Explanation of Amps

[tuxtlequino]

I am trying to learn more about electronic, but for some reason I
cannot fully Amps with the examples the book provided. I memorized
Ohm's law, and the formulas necesary to come up with the total
resistance in a circuit (either parallel or in a serie), but there is
still more questions that the book does not answer. Could somebody help
me with that please?

Here are my questions:

If E=IR can we just then increase the resistance and get as many volts
as we want?

The schematics in the book explains the number of volts (for example a
9 volt battery), but they never mention the amps in the battery, how
can we then figure out how much resistance do we need?

If I get a LED that says in the package 2 Volts, and .005amp. I
understand that I need to divide 7/.005 to figure out how much
resistance I need. But why .005? Where did it came from, why, what is
the resistance of the LED?

Now, if I have a system of 9 volts (I bought an electric kit with a
schematic explaning how it works!), and then there is a resistor for
4.7MOhms, what is my amperage? Would I use the same formula(e=ir) to
figure out? Does the use of capacitors change my amperage (I have seen
them in my circuit!)?

 

[John Popelish]

tuxtlequino wrote:

I am trying to learn more about electronic, but for some reason I
cannot fully Amps with the examples the book provided. I memorized
Ohm's law, and the formulas necesary to come up with the total
resistance in a circuit (either parallel or in a serie), but there is
still more questions that the book does not answer. Could somebody help
me with that please?

Here are my questions:

If E=IR can we just then increase the resistance and get as many volts
as we want?

If you have a current source (that provides as many volts as it takes
to push a fixed current through a resistance), then, yes.

That formula makes a lot more sense if you rearrange it to show that
R=E/I, or that ohms means 'volts per ampere'. A 100 ohm resistor is
one that requires 100 volts across it to force 1 ampere through it.
At any current or voltage, the ratio of volts divided by amperes is
always 100. That is what 100 ohms means.

The schematics in the book explains the number of volts (for example a
9 volt battery), but they never mention the amps in the battery, how
can we then figure out how much resistance do we need?

Divide the volts by the resistance connected across it to find the
amperes passing through that resistance.

If I get a LED that says in the package 2 Volts, and .005amp. I
understand that I need to divide 7/.005 to figure out how much
resistance I need. But why .005? Where did it came from, why, what is
the resistance of the LED?

An LED is not a resistance, because the ratio of volts across it to
the current through it does not stay constant, regardless of the
voltage and current. It is a non ohmic device. It is specified to
drop about 2 volts at one current, .005 amperes. You need some other
formula than ohm's law to figure out the voltage current pairs for
other currents. If you want ot run this LED at this current from a 9
volt source, you might put a resistor in series with it to burn up all
the extra voltage while 5 milliamps passes through both the LED and
resistor. So, by ohm's law, you need a resistor that drops (9-2)= 7
volts while .005 amperes passes through it and the LED. 7/.005= 1400
ohms.

Now, if I have a system of 9 volts (I bought an electric kit with a
schematic explaning how it works!), and then there is a resistor for
4.7MOhms, what is my amperage? Would I use the same formula(e=ir) to
figure out? Does the use of capacitors change my amperage (I have seen
them in my circuit!)?

If you can say that the circuit puts 9 volts across the resistor, then
ohm's law tells you how much current that voltage will push through
that resistance.
9V / 4.7M = 1.9 microamperes.

But if you put a capacitor in series with that resistor and connect
the battery across that pair, the situation is very different. The
rule that relates voltage across a capacitor ot the current through it
is I=C*dv/dt) or current in amperes equals capacitance in farads times
the rate of change of voltage in volts per second. Time is involved
in this relationship, but was missing from ohm's law. So the current
will depend on how long the voltage has been applied.

--
John Popelish

 

[tuxtlequino]

Wow guys, thank you very much for explaining everything so quickly and
in such a maner that it finally makes sense. Everybody apported
something new to help me understand Amp's. Part of my problem
understanding Amps had to do with the common knowledge out there. Most
of what I heard was as a little child being warned by his father, or
asking a silly question to an electrician. I heard many times that what
kills is not Voltage, but Current.

I also remember about how much static electricity we can accoumulate
and discharge (I used to get shocked everytime at my College's
library). So, through common sense I reached the conclussion that a
certain amount of Amps was storaged at the batteries(or cells). When I
first started to read the book in basic electricity and electronics my
common knowledge made everything harder to understand. Now I know that
everything has a certain resistance (and I supposed that is why a
lighting can kill us, since we offer resistance). And knowing that
batteries have current ratings explain why a car batery is more
dangerous than another one.

