

Since the cobber which I understood would be changed out with bulb 7, means it will have 4V.

In this case, I do not know the I or W, so I can't calculate the power usage here.
But my conclusion is that either the cobber will melt slowly or the battery will explode/melt.

It will still be direct between the pole, + and - with 4V, but I don't know how much I(A) it will go through this system, since I do not know the resistance of the bulbs...

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But bulbs 1 and 2 would use up the voltage, and have an amperage draw to them, so it wouldn't be like a direct ground fault.

Also no, bulb 1 and 2 won't use all the voltage, its equally spread out, but the last one is 5 bulbs kinda in a parallel(idk what you call it in english) so it will also equally be spread out with bulb 1 and 2

---

But bulbs 1 and 2 would use up the voltage, and have an amperage draw to them, so it wouldn't be like a direct ground fault.

Also no, bulb 1 and 2 won't use all the voltage, its equally spread out, but the last one is 5 bulbs kinda in a parallel(idk what you call it in english) so it will also equally be spread out with bulb 1 and 2

[roger_smith]

I disagree @tiszy

If you put a piece of copper in bulb 7, you're effectively cutting bulbs 3-6 out of the equation entirely. Those light bulbs all have a certain level of resistance (that's why they light up) and the copper replacing bulb 7 should be a lower resistance. Therefore, the electricity would follow that path, bypassing the additional legs of the parallel circuit. Meanwhile, bulbs 1 and 2 would still be in series, meaning that (assuming a 12v system) each bulb would use approximately 6v EACH. We of course making A LOT of assumptions here, things like batt voltage, the bulbs being the same, etc. Obviously if bulb 1 is an LED and bulb2 is incandescent, that changes a bunch of stuff.

[Tiszy]

I disagree @tiszy

If you put a piece of copper in bulb 7, you're effectively cutting bulbs 3-6 out of the equation entirely. Those light bulbs all have a certain level of resistance (that's why they light up) and the copper replacing bulb 7 should be a lower resistance. Therefore, the electricity would follow that path, bypassing the additional legs of the parallel circuit. Meanwhile, bulbs 1 and 2 would still be in series, meaning that (assuming a 12v system) each bulb would use approximately 6v EACH. We of course making A LOT of assumptions here, things like batt voltage, the bulbs being the same, etc. Obviously if bulb 1 is an LED and bulb2 is incandescent, that changes a bunch of stuff.

Regarding the original post:

"In the electrical circuit diagram above, If I were to replace bulb #7 with a piece of copper would that melt the battery due to the new route for the electricity having no resistance or would it simply make only bulb 1 and 2 have electricity and 3, 4, 5, 6 would just not have any power? This is fairly basic electricity stuff but I'm drawing a blank on this one."

In the electrical circuit diagram above, If I were to replace bulb #7 with a piece of copper
refer to you
If you put a piece of copper in bulb 7,

Just to make sure, we're replacing, not putting a piece of cobber into it.

He's questions are the following(which I parted in 2)

If I were to replace bulb #7 with a piece of copper would that melt the battery due to the new route for the electricity having no resistance?
OR
Would it simply make only bulb 1 and 2 have electricity and 3, 4, 5, 6 would just not have any power?

I'll come back to my answer soon to those, but to you Roger;

Even though its cobber, and we both agree, it have less resistance than a bulb, the electricity will go through there, as electricity choose the easiest path. But keep in mind, on a battery, there is two poles, + and -.

The electricity goes from MINUS(-) to PLUS(+), but this doesnt really matter here anyways, since its a full end on this elecricity, but anyways. We have no idea if the bulbs are equal but I guess so, even though its series + parallell, it can be different voltage due to the difference of the restitance of the bulbs, for example;


That is 3 bulbs in a serie, Bulb1 = 3, Bulb2 = 4, Bulb3 = MAINSUPPLY - BULB1 - BULB 2 = 12V - 3V - 4V = 5V.
So even though its serie, it just divides the total voltage on all the different or equal components.

Now back to the OP;


As you can see, I made it a little bit easier with my painting.
About my conclusion about it will melt, etc. It depends on the woltage, resistance, so my answer is kinda not accurate due to I have not enough information. If you use a cable thats able to transfer the amount of elecricity, ofc. it wont melt.

