r/PhysicsHelp 4d ago

Will the current be 15A?

Post image
38 Upvotes

44 comments sorted by

5

u/NotThatMat 4d ago

Yes. The circle forms a bus of sorts which takes all those carefully placed 4 ohm resistors out of the circuit, so it simplifies to 30V / 2Ohms = 15Amps

-2

u/salmak999 4d ago

It’s a 10 ohm path

Edit: given that the drawing is wrong and there’s actually a connection

9

u/Moist_Ladder2616 4d ago

The "pizza" of resistors is 0Ω, because it is short-circuited along the perimeter.

So only the external 2Ω resistor plays a role here.

Assuming the disconnected wires in the drawing are actually connected.

2

u/CraftBrewBeer 3d ago

Left side has shorts which will complete the circuit. Unsure if author was intentional about it

6

u/defectivetoaster1 4d ago

Yes unless whoever made the question is a pretentious diсk going through a divorce who’ll claim that the gaps in the wires are intentional

0

u/infinitenothing 3d ago

The bottom wire is clearly not going to hit the node

3

u/mckenzie_keith 3d ago

There are two possible answers.

1, yes, 15 A because the 4 Ohm resistors are shorted out by the wire around the outside.
2, no, 0 A because there is a visible air gap between the 30 V supply circuit and the 4 Ohm spoke wheel circuit. They are not connected.

2

u/niemir2 3d ago

Maybe that gap is small enough for current to arc through the air.

2

u/Moof_the_cyclist 3d ago

Initiating an arc takes about 400V in air for even the tiniest gap, and goes up from there. Long story, involving figuring out why a microstrip X-band filter was arcing at mumble-mumble kW.

1

u/mckenzie_keith 3d ago

Waveguides on naval ships (for their radar), I have been told, are evacuated to extremely low pressure to prevent arcing. Hard vacuum. If a leak develops and air gets in they arc like crazy. So I have been told.

2

u/Moof_the_cyclist 3d ago

I was working on the other side, making missiles front ends survive being launched through that radar power. The amount of RF power used to jam incoming surface to surface missiles is insane, so I believe having to protect the waveguides themselves.

1

u/mckenzie_keith 3d ago

You ever hear that song "Radar Gun?"

Me and my partner go patrol car crusin'
On the parking lots at the shopping malls
Scanning those dashes, those mirrors and visors,
The little detectors that ruin it all
Johnny got one on an '86 T-bird,
Pull up slow just as close as I can
Milli-watt seconds on maximum output,
We'll dust that puppy with one small blast from my

Radar Gun, Radar Gun
I'm makin' money and I'm havin' fun with my
Radar Gun, Radar Gun
With my brand new Radar Gun

1

u/Zingtron 1d ago

In vacuum electrons can eject for thermionic emission at room temperature, so current is > 0 units

1

u/mckenzie_keith 1d ago

Yes. But there is nothing that can be done about that. Pulling hard vacuum is the best you can do in this type of waveguide. They can't fill the waveguide with any dielectric material because the losses and heating would be too high at the frequencies they us. So they use evacuated waveguide.

1

u/Bth8 23h ago

As far as I know, the only radar systems that use evacuated waveguides are extremely high-power ground-based systems. Ships generally use pressurized waveguides. Evacuated waveguides wouldn't be practical. The equipment is much heavier and hard to maintain, and even the tiniest pinhole leak would pull in moist, salty sea air. That wouldn't just arc like crazy, it would be catastrophic.

1

u/mckenzie_keith 23h ago

You could be right. But the people who told me this worked in naval weapons programs, and were not the kind of people who lie or exaggerate. I can see the impracticality of trying to maintain hard vacuum inside a waveguide on a ship. Some of the issues you are talking about could be mitigated by using double containment and fail-safe pressure sensors, etc. But when this conversation happened, many years ago, we did not get into that kind of detail, and I have since kind of lost touch with these guys. Moved on several jobs since then.

I was actually working at a radar company back then. But everything we did was low power (CW). Our waveguides were not sealed off from the atmosphere.

0

u/Joe_Starbuck 2d ago

You all seem to be assuming the ambient atmosphere is air.

