# Amps and Volts

#### Pikester

Joined Nov 27, 2005
13
We're currently learning about amps and volts, and I'm having problems understanding the 2...
I know volts is the "possible" or "potential" energy(correct me if I'm wrong) and the amp is the current. I don't get how amps can be the current, and if amps are the currents, shouldn't most electrical appliance(lightbulb for example) should say the amount of amps not volts. Thats another thing I don't get, how is a ,oh let say, christmas lights "is" 2.5, or 5 volts, how can an object have "potenial" energy?

#### mozikluv

Joined Jan 22, 2004
1,435
hi

"shouldn't most electrical appliance(lightbulb for example) should say the amount of amps not volts."

in light bulbs they use volts and watts. you don't see the amps, but its there. use ohms law - I = P/E

"Thats another thing I don't get, how is a ,oh let say, christmas lights "is" 2.5, or 5 volts, how can an object have "potenial" energy?"

the 2.5v or 5v indicates the maximum amount of voltage that should be applied to each bulb. that's why you string up those christmas lights. if your source is 120vac and you use the 2.5v bulb, then you have 48. or 50 in which the applied voltage per bulb is 2.4v

the object does not have potential energy.

moz

#### Pikester

Joined Nov 27, 2005
13
Thx for the reply, but I still kinda fuzzy with the "Volts" and "Amps" actually are. For example, a 9v battery has the POTENTIAL of giving 9 volts, but the ACTUAL electricty running through a circuit is the amps(thats my understanding right there). Is this correct?

#### dougp01

Joined Dec 6, 2005
31
I always like to make an analogy to plumbing. I know it won't help on homework or a test but it does help you to conceptualize the idea. Think of voltage as water pressure (or potential) and amperage as water flow (or current).

You can have pressure on a pipe and no flow because the spigot is turned off. This is analogous to voltage or potential. The spigot acts like a resistance to flow; in this case it is infinite resistance and there is no flow.

When you begin opening the spigot, you reduce resistance the to flow and you get water current. This is is analogous to amperage or amps.

When we think of the two together, we have work. If you maintain pressure and flow at the same time, you then have work, or energy, or power.

We typically assign letter abbreviations to these terms as this:

E = Voltage (Volts, potential)
I = Amperage (Amps, current)
R = Resistance (Ohms, Ω)
P = Power (Watts)

Now:
The Ohms Law relationship is this: (E/R) = I
And by extension we have: (E x I) = P

-Doug

#### Pikester

Joined Nov 27, 2005
13
Ok, almost got this thing down, but i still have one more question.

If I use a peice of wire to connect a 9v battery to itself(with nothing else to the circuit) would there be an Amperage within the wire? From what I understand, since there is nothing blocking the "Potential energy", the Amperage would be 0.(E/R)=I so it's (9/0<- no blocking it, thus no resistance,) =0. And if it's 0 amps theres no current. Then why does the wire eventually get hot?

#### Firestorm

Joined Jan 24, 2005
353
the wire would probably carry a very very small amount of resistance but nothing worth measuring. On your math, you will realize that 9/0 isn't actually possible, because you can't divide 9 by nothing. In reality, it does have a lil' current and "eventually" get hot.

As for the potential energy. The lights have potential energy is the are on the tree only because they are elevated. If they fall, they will be using kinetic energy until it stops lol.

#### mozikluv

Joined Jan 22, 2004
1,435
hi,

moz

#### dougp01

Joined Dec 6, 2005
31
You can put a short wire across the battery and you already know what will happen. But let's talk about that.

We already know that it's impossible to actually achieve 0.0000... Ohms. So there is a very small but real resistance in the wire. Also the battery has an internal resistance that limits current flow.

So, if I = E/R you might discover that 9/0.001 equals 9000 Amps. Probably not true because the internal resistance of the battery is higher than this. But let's go with the 9000 Amps, or 9 kiloAmps.

Power is P x I or 9 x 9000, in this case. So theoretical power would be 81,000 Watts! Once again, not very likely, but it does explain why the battery overheats.

By the way someone mentioned that division by zero is not possible. This is true for your calculator, because if you divide any number by zero, the result is infinite.

-doug

#### kantgetslicker17

Joined Dec 7, 2005
6
Originally posted by Firestorm@Dec 6 2005, 05:51 PM
the wire would probably carry a very very small amount of resistance but nothing worth measuring. On your math, you will realize that 9/0 isn't actually possible, because you can't divide 9 by nothing. In reality, it does have a lil' current and "eventually" get hot.

