Volts, Amps, Ohms, and Capacitors!

[xeluc]

Hello, first off, I will warn you that my questions are really fundamentals of Physics that I probably should already know, but EVERYONE doesn't know something until they learn, so here's my chance.

 

First off, I need clarification on some things. I know what an amp is. It's how fast the current is moving. Great. Now Voltage. I know the literal definition. It states that Voltage is the Potenital difference of charges or something ot that effect. What I may or may not understand is it's relationship with Amperage. I know that you can convert high voltage low amperage currents to low voltage high amperage currents, but I don't understand how voltage is measured. It's easy to understand that current moves faster in higher amperage currents so that means high amperage can kill you easily. thing is, If I have a high and low voltage of the same amperage going throuh me, what would be the difference. I realize that the higher voltage could be converted to a higher amperage, then again so coudl the lwoer voltage. So like I said, whats the relationship here, other than jsut sayuing voltage is lie kthe pressure in a water pipe. that doesnt hlep. What difference does a high and low voltage have when both currents have the same amperage? So that's that. Now we go to resistance. I understand the equation I=V/R . It's all good. Here's what I dont get. If you have a current thats 9 volts flowing at 1 amp, the resistance would be 9 Ohms, correct? That would satisfy the above equation, but then you have to deal with the resistance of the wire the current is traveling through. I understand that the resistance would be higher (since the equation could be re-arranged as IR=V) so wouldn't the voltage continually rise as it flowed through the wire? I know it doesn't SEEM correct, but that's what the equation states. Of course some power would be lost due to heat from the resistance, but I'm not sure if that's relevant. So here comes my last 2 questions. Sorry again. If I have 2 capacitors; one the size of a marble and the other the size of a 2-liter bottle, and they were both charges at 1.5 Volts; What would happen? Would they both be charged at 1.5 votls and the bigger capacitor have a larger amperage? Would they both have the same amperage and the bigger capacitor have a larger voltage? Would the larger capacitor have a larger amperage AND voltage? If so, what determines the ratio of power between voltage and amperage. So thats one quesiton ; here is the last one. If I have 9 volts flowing at 1 amp from a capacitor, and in the "circuit" is a mor eresistive material, like graphite; would that raise the voltage since resistance was added? That ties in with a previous question I guess.. So thats everything I'd lie kto know. I'm sorry if all this has been answered, i looked around for a little while.. And I'm sorry that this was very long also. SAo thanks in advance for anyone who helps me!

[Bio-Hazard]

I usually like it when people space out their paragraphs......

 

Anyways, the voltage can kill you. You must remember that resistance comes from somewhere. Even the human body has a resistance to electricity.

 

So, if you have a super high voltage, (pressure of water) and you touch it with your finger (making a dam for hurricane katrina that wasn't good enough) you are going to get shocked like hell.. Because resistance against a voltage will create a current.

 

Potential difference is just a word. I like to think of how much of a wallop it can have when applied to a resistance. Negative voltage isn't going to do anything to a piece of rubber. It's like how hard someone is going to hit. When the person gets hit, they have a muscle resistance. Their pain will be the amperage. Either they'll feel so much pain they get knocked, or not.

 

I tried to answer a few questions. I think I messed up on the human analogy, but eh.

 

These people are my friends:http://www.dutchforce.com/~eforum/index.php

[Externet]

....."First off, I need clarification on some things. I know what an amp is. It's how fast the current is moving. Great.

 

 

------> WRONG. First, it is not amp. It is Ampere, and second, is the amount of electrons flowing per second. EXACTLY as water flowing in a pipe at a constant velocity.

 

 

Now Voltage. I know the literal definition. It states that Voltage is the Potenital difference of charges or something ot that effect.

 

--------> your "something ot that effect" Is like the PRESSURE on a water pipe.

 

What I may or may not understand is it's relationship with Amperage.

 

--------> More Volts, -> more Amperes. As if the water has more pressure, more amount of water will flow.

