Why doesn't this circuit work? What am I doing wrong here?

Why doesn't this circuit work? What am I doing wrong here?

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but that's not a short circuit

If you draw a circuit like that, then you may want to be studying a little bit more.

define work.
also, try using kirchhoff's laws to do analysis on the loops and junctions

I think it is, user

how?

You're short-circuiting that one resistor.

how is that a problem for the entire circuit?

Why doesn't this circuit work?

You're using a capacitor in a DC circuit.

What am I doing wrong here?

Literally everything.

I don't understand why it wouldn't still work, I'm at a loss here

That's precisely why it doesn't work. You don't understand some basics. Start over and follow the foundation logic.

you'se using MS paint

I can't see anything wrong with it.

That isn't how it works

And what? What if it's a ceramic disk capacitor? And why can't you use a cap in a DC Circuit, if you check the - and + correctly?

What's the load? Won't this just heat up and explode?

It blocks DC voltage.

and what? the circuit still works, the part with the capacitor is just not powered since the cap is filled

This has gone from noob to bait territory.

Reported, saged, hide, ip hacked, and I am coming for your family.

there will be a power loss

What should it even do? Define work as
said.

Why doesn't this circuit work?

What do you mean, "doesn't work" - what did you expect to see that you aren't seeing?

No, the cap will drain the battery in seconds. This is how the circuit should look like.

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is this loss?

Fuck

Brainlet here, why isn't the fact that there's a resistor in parallel with a wire not a problem? Wouldn't no current flow through the resistor?

Not strictly a problem, just a resistor that would be ignored. But that still doesn't really answer OP's shitty vague question.

It blocks DC voltage.

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the cap will drain the battery in seconds

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Define "work"
Right now you have current through the two loads in the bottom right

wait a second op...

force times distance

distance raptor/time raptor = velociraptor

you're right

I don't get it

I guess the joke is lost on you.

The impedance of a capacitor is $\frac{1}{i\omega c}$, for DC $\omega$ is 0 so the impedance is infinite and thus no DC can pass

Are you trolling or something?

It's just an RC circuit. It's not really going to DO anything without some kind of active element like a transistor or op-amp or something. Also, you have a resistor with both ends connected to the same node so it won't do anything because there is no voltage across it.

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using i instead of j for a circuit

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It seems that you're at a loss of power

This is a great IQ test.
I'm gonna use it next time I interview someone.

this
all you guys didn't get it holy shit

So what field is this? Who studies this stuff? Electricians?

T.brainlet

/v/

Capacitors are short circuit for DC signals.

That's not entirely true - if it's discharged yes it's a short circuit but if it's fully charged it's an open circuit

That isn't how it works
It's common to say that when a component is bypassed with a wire, the component is shorted.

Electrical engineers
...with a mimetic behavior bend.

Specify what you mean by "doesn't work".

Regardless of your answer, that circuit in steady-state is essentially a single load, a purely linear resistor. The upper resistor is shorted and the capacitor does nothing, so you have two resistors in series and that's it.

all you guys didn't get it holy shit

"I showed a meme image that looked like something board relevant on a board that isn't drowning in that meme, and people thought it was relevant instead of a meme" T. user, 2018, University Press Department of Auto-Ethnographies.

Why are there two positives on the bottom

It's a capacitor.

Case closed.

j is for engineering plebs

effective circuit, we see that the circuit actually works

What's it supposed to do? I see a directional capacitor, a capacitor and three objects of unknown properties, possibly resistors.

you need to learn how to draw and read circuits, that IS a short circuit, it is a path of virtually no resistance between two points
study harder before posting your homework here

putting a capacitor in parallel with resistors does not charge the capacitor even a little?

I find that hard to believe

I check for secret doors

And how is that circuit even supposed to work? What should it do?

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I thought short circuits(non trivial) were paths from one end of the power source to the other without resistance

No the fuck it is not, it's for anyone who knows how to properly analyze circuits. Granted those are usually engineers.

Is this Pentium processor?

it does
his image only analyses the circuit for the steady state when the capacitor is already charged

Bypassing any kind of element in the circuit via 0 resistance wire (or for example copper wire in practice) is considered to be short circuiting.

R1 and R2 make a voltage divider. the capacitor is irrelevant because it's dc, and the other resistor is parallel with a short circuit.

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what the fuck? in which shithole country do you draw the voltage like a capacitance?

not OP but
Symbol most likely specifies a battery rather than simply any voltage source. I was told to use circles with + and - for sources that can be assumed to be infinite for the purposes of the circuit, like a wall outlet and the battery symbol ones for ones that are not, like batteries.

Hey since this is the Electronics for Retards thread can I replace this IF transformer with a differential amplifier?

IF cans are hard to find and expensive compared to a few decades ago and I hate winding my own transformers. I still want good isolation between the LO and IF. Diff amp is high impedance and it allows me to impedance match easily. Throw a buffer on the output and throw a 50 ohm resistor on the output. Easy 50 ohm output impedance.

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