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Current, Voltage, and Batteries

Passions

Diamond Member
I am a bit confused here. From what I understand, current is the rate which electric charge moves through
a conductor. And Voltage is force that causes that electric charge to move in the first place. So,

high current, low voltage is where speed of electrons is moving very fast with little amount of force.
high voltage, low current is where alot of force is being applied but no electrons are moving very fast.

Is that correct? And how come on batteries, they do not tell you the amperage. Is it not important regarding
batteries? A high voltage on a radio means a stronger signal and on a flashlight it means a brighter glow,
what would a high current do to those things? Blow them up? Thanks.
 
Realize that what a battery is is just a capacitor, not a conductor. When you attach it to an electrical device, it then becomes a circuit. That electrical device can have vastly different resistance, like a piece of metal may only have 1 ohm resistance while a cd player may have 500 ohms resistance therefore the current (ampere) the battery will induce will be drastically different depending ont he device you put it in. There is a measurement of how many ampere-hours (average) that the battery will last before the chemical properties of the batteries wear out.
 
Ohm's Law my friend: V=I*R
You can't just talk about voltage and current, without including resistance.

Your statement should actually be this:
high current = high voltage and low resistance
low current = low voltage and high resistance

With the battery issue, it's conventional to just provide voltage information. Like imgod2u mentioned, depending on the type of application, the current drawn is mostly dependent on the total resistance of the complete circuit, and the chemical integrity of the battery. A higher voltage generally means stronger performance in electronics because a bigger current can flow. Current flow is the key, not voltage. When people get electrocuted, it's the CURRENT that causes the injuries, not the voltage.
 
Think of this analogy: You have a water tower and a pipe at the bottom of the tower that feeds a town. The voltage potential is related to how much water is in that tower. The resistance is inversely related to the diameter of the pipe. Current is how much water flows out of that pipe. Now, if you reduce the diameter of that pipe (thus, increasing the resistance), there is going to be less water flow. I know this is a crude analogy but it's what helped me understand Ohm's Law.
 
High current can also be a large amount of electrons moving very slowly as opposed to a small amount moving very quickly.

With batteries, the reason they provide only voltage information is because for all intents and purposes, the battery you buy at the store is a constant voltage source. It'll provide as much or as little current to maintain a certain voltage. Up until it runs out of charge, of course.
There are also power supplies that will provide constant current. Same principle, just swap voltage and current.
 
Originally posted by: blahblah99
Think of this analogy: You have a water tower and a pipe at the bottom of the tower that feeds a town. The voltage potential is related to how much water is in that tower. The resistance is inversely related to the diameter of the pipe. Current is how much water flows out of that pipe. Now, if you reduce the diameter of that pipe (thus, increasing the resistance), there is going to be less water flow. I know this is a crude analogy but it's what helped me understand Ohm's Law.

you got voltage wrong - voltage is the height of the tower.
 
> high current, low voltage is where speed of electrons is moving very fast with little amount of force.
> high voltage, low current is where alot of force is being applied but no electrons are moving very fast.

Unfortunately, this is not quite correct. Voltage is the difference between two potentials. When an electron is accelerated through a potential difference, from say a battery, the increase in the electron's kinetic energy is directly proportional to that potential. That is, larger voltages result in faster-moving electrons. This principle is used in particle accelerators, X-Ray devices, even your computer CRT.

For example, a 7eV electron (that is, an electron that has been accelerated from 0 through a 7 volt potential) has a speed of a little less than 1/2 of 1% the speed of light...about 3.5 million miles/hour.

But this isn't the whole story. When electrons pass through a conductor, say copper wire, they don't pass freely. They bounce around scattering off the copper atoms. At room temperature, the so-called "mean free path" of copper is on the order of 150 times the size of a copper atom. This is the average distance a conduction electron in the copper will travel before scattering off a copper atom -- bouncing off in a different direction. Given the aforementioned electron speed, you can imagine not much time passes between collisions. There are a boatload of collisions each second.

Now, if the ends of the wire aren't connected to a battery, there is no potential difference and thus these conduction electrons are bouncing around in random directions within the wire. The net result is no current flows which is what we'd expect. If you were to discover electrons "pouring" from the end of a wire without the presence of a potential difference, you'd become a wealthy person 🙂

Now, attach ends of the wire to a battery. Now there's a potential difference (and thus, an electric field) in place and this will affect the overall direction of these bounces. Instead of scattering off in random directions, the electric field will cause the conduction electrons to experience a force, attracting them to one end of the wire. So while they're still bouncing into copper atoms, if you were to track the horizontal position of a single conduction electron, you'd see it very slowly make its way to one end of the wire. This gives rise to the electron "drift speed." In a typical copper wire with a 1amp current, the drift speed might be as little as a dozen cm/hour. The reason you see lights come on almost immediately when you flip the switch is that the wire is full of conduction electrons distributed uniformly throughout and an applied electric field travels roughly at the speed of light. A rough analogy would be like when you turn on a garden hose...if the hose doesn't contain any water, you'll have to wait a few seconds for water to start pouring out the end; if the hose is already full, you see water come out almost immediately.

