# Explanation of Amps

T

#### tuxtlequino

Jan 1, 1970
0
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

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!)?

J

#### John Popelish

Jan 1, 1970
0
tuxtlequino said:
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

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.

J

#### Jonathan Kirwan

Jan 1, 1970
0
If E=IR can we just then increase the resistance and get as many volts
as we want?

If you already know the I and you already know the R, then it allows you to
compute the voltage which must be present across the two terminals in order to
make that particular current through that particular resistance.

However, it is the more usual case in simple circuits that you have a given
voltage and you either want to calculate the expected current, given some
particular resistance... or else you want to calculate the desired resistor to
use in order to achieve some given current value.

Your next question gets to this...
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?

The battery supplies whatever current is required by the circuit. Most circuits
will specify the desired voltage (say, 9 volts) and will have some particular
effective resistance (not usually specified, though.) When you connect up the
circuit, the battery presents a particular voltage to the circuit and the
current simply is what it is, based on what the circuit requires.

If you hook up a 1k Ohm resistor to your 9V battery, 9 milliamps will flow. If
you hook up a 10k Ohm resistor to your 9V battery, 0.9 milliamps will flow (or
900 microamps, as you prefer to imagine it.) The amount of current flowing
depends on the "load" resistance. But the voltage at the terminals remains the
same, with a battery -- 9 volts, in this case.

Batteries actually prefer it if the circuits require less current. They last
longer and they provide the correct voltage for longer, as well. If the battery
were an "ideal battery" it would provide 9 volts no matter what current is, but
real batteries are limited.

A more useful specification for a battery, other than its voltage, is the total
amount of energy it holds. Batteries don't really "hold amps" in them, they
hold potential energy. As the energy is consumed, by way of the voltage and
amps required over time, they gradually expire. While that takes place, their
ability to maintain the specified voltage degrades somewhat and their ability to
supply higher levels of current suffers. But it is the circuit that determines
the "amps" that the battery needs to supply it. Not something inside the
battery.

Still, batteries *do* have limitations in current. A car battery has a very
high current rating and is able to supply very high currents into very tiny
resistances, while still maintaining their voltage. A small hearing aid battery
(a so-called button battery), on the other hand, may not be able to supply more
than a milliamp or two and still hold their proper voltage, too. Something will
suffer, if the circuit demands more than some tiny current limitation for those
button batteries.

The voltage rating may be the same, but bigger batteries are usually able to
handle more current for longer times, because they store more energy and because
they are designed for higher current requirements. For example, a D-cell has
the ability to supply much higher current than a AAA-cell does, while still
holding a proper 1.5 volts. If the circuit requires too much for the AAA, for
example, the AAA voltage itself may suffer *and* the battery may also warm up
and the total lifetime of use will also suffer. In the same situation, the D
cell may be perfectly fine supplying the higher current, not getting warm at
all, and holding its proper voltage the entire time.
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?

The LED "requires" about 2V to light up. With .005 amps (5 milliamps), the LED
will be reasonably visible. Since your 9V battery has too much voltage for the
LED to operate properly, you need to "throw away" part of that. To do so, you
take the difference of what you have in the battery and what you need at the LED
to compute the part to throw away. This is the 7 volts, as you said. Now,
since you also want about 5 milliamps in the LED at the same time, and since the
resistor is to be in series with the LED, whatever current the LED requires must
come through the resistor. So the current through the resistor must be equal to
the desired current through the LED. (Otherwise, the charge would have to "pile
up" somewhere.) If you compute your resistor to be 7V/.005, or 1400 Ohms, and
use that value then it will be true that the voltage taken away by the resistor
(opposing the battery voltage) will be 7 volts *if* it turns out that 5
milliamps occurs. But if the LED takes a little less than 2 volts then the
actual current through the resistor will be a little more and if the LED takes a
little more than 2 volts, then the actual current will be a little less.
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

If the resistor is hooked directly across the two battery terminals, then the
current through the resistor will be 9V/4.7Meg or slightly less than 2
microamps.
Would I use the same formula(e=ir) to figure out?

Yes, after some algebra to solve for I.
Does the use of capacitors change my amperage (I have seen them in my circuit!)?

It makes your current time-dependent. That's too complex for you to worry

Jon

M

#### Miles Harris

Jan 1, 1970
0
If you have a current source (that provides as many volts as it takes
to push a fixed current through a resistance), then, yes.

I appreciate where you're coming from on this one, John. Also, JK's
explanation was outstanding in detail and clarity.
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. 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?

J

#### John Popelish

Jan 1, 1970
0
Miles said:
I appreciate where you're coming from on this one, John. Also, JK's
explanation was outstanding in detail and clarity.
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. 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?

I don't like to try to explain everything in a single post, but try to
guess what the op is ready to soak up in one dose. If he asks further
questions, I will try to guess what he is ready for, then. I didn't
think that the question of source resistance was needed at this place
in his education, but I will be watching for the opportunity to inject
it. If your judgment differs, by all means jump in with both feet.

T

#### tuxtlequino

Jan 1, 1970
0
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

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.

J

#### John Popelish

Jan 1, 1970
0
tuxtlequino said:
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.

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

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
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

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.

A

#### Andrew Holme

Jan 1, 1970
0
John said:
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."

J

#### John Popelish

Jan 1, 1970
0
Andrew said:
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.

B

#### Bob

Jan 1, 1970
0
Wow guys, thank you very much for explaining everything so quickly and
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.

Yes!! This is a great way to learn. You can only learn so much from books.
Experimentation is the key to a long, successful, career in electronics.

As a great philosopher once said:

"Give a man a fire and he'll be warm for a day. Set a man afire and he'll be
warm for the rest of his life."

Bob

M

#### Miles Harris

Jan 1, 1970
0
Yes!! This is a great way to learn. You can only learn so much from books.
Experimentation is the key to a long, successful, career in electronics.

ROTFLMAO!!
Thanks, Bob. I can always use a bit of humor to lighten an otherwise
dull day.

J

#### John Fields

Jan 1, 1970
0
I appreciate where you're coming from on this one, John. Also, JK's
explanation was outstanding in detail and clarity.
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.

---
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.
---

M

#### Miles Harris

Jan 1, 1970
0
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.

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

J

#### John Fields

Jan 1, 1970
0
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
inventory of devices you've demonstrated you know little about.
---

M

#### Miles Harris

Jan 1, 1970
0
---
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
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.

Nil points for observation and comprehension again, Jr.

J

#### John Fields

Jan 1, 1970
0
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:

---
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?

M

#### Miles Harris

Jan 1, 1970
0
So you retract your earlier mis-statement? That's a welcome change.
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.. :-(
I won't bother to request an apology for obvious reasons...

J

#### John Fields

Jan 1, 1970
0
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?

K

#### keith

Jan 1, 1970
0
---
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.

....zipless ****. Though Miles' parents apparently mastered it.

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