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/sci/ - Science & Math

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5696022 No.5696022 [Reply] [Original]

I'm a little confused on how electronics work. If you look at ohms law, then you should be able to get 9 amps from a 9 volt battery if you have something with 1 resistance. This is obviously not true. As I understood it, all sources have a limit of how much current they can deliver. Question is, will the current just be at the limit if you try to go over it, or is the limit itself part of the equation, causing it to be somewhere else? Also, if you turn up the amp limit, will the amp in the curcuit remain the same because res and volt is still the same?

>> No.5696034

That's because ohms law is bugged.

>> No.5696036

>yfw if resistance is 0 then unlimited current

>> No.5696051

>>5696022
this question is very broad if you really want details. The short story is that
1) you can't get out more than you put in, and
2) though ideal voltage sources have zero internal resistance, real voltage sources have non-zero internal resistance
3) internal resistances are themselves a complicated function of the device and current operating conditions

When you use things "within reason" you can ignore a lot of these factors and just pretend things are ideal.

>> No.5696079

>>5696051

but that's not related to what he asked at all

>> No.5696082

>>5696079
except it is

>> No.5696089

>>5696079
It's related because he is explaining some facets of electronics necessary to know in order to understand why things can't be ideal in the real world. The same as newtons laws of motion don't help when you really want to know how high a ball will go into the air because you don't take into account air resistance, temperature, pressure. You just look at ideal situations for equations to make life easier.

>> No.5696098

>>5696051

Also:

Ohm's law implies uniform charge density across the conducting surface. Current is really equal to <span class="math"> I=nAvQ [/spoiler] where n is the number of charge carriers per volume, A is the cross sectional area, v is the velocity of the charge carriers due to an electric field and Q is the charge. The limit is going to be how fast the charge carriers are going to move due to a voltage potential which is essentially going to equate to I=V/R. However, the large the cross section, the more charges can actually move through the surface, the higher the current. Of course, if you have no resistance, current is going to be limited by the fact that charge carriers have mass and can't exceed the speed of light. Wikipedia says that electrons move through cathode ray tubes at 1/10th the speed of light.

>> No.5696101

>>5696089

He was asking about limits of amps and how it works, and you start babbeling about internal resistances and ideal situations...

>> No.5696104

>>5696101
you understand the relationship between current and resistance, right

>> No.5696106

>230v power socket in house
>touch wires
>can die from the shock

Human resistance: ~500kohm
230/500000=0,46mA

How does that make any fucking sense?

>> No.5696113

>>5696104

The more res, the less current.

But fuck, let me put the question differently. Say a power source can deliver 100 volts with 10a. If you hook it up to a device with 5 resistance, it "wants" to get 20 amps through the circuit, right? Will that situation cause it to cap at 10?

>> No.5696116

>>5696106
Where the jesus did you get 500,000Ohms for human resistance?
There are so many factors involved. Electricity breaks down human skin pretty quickly. Perspiration is a factor. Your body type, what you are wearing, the time of day, humidity, pressure. Nothing even close to 500kOhms though.

>> No.5696119

>>5696116

If I put my fingers up to a multimeter I get around 500kohm.

>> No.5696122

battery has ist own resistance
its obvious that short circuit on any bat is not inf amps

>> No.5696128

Another question, say I have a power supply with variable amps.

The amp is set to 1, volt to 10, and the resistance of the circuit is 10. 10/10 = 1 so you'll get 1 amp. BUT, if you turn up the amp to 2, will it still stay at 1 because of the volt/res being the same?

>> No.5696147

>>5696128

Volts would change OR the total resistance would chance due to internal resistance in the power supply OR it would stay at 1.

>> No.5696160

>>5696122
>you should be able to get 9 amps from a 9 volt battery if you have something with 1 resistance.

In theory, yes. But 9 volt batteries are tiny. It can't push 9 A. There's not enough chemicals inside and it can't react quick enough to sustain 9 Amps. You'd be lucky to get half an Amp.

The way they rig the maths to explain this is that the battery has an "Internal resistance". The more current you try and pull, the less and less like 9 V it actually looks.

Say the internal resistance is 1 ohm. You measure after this resistance because all you can measure is the terminals of the battery. When nothings connected you measure 9V. When something with high resistance is connected you measure a fraction under 9V.

What happens if you connect something with extremely low resistance though? 1 ohm? Well now you have 1 ohm internal and 1 ohm external, so the voltage you measure over the battery (over the internal resistance) is only 4.5 V now. It drops dramatically when you try and draw a big current. You also get lots of heat (and indeed, try shorting a battery and you'll feel it get hot quickly).

>> No.5696161

>>5696113
You'd have to learn about how power supplies are constructed.
My guess is that the internal resistance of the power supply would be 10 ohms, and that's why it couldn't output more than 10 amps.

>> No.5696182

>>5696113
> will that situation cause it to cap at 10?
Can't say in general what will happen. Most actual power supplies (e.g. you buy one from a large manufacturer) can supply "transient" currents well above their nominal rating, and their datasheets will tell you both how long and how often this can happen. If you're talking about slamming some bridge-rectified AC into a linear regulator, the matter is considerably different. And batteries are different still.

As I tried to say in (3) above the question is very deep.