So, while the drift speed of electrons in the "high current, low voltage" scenario is higher than those of the "high voltage, low current" scenario, the individual high voltage electrons are moving faster than the low voltage electrons. And in neither case would you consider the drift velocity to be "very fast": a 100amp current in a 16gauge wire would yield a drift speed of around 10 meters/hour while a 1amp current in the same wire yields a drift speed of around 10 cm/hour.

 
Originally posted by: blahblah99
Think of this analogy: You have a water tower and a pipe at the bottom of the tower that feeds a town. The voltage potential is related to how much water is in that tower. The resistance is inversely related to the diameter of the pipe. Current is how much water flows out of that pipe. Now, if you reduce the diameter of that pipe (thus, increasing the resistance), there is going to be less water flow. I know this is a crude analogy but it's what helped me understand Ohm's Law.

This is a very good analogy. Voltage is just like the pressure in a water system whose potential energy is expressed in 'head feet'. Current is like the rate of the water coming out of the water pipe in volume per unit time like liters/minute.

 
Hay. I was going to post something related to electricity too but i found this so i'll add mine. BTW, the analogy here is quite good.

You guys were talking about how increasing the resistance will decrease the current. Ok. So does that mean that if i put a 500 Ohm resistor, then compared it to a 50 Ohm resistor on a different circuit with the same voltage, the 50 ohm resisotr would dissapate more heat? This is because i read that, in both circuits, the electron must give up all energy to the resistor since there are no other places to give it up. (assume that wires ofer no resistence.) Since in the 50 Ohm resistor there would be a large current flowing, more electron give up their energy per unit time, therefore, more energy dissapated.

Now... I find this puzzling..... Because, the less resistence, the more energy dissapated. Does this mean that a wire.... perhaps with 0.001 Ohm would dissapate LARGE amounts of heat?. Actually you notice this when you short circuit a battery. the battery gets hot. BUT, still, connect a 1 OhM resistor to a cuircit, shouldn't the energy dissapated be huge?

If i could try it i would test if a lower Ohmic resistor will be hotter or not...
My dilema is that i always thought that a large resistance will appose electron flow more and therefore dissapate more heat.
I think my thinking is wrong but to me it is theoretically logical, yet impractical...




Oh and a quick problem i found too...

If a device has a resistance of 100Ohm
What would be the resistanec of another device that had twice the power rating as the first, provided that is recieves that same voltage?

you see, using my thinking, the device with more power rating should have less resistance so it will dissapate more energy given unit time.... I am confused indeed.
 
So does that mean that if i put a 500 Ohm resistor, then compared it to a 50 Ohm resistor on a different circuit with the same voltage, the 50 ohm resisotr would dissapate more heat?

Yep. The "heat" you're speaking about is the "power dissipated by the resistor." To find that value, you multiply the current by the voltage.

Let's say you have a circuit carrying 100V and no resistance. Insert your 50 ohm resistor and calculate the resulting current from Ohm's Law. The circuit will carry 2 amps. That means the 50ohm resistor is dissipating (100V * 2A) 200 watts. Replace the resistor with your 500 ohm resistor and recalculate. The circuit now carries 0.2 amps. That means the 500 ohm resistor is dissipating (100V * 0.2A) 20 watts. Thus, the 50 ohm resistor will get much hotter.

What would be the resistanec of another device that had twice the power rating as the first, provided that is recieves that same voltage?

You're looking at it wrong. The power rating for a resistor is related to the maximum amount of power (ie. heat) the component can safely dissipate. If you have 2 100 ohm resistors: a little 1/4 watt carbon resistor, and a big 20 watt wire-wound resistor, they both have the exact same resistance and would initially behave identically in a circuit. That is until you exceed the carbon's power rating. If you stick them in a circuit where the resistor will be forced to dissipate say, 10 watts, the little carbon resistor is likely to explode or cause a fire while the wire-wound resistor will continue to function happily.

 
I think they should include the amp rating on a battery. Imagine some poor unknowing person trying to start his/her car with 8 AA batteries!
 
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