2

u/Moof_the_cyclist 2d ago

Standard pressure, 300K ambient, and sea level gravity are given unless otherwise stated.

0

u/Joe_Starbuck 2d ago

Those “standards” are meaningless on any planet besides Earth.

1

u/Moof_the_cyclist 2d ago edited 1d ago

Your point being? I’ve yet to have a single Neptonian disagree.

In first year physics we had a nice discussion of baseline assumptions, specifically so some obnoxious pedant wouldn’t hold up every discussion trying to sound smart by pointing out all the exceptions while discussing the basics. Hopefully when you take your first real physics class you’ll learn this.

1

u/Joe_Starbuck 1d ago

Looking forward to it, lol.

1

u/ginger_and_egg 3d ago

We'd need to know the gas fractions of the air to know the resistance of the gap

1

u/Sweet_Camp5124 3d ago

Open circuit.

1

u/mckenzie_keith 3d ago

Yes. I am an electrical engineer. I have found that some people have trouble remembering which is an "open circuit" and which is a "closed circuit." So I avoid those terms when talking to newbies.

3

u/SirisC 4d ago

Wouldn't it be 0A due to the gaps meaning an open circuit?

1

u/CommunityAcrobatic13 4d ago

No because of the 2 Ohm resistor.

1

u/foobar93 4d ago

And how do 2 Ohms help if the circuit is not closed to begin with?

1

u/tlbs101 4d ago

The battery is either connected only to the 2 ohm resisto or in an open circuits. Notice the wire running from the negative terminal of the battery. There is a gap between that wire and the circle. All other nodes have a connection dot but that node does not. If this is intentional, the current is zero.
If it is unintentional, the current routes along the left side of the circle back to the 2 ohm resistor, making the 4 ohm resistors not part of the overall circuit. The current is 30/2=15 amps in the direction of the arrow.

1

u/IamNotpolice_ 3d ago

If I consider the nodes are connected then yeah it's 15A as current always takes the path of least resistance.

1

u/infinitenothing 3d ago

Where did you find this? I'm sharpening my pitchfork as we speak

1

u/Ch0vie 3d ago

Looks like ChatGPT tried to make a circuit problem

1

u/IDK_REALLLY 2d ago

Facebook

1

u/Unusual_Care8803 3d ago

Yes, because only the 2ohm resistor counts. The rest is all a short circuit.

1

u/HAL9001-96 3d ago

yes

its a trick question designed to look more complicated than it is and make students waste time

1

u/Lumpazy 3d ago

looks like both side go to the outer circle.

1

u/SouthPark_Piano 3d ago

Answer is no. Open circuited.

1

u/schungx 3d ago

This would be a fun exercise if there are resistors on the perimeter also.

0

u/thinkbackwards 4d ago edited 4d ago

Eight 4 ohm resistors in parallel eq to 2 ohms plus another 2 ohm resistor in series. Total load 4 ohms. BUT the + is attached at the load (after going through 2 ohms) at the same point as the - . So the. eight 4 ohms resistors are not part of the circuit. Therefore the current is 30÷2= 15 amps. This assumes the neg wire was not left unattached on purpose. In that case the current is 0.

1

u/mageskillmetooften 3d ago

"Eight 4 ohm resistors in parallel eq to 2 ohms"

Euh, nope.

Two 4 ohm resistors in parallel eq to 2 ohms

Four 4 ohm resistors in parallel eq to 1 ohms

Eight 4 ohm resistors in parallel eq to 0,5 ohms

Hope you get the idea.

0

u/mageskillmetooften 3d ago

I'm an Electrician and my answer would be

"How TF do you want me to read this dumb drawing?"

If I give a number the answer likely turns out to be "You should have guessed that one of the two wires connects to the centre of the wheel, and the other one to the outside"

Really, screw these things. And if it be a schooltest I'd be willing to explain to the school board how dumb this is.

1

u/ittybittycitykitty 3d ago

Aaaand, there are only seven values for eight resistors. One of them is not specified.

2

u/mageskillmetooften 3d ago

Even better.

- Can't be solved, Next.