The rest of your post is a salad of confusion, set it aside until you understand the concepts clearly, and then we can continue ONE question at a time

 

 

I know that you can convert high voltage low amperage currents to low voltage high amperage currents, but I don't understand how voltage is measured. It's easy to understand that current moves faster in higher amperage currents so that means high amperage can kill you easily. thing is, If I have a high and low voltage of the same amperage going throuh me, what would be the difference. I realize that the higher voltage could be converted to a higher amperage, then again so coudl the lwoer voltage. So like I said, whats the relationship here, other than jsut sayuing voltage is lie kthe pressure in a water pipe. that doesnt hlep. What difference does a high and low voltage have when both currents have the same amperage? So that's that. Now we go to resistance. I understand the equation I=V/R . It's all good. Here's what I dont get. If you have a current thats 9 volts flowing at 1 amp, the resistance would be 9 Ohms, correct? That would satisfy the above equation, but then you have to deal with the resistance of the wire the current is traveling through. I understand that the resistance would be higher (since the equation could be re-arranged as IR=V) so wouldn't the voltage continually rise as it flowed through the wire? I know it doesn't SEEM correct, but that's what the equation states. Of course some power would be lost due to heat from the resistance, but I'm not sure if that's relevant. So here comes my last 2 questions. Sorry again. If I have 2 capacitors; one the size of a marble and the other the size of a 2-liter bottle, and they were both charges at 1.5 Volts; What would happen? Would they both be charged at 1.5 votls and the bigger capacitor have a larger amperage?

 

--------> The charge on a capacitor is not Volts nor Amperes, it is Coulombs, equal to Volts times Farads.

 

Would they both have the same amperage and the bigger capacitor have a larger voltage? Would the larger capacitor have a larger amperage AND voltage? If so, what determines the ratio of power between voltage and amperage. So thats one quesiton ; here is the last one. If I have 9 volts flowing at 1 amp from a capacitor, and in the "circuit" is a mor eresistive material, like graphite; would that raise the voltage since resistance was added? That ties in with a previous question I guess.. So thats everything I'd lie kto know. I'm sorry if all this has been answered, i looked around for a little while.. And I'm sorry that this was very long also. SAo thanks in advance for anyone who helps me!

[Flunch]

Externet is right, you need to master the concepts of voltage, amperage and resistance first.

 

Think of voltage (pressure) as what causes electrons (or water) to flow. The flow is the current. As current or water flows through conductors (or pipe) the voltage (pressure) drops because of resistance (electrical or frictional). So across any resistor we can measure a voltage drop.

 

Voltage will continually DROP as current passes through increasing resistance of a circuit. The total of the drops across all resistors (including the small resistance of the conductors) will be the voltage of your battery. For elementary purposes you may decide to ignore the resistance of the conducting wires because they are very small reletive to say a 3 ohm resistor in your simple circuit.

 

Also remember that Power (P) = Voltage (V) x Current (I). A 1 amp current flowing between a potential difference of 9 volts constitutes 9 watts of power. If the same current flowed between a potential difference of 5 volts, you'd have 5 watts.

[insane_alien]

err voltage doesn't kill. its current that kills. i got a 20,000 Volt van de graaf and i if i can die from 230V mains voltage then this thing should completely fry me if it was voltage. but the current it can produce is tiny so i live when i touch it.

[5614]

err voltage doesn't kill. its current that kills. i got a 20,000 Volt van de graaf and i if i can die from 230V mains voltage then this thing should completely fry me if it was voltage. but the current it can produce is tiny so i live when i touch it.

Yep, the voltage is what makes the current happen.

 

So there's a voltage difference between you and a charged object, this will not kill you, but if you touch the charged object then when you touch, because of the voltage, a current will flow through you, this current flowing through you (if it is high enough) will kill you.

[insane_alien]

but never the less it is the current

[5614]

I totaly agree.

 

At the same time it is important to note that if there were no voltage then there would be no current.

 

It's totaly true and everyone was stressing it's the current which kills you, I was just pointing out that to make that lethal current you need a voltage.

 

The voltage makes the current, the current kills.

[danny8522003]

There's something ive always been confused about. In a DC circuit, the electrons flow from the negative terminal to the positive terminal. Does the current follow this same direction or does it go the other way?

[insane_alien]

current is treated as flowing from the positive end even though the actual current is flowing from the negative end. Who started this convention to have current flow from the positive end? the americans....

[Bio-Hazard]

current is treated as flowing from the positive end even though the actual current is flowing from the negative end. Who started this convention to have current flow from the positive end? the americans....

 

Ey. That's a touchy subject. Don't be starting about electron flow vs. conventional flow.

 

I believe through science people have tracked that electrons come from the negative side of a DC power source. Inside of the DC source however, the flow is reversed.