[post=12248]Quoted post[/post]​
ARE YOU KIDDING ME? a LITTLE CURRENT? What would happen when you short a battery, by putting a wire across it, you would get maximum current. The current is determined by the internal resistance, and yes you will burn your hand if you happen to touch the wire that's shorting the terminal of the battery.

[attachmentid=1028]

In calculus, as the denominator approaches 0, your answer approaches infinity. If you have a graphing calculator, you can graph y=9/x. You can see as x is reaching closer to zero, y becomes infinitely large. Which means are current is reaching its maximum.

And don't worry Pikester, once you get to connecting a couple of circuits, you will start understanding the difference from voltage and current. That's if you have a good teacher to explain it. My teacher use cars traveling through the wire as if it was traveling on a freeway, to represent current.

#### Pikester

Joined Nov 27, 2005
13
My teacher use cars traveling through the wire as if it was traveling on a freeway, to represent current.
Your lucky... My teacher sucks.. He teaches us the basics, and just tell us to built circuits after circuits. If we didn't get how something worked, he'll just tell us to go look in the textbook. #### Firestorm

Joined Jan 24, 2005
353
sounds like my ap chem teacher . Thats why we have this forum #### Brandon

Joined Dec 14, 2004
306
Another way to think of it. Son of a plumber here, so the concept came very easy for me.

Voltage. Don't think of just pressure, rather, amount of water, or in EE, amount of electrons. If you have more at 1 place than another, then there is a difference, obviously.. this difference is voltage. Its this difference that creates the POTENTIAL to cause work in the form of releasing that potential as a FLOW of electrons called current.

Raising a bucket of water into the air is the same concept at pulling electrons from one place and moving them another. If u put a hose in the bucket while it is floor level, doesn't really drain. i.e. 0 volts. You raise the bucket in the air, the voltage goes up, the water flows. The higher you raise it (voltage) the faster it flows (current). ALSO, if you use a large hose (low resistance) thw water flows faster or a small hose (high resistance) the water flows slower.

Its comes down to a relationship which describes a difference in the amounts of something in 2 places, how fast this something goes between the difference, and how much of this something permitted to flow.

Voltage - Difference of the electrons/charge
Current - The amount of electrons/charge flowing between a difference (voltage)
Resistance - The 'force' slowing down the current.

Its not like a hard formula. Its very dynamic. If one thing changes in the equasion, something else will chage to compensate for it. It must maintain the V=IR relationship.

#### n9xv

Joined Jan 18, 2005
329
Originally posted by Pikester@Dec 8 2005, 08:57 PM
My teacher sucks.. He teaches us the basics, and just tell us to built circuits after circuits. If we didn't get how something worked, he'll just tell us to go look in the textbook. [post=12315]Quoted post[/post]​
Sad state of affairs in the education relm these days. Sounds like he does'nt really enjoy the subject matter he's teaching. One of my tech school teachers would practically write something on the chalk board with the right hand and 2-feet away was the left hand with the erraser! - Talk about keeping up!!!

Voltage sets up the condition & the current does the work.

#### JoeJester

Joined Apr 26, 2005
4,390
One of my tech school teachers would practically write something on the chalk board with the right hand and 2-feet away was the left hand with the erraser! - Talk about keeping up!!!
And when the teacher wonders about a high failure rate ... all they have to do is look in the mirror.

That's what happend to me and another instructor back in 1981 ... the students spent most of their time concentrating on transcribing rather then listening and paying attention. There are better ways to deliver such things.

#### lv91t

Joined Feb 9, 2006
1
OK I've got a question about the comment of resistance going to infinity and the divide by 0.

I'm doing an auto electricity course and we are playing with simple ole ohms law...

E = IR

So you have a circuit with a battery, a lamp, and a ground.

My teacher says voltage measured after the lamp will be 0, however current will always remain constant at whatever the total circuit amperage was measured as, lets just say 2.3 A.

So explain to me, how can you have 0 = 2.3 x R... it would seem to me amperage would have to go down, or resistance in some crazy way would skyrocket... somehow I feel like I should be doing some calc here to show the teacher up... but I am really rusty so I can't remember.