More questions will come as I learn more, but thank you for helping me
with my questions about Amperes.

I think I know that what Milles was trying to explain is that since
every batery has a different resistance, they will have a certain
amount of amp's (although not significant enough!). What I will like to
know is how does the electrical companies carry electricity from far
places. I mean, they have to fight all the resistance from the long
cables, so the voltage has to be greater, and therefore we will have a
very good amount of Amps in the lines. Does our AC electricity comes
with Amp's?, or they are somehow wasted at the transformers?

Sorry about the silly questions. I am a computer programer, so if you
have questions about your computer I can help, but my knowledge in
electricity is very limited.

---
Victor Gutierrez

 

[John Popelish]

tuxtlequino wrote:

Wow guys, thank you very much for explaining everything so quickly and
in such a maner that it finally makes sense. Everybody apported
something new to help me understand Amp's. Part of my problem
understanding Amps had to do with the common knowledge out there. Most
of what I heard was as a little child being warned by his father, or
asking a silly question to an electrician. I heard many times that what
kills is not Voltage, but Current.

There is a way of thinking about this that makes sense. If you touch
a source of voltage (with respect to the Earth, let us say) but you
are not also touching the Earth, your body potential rises to match
the potential of what you are touching, but, since there is no path
for current to leave your body through a second point, you will not
receive a shock.

Grab the water faucet with the other hand, and that potential will
drive current through you on its way to ground, and you will be
shocked by the current passing through you. But it was the voltage
difference between the two things you touched that pushed that current
through you.

I also remember about how much static electricity we can accoumulate
and discharge (I used to get shocked everytime at my College's
library). So, through common sense I reached the conclussion that a
certain amount of Amps was storaged at the batteries(or cells). When I
first started to read the book in basic electricity and electronics my
common knowledge made everything harder to understand. Now I know that
everything has a certain resistance (and I supposed that is why a
lighting can kill us, since we offer resistance).

Wrap yourself in metal, and the lightning will mostly choose that
material and avoid you almost completely.

And knowing that
batteries have current ratings explain why a car batery is more
dangerous than another one.

A car battery does not produce enough voltage to push a dangerous
current through your skin resistance (though it will give your tongue
an unpleasant sensation). But it is dangerous because of the energy
it can release into a low resistance load that will get hot enough to
burn you. You can weld metal with a car battery.

More questions will come as I learn more, but thank you for helping me
with my questions about Amperes.

I think I know that what Milles was trying to explain is that since
every batery has a different resistance, they will have a certain
amount of amp's (although not significant enough!).

It will have a short circuit current limit that occurs when its
internal voltage is all used up across its internal resistance.

What I will like to
know is how does the electrical companies carry electricity from far
places. I mean, they have to fight all the resistance from the long
cables, so the voltage has to be greater, and therefore we will have a
very good amount of Amps in the lines. Does our AC electricity comes
with Amp's?, or they are somehow wasted at the transformers?

About half of all the electrical energy that is generated is lost
heating the resistance of the distribution grid. They jack the
voltage up very high, since doubling the voltage allows the same power
(volts times amperes) to be delivered with half the amperes. And half
the amperes drops half the voltage across a given piece of wire
resistance. This reduces the wire losses to 1/4. Etc. But there are
practical limits to how high they can raise the voltage. Stand under
any high tension transmission line, and you can hear the crackle of
corona discharge taking place, draining energy from the wires.

Sorry about the silly questions. I am a computer programer, so if you
have questions about your computer I can help, but my knowledge in
electricity is very limited.

But increasing.
--
John Popelish

 

[John Popelish]

Andrew Holme wrote:

John Popelish wrote:
About half of all the electrical energy that is generated is lost
heating the resistance of the distribution grid.

Are you sure about that, John? I find that hard to believe. Googling for a
figure, I found this:

http://healthandenergy.com/electric_power_grid.htm

It says:

"Seven percent of the energy is lost in transmission," said George David,
chairman and chief executive of United Technologies, which is based in
Hartford. "The solution is to put power generation much closer to where the
electricity is consumed."

I hope you are right. I heard the 50% figure a long time ago and
don't know where to look it up.