 

Electrons are negative.

[xeluc]

"--------------------------------------------------------------------------------

 

....."First off, I need clarification on some things. I know what an amp is. It's how fast the current is moving. Great.

 

 

------> WRONG. First, it is not amp. It is Ampere, and second, is the amount of electrons flowing per second. EXACTLY as water flowing in a pipe at a constant velocity."

I know it an Ampere, excuse me for abreviating.

 

Next, I have "mastered" the relationship between current, voltage, and resistance. I=V/R takes care of that.

 

I also understand that Capacitors are rated in Farads. my questions are about the current produced when the two wires of a charged capacitor are brought together. Since noone is touching my huge paragraph, ill structure two questions I have in Paragraphs.

 

Question 1: If I have two different sized capacitors and they are both charged using 1.5 Volts DC; What difference would there be in the Capacitors. The larger one would obviously Have more electricity in it, but would the larger capacitor have more volts or Amperes in the current it produced when connected to itself.

 

Question 2: Do capacitors dump out all of their energy at once or is it relativly slow (by that i mean a few seconds/minutes instead of instantaniously)? It doesn't make much sense for a computer to use a capacitor if all energy it had is expelled instantaniously, it woul have to be able to slowly draw power from the capacitor. So either the capacitor dumps it's energy slowly or there are other components that conserve the charge. Can soemone please tell me which? Because Camera flashes use capacitors that dump their energy all at once..

[danny8522003]

It all depends upon the resistance of the circuit, the lower the resistance the faster they discharge.

 

If you charge both the large and small capacitor with 1.5V DC then they will both have a p.d. of 1.5V across the terminals, the current upon discharge depends on the capcitance and resistance of the curcuit.

 

You maybe interested to know that the rate of decrease in V and Q are exponential functions.

 

X=X0 * e^(-t/CR)

[xeluc]

So, if I had an extremly large capacitor, say The size of a computer tower, and hooked up a variable resistor to it, I could make a low voltage that would last a while or a high voltage that would go by quick.... right? If so, this serves my purpose exactly.

 

Lastly, The two capacitors are charges with 1.5 volts. I'm just assuming that you "short" the capacitor. So we'll jsut say that the resistance is negligible. SO that's all there is in the circuit. My question would be, if i hooked up a voltmeter to it would it read 1.5 volts or would it read a higher voltage the bigger the capacitor is. Same thing with Amperage. I thought that capacitors kept a constant voltage as the amperage dereased and an inductor keeps the same amperage while the voltage increases and decreases depending on the power going to it. Am I right or wrong in saying this...

[danny8522003]

With capicitors, it isnt physical size that matters. What you're saying though is true,capacitance is the amount of charge required to increase the p.d. across the terminals. So yea, "bigger" = longer.

 

No, if you charge it with 1.5V, it will discharge from 1.5V. I could give you a huge analogy to explain like my physics teacher did but it would take too long on a forum though. The current around the circuit decreases exponentially as the capacitor discharges.

[xeluc]

So what your saying is that If i charge the capacitor at 1 volts. It will discharge at 1 volts with a varying amperage? So it could be said that a larger Capacitor would hold a higher "start" amperage, therfore holding the voltage longer... right? Does that also mean that you can store more energy in a capacitor at a higher voltage? If you can charge a capacitor 1 Ampere at 1 volts and 1 Ampere at 20 volts, then the only real limit to the amount of charge the capacitor can achieve is either when theres so many electrons in the capacitor that it physically cant fit more in it or if the voltage gets so high that it spontaniously arcs and shorts itself out... I think this is all right.. yes? no? If all this is true then my questions are answered!

 

EDIT: I guess when I say 1 ampere I really mean that the capacitor is rated at one farad, so the capacitor would discharge 1 amp in 1 second at one volt. But if the capacitor were charged at 9 volts, the nthe capacitor should discharge 1 amp at 9 volts in a second. OF COURSE this does not take resistance into effect; All I'm trying ot do here is understand capacitors fully, I'll add resistance to the equation later

[danny8522003]

No, the current is only dependant on V=IR. A capacitor charged to 1.5V will produce a certain current regardless of its 'size'. The higher the number of farods, the longer it will take for its voltage and current to fall.

[xeluc]

Ok, that's great. So is there any way of... inferring what the voltage and current would be? There has to be some ratio or equation..

