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!)?
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.
Divide the volts by the resistance connected across it to find the amperes passing through that resistance.
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.
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.
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 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.
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.
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.
Yes, after some algebra to solve for I.
It makes your current time-dependent. That's too complex for you to worry about, just yet.
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.
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.
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.
Wrap yourself in metal, and the lightning will mostly choose that material and avoid you almost completely.
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.
It will have a short circuit current limit that occurs when its internal voltage is all used up across its internal resistance.
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.
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?
Are you sure about that, John? I find that hard to believe. Googling for a figure, I found this:
formatting link
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."
Yes!! This is a great way to learn. You can "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."
--
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.
--
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.
--- 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?
--
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?
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:
Nil points for observation and comprehension again, Jr.
So you retract your earlier mis-statement? That's a welcome change.
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...
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