--
John Popelish

 

[John Fields]

On Tue, 18 Jan 2005 14:41:26 GMT, Miles Harris <mazzer@yahoo.com>
wrote:

On Mon, 17 Jan 2005 20:26:05 -0600, John Fields
jfields@austininstruments.com> wrote:

Yes, it does. an _extremely_ common example is a high voltage being
fed, through a high value of resistance, to a load which requires a
constant current. An arguably less common example is a current
regulating diode.

I'm talking about a _perfect_ current source, Junior. Surely even
*you* must have worked that out.

---
Well, maroon, in the context of the thread, everybody knows there's no
perfect current source, so when you don't explicitly state that that's
what you're talking about, the default becomes real current sources
which, like lead-acid batteries and diodes, we can add to the
inventory of devices you've demonstrated you know little about.
---

No one - to my knowledge - has mentioned the essential missing ingredient
here: the battery's _internal resistance_. This is what the OP really
needs to be informed about, but I'm not apparently very good at
expounding the basics, or so I'm told. Can someone please elucidate in
a language the OP will understand?

---
Why don't _you_ give it a try and see how you do? If you make any
mistakes I'm sure _someone_ will correct you.

That "someone" will 'correct' me even if I don't make any mistake, it
seems. :-/

---
Apparently, we'll never know.

--
John Fields

 

[John Fields]

On Tue, 18 Jan 2005 21:36:14 GMT, Miles Harris <mazzer@yahoo.com>
wrote:

On Tue, 18 Jan 2005 09:38:16 -0600, John Fields
jfields@austininstruments.com> wrote:

On Tue, 18 Jan 2005 14:41:26 GMT, Miles Harris <mazzer@yahoo.com
wrote:

I'm talking about a _perfect_ current source, Junior. Surely even
*you* must have worked that out.

---
Well, maroon, in the context of the thread, everybody knows there's no
perfect current source, so when you don't explicitly state that that's
what you're talking about, the default becomes real current sources
which, like lead-acid batteries and diodes, we can add to the
inventory of devices you've demonstrated you know little about.
---

Wrong again, Junior. It was John Popelish that introduced the current
source in his first response on this thread. It was clear that he was
also talking about a perfect current source:

JP:
If you have a current source (that provides as many volts as it takes
to push a fixed current through a resistance), then, yes.

---
As far as I can tell, JP was expounding on the _concept_ of a current
source, which does assume ideal conditions, much like the
impossible-to-find massless lever, weightless string, and frictionless
air.

Moreover, John clearly defined his terms in order to, ostensibly,
remove any ambiguity from his argument, so his meaning was perfectly
clear.

You, on the other hand, with your sloppy:

"However, it's confusing to introduce a newbie to the concept of the
current source. It doesn't exist in the real world as we know."

Are claiming that the current source (any current source) doesn't
exist in the real world. You seem to be reasonably adept with the
language, and if you had meant otherwise I find it difficult to
believe that you would have deliberately omitted the 'perfect' from:

"However, it's confusing to introduce a newbie to the concept of the
perfect current source."

and in the process botched what you claim your intent was so
perfectly. But perhaps it was only an oversight?

--
John Fields

 

[John Fields]

On Tue, 18 Jan 2005 23:55:28 GMT, Miles Harris <mazzer@yahoo.com>
wrote:

On Tue, 18 Jan 2005 17:17:42 -0600, John Fields
jfields@austininstruments.com> wrote:

Wrong again, Junior. It was John Popelish that introduced the current
source in his first response on this thread. It was clear that he was
also talking about a perfect current source:

JP:
If you have a current source (that provides as many volts as it takes
to push a fixed current through a resistance), then, yes.

---
As far as I can tell, JP was expounding on the _concept_ of a current
source, which does assume ideal conditions

So you retract your earlier mis-statement? That's a welcome change.

---
I'll retract nothing, and it wasn't a mis-statement. A welcome change
from you would be something a little more substantial than footwork,
but I guess you're afraid to try it 'cause you might get nailed again?
---

Moreover, John clearly defined his terms in order to, ostensibly,
remove any ambiguity from his argument, so his meaning was perfectly
clear.

It certainly was. And yet again you failed to pick up on it, Junior.
Are you actually bunking off high school.... or _grade_ school? Don't
they teach English any more in American schools? I saw something on
the news tonight that indicated Darwin and his work was to be dropped
from the curriculum so I guess anything's possible these days in the
good ol' US of A..

---
Yup, we continually reinvent ourselves.
---

I won't bother to request an apology for obvious reasons...

---
Yeah, 'cause you know damned well you've got no grounds to ask for
anything but mercy!

--
John Fields