 

BTW, thanks for taking the time to help me out. It IS appreciated

 

edit: I think im understanding this. The size of the capacitor has nothing to do with what the voltage and current will be, the size will determin how long that charge will last?

[danny8522003]

Yea, as i said earlier there is the equation:

 

X=X0 * e^(-t/CR)

 

and

 

C=Q/V.... Baring in mind V=IR and Q=It

[Externet]

Hi xeluc. (This is long because you made it that way)

 

..... "my questions are about the current produced when the two wires of a charged capacitor are brought together. Since noone is touching my huge paragraph, ill structure two questions I have in Paragraphs.

 

-----------> The current produced joining the wires of a charged capacitor is : Amperes = charge in Coulombs ÷ time in seconds.

And the Amperes will be the Volts ÷ by the resistance of the wires.

Then,

 

Coulombs ÷ seconds = Volts ÷ Ohms

 

 

Question 1: If I have two different sized capacitors and they are both charged using 1.5 Volts DC; What difference would there be in the Capacitors. The larger one would obviously Have more electricity in it, but would the larger capacitor have more volts or Amperes in the current it produced when connected to itself.

 

----------->The larger CAPACITY one (not size !) will have more charge in Coulombs, (not volts nor amperes) and (not electricity)

 

Example: If the large capacitor is 1000 µF, charged to 1.5 V, it will contain 1500 µCoulombs.

If the small capacitor is 10µF, charged to the same 1.5V, it will contain 15 µCoulombs

 

Question 2: Do capacitors dump out all of their energy at once or is it relativly slow (by that i mean a few seconds/minutes instead of instantaniously)?

 

---------> the time to discharge to 66% is time in seconds = Charge in Coulombs ÷ current in Amperes.

 

It doesn't make much sense for a computer to use a capacitor if all energy it had is expelled instantaniously, it woul have to be able to slowly draw power from the capacitor. So either the capacitor dumps it's energy slowly or there are other components that conserve the charge. Can soemone please tell me which? Because Camera flashes use capacitors that dump their energy all at once..

 

---------> Yes, but you are still confused. A camera flash will discharge the capacitor in few milliseconds. And will take many seconds to recharge.

And a capacitor in a computer circuit is NOT meant to discharge and recharge cyclically; it is meant to discharge briefly a small amount of its charge while continuosly being charged ALSO.

 

 

So, if I had an extremly large capacitor, say The size of a computer tower, and hooked up a variable resistor to it, I could make a low voltage that would last a while or a high voltage that would go by quick.... right? If so, this serves my purpose exactly.

----------> It would not MAKE a lower voltage, it would discharge a CURRENT. A larger CAPACITY capacitor (NOT SIZE!) will discharge slower trough a larger resistor.

 

Lastly, The two capacitors are charges with 1.5 volts. I'm just assuming that you "short" the capacitor. So we'll jsut say that the resistance is negligible. SO that's all there is in the circuit. My question would be, if i hooked up a voltmeter to it would it read 1.5 volts or would it read a higher voltage the bigger the capacitor is.

 

-----------> both capacitors charged to 1.5 Volts, shorting the leads, the voltage will decrease immediately in microseconds.

 

Same thing with Amperage. I thought that capacitors kept a constant voltage

------------> NO. Voltage nor Amperage is NOT constant during discharge

nas the amperage dereased and an inductor keeps the same amperage while the voltage increases and decreases depending on the power going to it. Am I right or wrong in saying this...

------------> You are confused.

 

So what your saying is that If i charge the capacitor at 1 volts. It will discharge at 1 volts with a varying amperage?

-------------> No. The voltage will start decreasing immediately when current starts flowing.

 

So it could be said that a larger Capacitor would hold a higher "start" amperage, therfore holding the voltage longer... right?

-------------> Yes.

 

Does that also mean that you can store more energy in a capacitor at a higher voltage?

-----------> Yes. Energy in Joules = voltage x capacity. More Volts = more energy

 

If you can charge a capacitor 1 Ampere at 1 volts and 1 Ampere at 20 volts, then the only real limit to the amount of charge the capacitor can achieve is either when theres so many electrons in the capacitor that it physically cant fit more in it or if the voltage gets so high that it spontaniously arcs and shorts itself out... I think this is all right.. yes? no? If all this is true then my questions are answered!