#### n9xv

Joined Jan 18, 2005
329
Voltage measured "after" the lamp would be at the grounded point. Current is still flowing to ground but no voltage can be measured from "that point" to ground.
You could measure voltage across the battery & voltage across the lamp. At the point your attempting to measure a voltage, there is effectively no resistance in the "miniscule" length of wire in the simple circuit. So, in effect R = 0. Therefore,

0 = 2.3 X R

0 = 2.3 X 0

Because in your example, R = 0. No resistance measured "after" the bulb to ground.

#### alim

Joined Dec 27, 2005
113
Originally posted by n9xv@Feb 9 2006, 08:26 AM
Voltage measured "after" the lamp would be at the grounded point. Current is still flowing to ground but no voltage can be measured from "that point" to ground.
You could measure voltage across the battery & voltage across the lamp. At the point your attempting to measure a voltage, there is effectively no resistance in the "miniscule" length of wire in the simple circuit. So, in effect R = 0. Therefore,

0 = 2.3 X R

0 = 2.3 X 0

Because in your example, R = 0. No resistance measured "after" the bulb to ground.
[post=13948]Quoted post[/post]​
Hi I follow n9xv reply and agree with it, but cannot find areason why the teacher would measure the voltage at that point,both meter probes essentially at the same point.What was the lesson about?

#### chesart1

Joined Jan 23, 2006
269
Originally posted by lv91t@Feb 9 2006, 04:29 AM
OK I've got a question about the comment of resistance going to infinity and the divide by 0.

I'm doing an auto electricity course and we are playing with simple ole ohms law...

E = IR

So you have a circuit with a battery, a lamp, and a ground.

My teacher says voltage measured after the lamp will be 0, however current will always remain constant at whatever the total circuit amperage was measured as, lets just say 2.3 A.

So explain to me, how can you have 0 = 2.3 x R... it would seem to me amperage would have to go down, or resistance in some crazy way would skyrocket... somehow I feel like I should be doing some calc here to show the teacher up... but I am really rusty so I can't remember.
[post=13944]Quoted post[/post]​

#### chesart1

Joined Jan 23, 2006
269
Originally posted by lv91t@Feb 9 2006, 04:29 AM
OK I've got a question about the comment of resistance going to infinity and the divide by 0.

I'm doing an auto electricity course and we are playing with simple ole ohms law...

E = IR

So you have a circuit with a battery, a lamp, and a ground.

My teacher says voltage measured after the lamp will be 0, however current will always remain constant at whatever the total circuit amperage was measured as, lets just say 2.3 A.

So explain to me, how can you have 0 = 2.3 x R... it would seem to me amperage would have to go down, or resistance in some crazy way would skyrocket... somehow I feel like I should be doing some calc here to show the teacher up... but I am really rusty so I can't remember.
[post=13944]Quoted post[/post]​

#### chesart1

Joined Jan 23, 2006
269
Originally posted by chesart1@Feb 9 2006, 04:01 PM

[post=13966]Quoted post[/post]​
If you want proof that current is the same in a series circuit, you should use an ammeter. You will note that no matter where you hook up the ammeter in the circuit, you get the same reading. Forgive me if I am getting too elementary, but remember that an ammeter is always hooked up in series. Never connect an ammeter across a component.

Here is an explanation of why the teacher assumes the resistance of wire is zero ohms:

First lets consider a circuit with a 1 volt battery connected to a 1 watt light bulb and assume the wire has no resistance.
p = e * i
1 watt = 1 volt * 1 amp
e/i = r
1 volt/1 amp = 1 ohm

Now lets consider the same circuit with a 1 watt light bulb connected to a 1 volt battery. Except this time lets consider the resistance of the wire. Lets assume the resistance of the wire between the battery plus terminal and the bulb is 0.00000001 ohms. Lets also assume that the resistance of the wire between the battery minus terminal and the bulb is 0.00000001 ohms. We know that the total current through the circuit equals the total voltage divided by the total resistance. And since there is resistance throughout the entire circuit, the current must flow through the entire circuit.

As you can see, the total resistance of the wire [0.00000002 ohms] is insignificant when compared to the 1 ohm resistance of the bulb and will not have any significant effect on the operation of the circuit. This is why the teacher assumes that the wire has no resistance.

When I studied basic electricity, I just assumed the rule "current flow is constant throughout a series circuit" without giving it much thought. That was a mistake on my part. There were so many things that I assumed was right simply because I was told it by a teacher who "should know."

It is good that you are questioning these issues.

John