Anyways, if you change out the last bulb with a wire, it will look like this:


But to make it even easier for us, we take
bulb 1 and 2, and make it A
bulb 3 and 4, and make it B
bulb 5 and 6, and make it C

Will look like this:


And we can also remove the last wire, as it has no sense, will look like this:


We can also add B and C into BC or something, and we can calculate some numbers from there, but I guess I wont go so deep.

I hope the last image I showed you was a bit easier to read instead of your first.
So;

If I were to replace bulb #7 with a piece of copper would that melt the battery due to the new route for the electricity having no resistance?

The route, well, the electicity always take the easiest path, but keep in mind, you have both bulbs in series and parallells. If you change out the bulb with copper wire, as I've shown you above, it will look like this. See anything wrong? Neither do I

It can, and can not, melt the battery, it really depends on many factor, what bulbs(W and R), what size of cable(which can carry the amount of A), etc. So I can not give you a set of answer, but from this, and the typical bulbs, etc. I'd say no.

And yes, the electricity have a resistance even though you remove the bulb. It gets LESS resistance;
I = AMP
W = WAT
V = VOLTAGE
R = RESISTANCE

Take a look, if you remove some resistance(I bet the bulb have more resistance than the copper, if not its a giant cobber wire though)

I = V / R

As you can see, R is a big factor, which is divided by V, which will result in I(A)

So, no, it will not have 0 resistance, but less resistance.

OR
Would it simply make only bulb 1 and 2 have electricity and 3, 4, 5, 6 would just not have any power?

No, if you take a look at my last picture, or the others, you'll see that bulb 1,2,3,4,5 and 6 will still light.

Anyways, I see you asked for the basic, I think I gave you a little bit more, hehe

[roger_smith]

Your last picture is wrong. If you add a piece of copper in at 7, that would take bulbs "B" and "C" out of the mix. You're removing the parallel circuit from the equation by doing so. If the electricity follows the path of least resistance, it would flow through bulb 1 and 2, then go through the now-replaced bulb 7 that is instead a jumper wire. BECAUSE OF THIS, THERE MAY BE MINIMAL VOLTAGE AT THE REMAINING LEGS OF THE PARALLEL CIRCUIT, HOWEVER THE BULBS WILL NOT LIGHT, BECAUSE THERE ISN'T ENOUGH VOLTAGE TO LIGHT THEM.

Remember, voltage = electrical pressure. The lights require a certain amount of "pressure" to work, kinda like a garden hose or an air tool. If you don't have enough pressure, the air tool won't run, the hose won't spray.

I took automotive electrical and held electrical certifications for a while. We did nearly this exact lab in my class. The lights didn't light.

[Tiszy]

Your last picture is wrong. If you add a piece of copper in at 7, that would take bulbs "B" and "C" out of the mix. You're removing the parallel circuit from the equation by doing so. If the electricity follows the path of least resistance, it would flow through bulb 1 and 2, then go through the now-replaced bulb 7 that is instead a jumper wire. BECAUSE OF THIS, THERE MAY BE MINIMAL VOLTAGE AT THE REMAINING LEGS OF THE PARALLEL CIRCUIT, HOWEVER THE BULBS WILL NOT LIGHT, BECAUSE THERE ISN'T ENOUGH VOLTAGE TO LIGHT THEM.

Remember, voltage = electrical pressure. The lights require a certain amount of "pressure" to work, kinda like a garden hose or an air tool. If you don't have enough pressure, the air tool won't run, the hose won't spray.

I took automotive electrical and held electrical certifications for a while. We did nearly this exact lab in my class. The lights didn't light.

Yes they will still light if you have the wire at the end, because all the way its 12V/6V since its both series and parallell...

And thats weird, it should light, I think I'll setup this myself tomorrow

[Eclipse]

I have an idea. Why don't you try it and find out?

[roger_smith]

Yes they will still light if you have the wire at the end, because all the way its 12V/6V since its both series and parallell...

And thats weird, it should light, I think I'll setup this myself tomorrow

adding the wire at bulb 7 cuts the parallel circuit out of the equation is my suspicion.

[Inori]

too lazy to read the above posts so if this has already been answered, oops.

The bulbs would all stay on and the source (battery) would be fine for a few reasons.

1. the bulbs are connected in a parallel block, making the copper joint a resistor
   
2. because the copper acts as a resistor, it acts like a load
   
3. Voltage stays the same, even when a parallel circuit is the type in question, meaning that if your source is 6 V, 6 V is what's flowing through each joint, dividing itself among each bulb in the joint.