----------> No and yes and no. You do not charge Amperes to a capacitor; you charge Coulombs. The limit is when its capacity is filled. It will arc if its voltage ratings is exceeded.

 

your EDIT: I guess when I say 1 ampere I really mean that the capacitor is rated at one farad, so the capacitor would discharge 1 amp in 1 second at one volt.

-------------> If you meant 1 Farad charged INITIALLY to 1 Volt,( = 1 Coulomb) it will discharge 1 Ampere during 1 second.

 

But if the capacitor were charged at 9 volts, the nthe capacitor should discharge 1 amp at 9 volts in a second. OF COURSE this does not take resistance into effect; All I'm trying ot do here is understand capacitors fully, I'll add resistance to the equation later

---------> If 1 Farad is charged to 9 Volts, (9Coulombs) it can discharge 1 Ampere during 9 seconds or 9 Amperes during 1 second

 

Ok, that's great. So is there any way of... inferring what the voltage and current would be? There has to be some ratio or equation..

----------->The current is = Coulombs ÷ time in seconds

 

BTW, thanks for taking the time to help me out. It IS appreciated

 

your edit: I think im understanding this. The size of the capacitor has nothing to do with what the voltage and current will be, the size will determin how long that charge will last?

-----------> You are getting closer.

Suggestion : Print this and read and re-read slowly.

Miguel

[5614]

There's a lot of posts since I last read, I haven't read them all, all I'm saying is the current vs electron flow.

 

Before the discovery of electrons scientists knew there was a current in an electrical circuit. They thought it logical that this current should start at the positive end and travel to the negative end.

 

Then after the discovery of the electron it was realised that electrons must flow from negative to positive, that is the nature of an electron.

 

So now we have a combination, current flows from positive to negative whereas current flows in the opposite direction.

[xeluc]

Wow, that was an extremely thorough explanation, thanks alot. I understand everything you said; You said a 1 Farad capacitor charged with 9 volts can discharge 1 Ampere in 9 seconds or 9 Amperes in 1 second. How would you pick which one you wanted ot happen, a resistor?

[Externet]

Hello xeluc.

 

...."I understand everything you said; You said a 1 Farad capacitor charged with 9 volts can discharge 1 Ampere in 9 seconds or 9 Amperes in 1 second. How would you pick which one you wanted ot happen, a resistor?"

 

--------> Yes, the resistor . But your wording should say {Charged TO 9 Volts, not WITH}

 

A 1 Farad capacitor charged to 1 Volt would have:

 

1 F x 1 V = 1 Coulomb.

 

If the wires discharging it have 1 Ohm resistance:

 

1 V ÷ 1 Ohm = 1 Ampere

 

and the time to discharge is

 

1 Coulomb ÷ 1 Ampere = 1 second

 

The other case, with the voltage being 9 Volts:

 

9 V ÷ 1 Ohm = 9 Amperes

 

and the time to discharge is

 

9 Coulombs ÷ 9 Amperes = 1 second

 

Or; if you increase the resistance of the discharging wires

With the resistance being 9 Ohm instead of 1 Ohm,

9 Volts ÷ 9 Ohms = 1 Ampere

 

The discharge time is :

9 Coulombs ÷ 1 Ampere = 9 second

 

[if you increase the resistance, the current will discharge in a longer time]

 

Remember:

 

Coulombs ÷ seconds = Amperes = Volts ÷ Ohms

 

And imagine a swimming pool (large CAPACITY capacitor) filled with 2 inches of water (~2 Volts) and a frying pan (small capacity capacitor) filled also with 2 inches of water (same ~2 Volts)

The time needed to empty each trough equal drain pipe size (resistor) will be different.

Changing the size of the drain pipe (resistor) will change the draining times.

 

The water depth is the voltage; the size of the container is the capacitor capacity, the restriction by the drain is the resistance.

 

I want to correct MY mistake in post #20 above.

Where says " Energy in Joules = voltage x capacity "

Should say "Energy in Joules = voltage x charge"

 

 

Miguel

[xeluc]

Excellant! So one last thing. Is there any formula to determin the capacity of a capacitor? Like say I had a capacitor that had a 2 70 square feet plates seperated by wax paper. Could I find it's capacity?

[Externet]

Hello.

The formula is:

C in Farads = Plate area in cm² x k (dielectric constant) ÷ distance between plates in cm.

The constant for waxed paper is around 2. And good for up to about 300Volts.

Miguel