OT: voltage vs. current, the basics.

For those who got a B or better in physics?
I still don't get the relationship between volts and current. Can
someone give me an analogy? I understand the voltage is a force with
the potential to do work. Does this explain it roughly? Two
cartridges of the same size with the same amount of powder in each (so
they have the same potential to do work). One cartridge is necked
down for a .223 40gr bullet and the other is .375 200gr. Not
accounting for the difference in fps (electricity is at the speed of
light?), then is the bullet mass and the energy has when it arrives a
stand-in for the Amps at the same voltage (amount of powder)in this
analogy??? Does that hold for two identical .270 bullets (same amps)
driven by different grains of powder (voltage), so that of the two,
the amps driven by lesser voltage crossing the potential are less
powerful? Any better and SIMPLE analogy is appreciated?
TIA
Reply to
John
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I think that water in a garden hose is a better analogy. Electrical voltage is like the pressure in a hose and electrical current is like the amount of water that is flowing. If there is a partially opened nozzle on the end of the hose, that's analogous to an electrical resistor.
For electricity, Ohm's law is: Voltage = Current x Resistance
The same relationship holds for the water example above.
Andy S.
Reply to
andy
Perhaps a hydraulic analogy will help? Voltage is pressure while current is flow. Let's say you have a pump putting out a steady 100 PSI to a hose that runs from the output of the pump back to the reservoir. In the hose there is a valve. The valve would be resistance and would control the flow.
If you open the valve all the way you have zero resistance and maximum flow. The electrical analogy to this would be called a short circuit.
If you close the valve you have essentially infinite resistance and zero flow. The electrical analogy to this would be an open circuit.
With the valve open partially you have a value of resistance that controls the amount of flow. In the world of electricity the equation to calculate the amount of current or flow is...
Voltage (Volts) divided by Resistance (measured in Ohms) equals Current (Amps).
That equation is Ohms Law. V is used to represent voltage, R to represent restistance and I to represent current. I can't remember why I is used so someone else will have to answer that one. ;-)
V = I * R I = V / R R = V / I
Best Regards, Keith Marshall snipped-for-privacy@progressivelogic.com
"Even if you are on the right track, you'll get run over if you just sit there." - Will Rogers (1879-1935).
Reply to
Keith Marshall
Maybe because C is used for Capacitance. Annoying that they chose to use I for current, though - that, in turn, may be why L is used for Inductance! :-)
ABS
Reply to
Alaric Snell-Pym
Actually I was thinking more of A for Amps. :-)
Best Regards, Keith Marshall snipped-for-privacy@progressivelogic.com
"Even if you are on the right track, you'll get run over if you just sit there." - Will Rogers (1879-1935).
Reply to
Keith Marshall
The water analogy is a good tool. With arc welding there is a difference since you are dealing with an electric arc jumping a gap. The characteristics of the power supply control just what happens when the arc distance is increased or decreased. Generally with stick machines ( constant current ) the transformer holds the amperage fairly steady. If you draw back increasing the distance you need more water pressure (voltage) the transformer is designed to give more voltage and the amperage drops slightly. If you hold a close arc the voltage drops and the amperage slightly increases. On some units this relationship or slope can be varied for different applications. With constant voltage power supplies such as wire feed systems the current tends to vary with an increase in amperage when stickout is decreased and a decrease if wire stickout is lengthened. The voltage varies only slightly. The design of the transformer has a lot do with the ease of welding and the particular application. Randy
Reply to
Randy Zimmerman
yes !!!! I really like that feature on a MIG(wirefeed) being able to momentarly adjust "heat " just by moving the gun closer or away from the work piece , specially on thin stock.
Miller defeted this" ability of momentarly temporary adjustment " , by adding, wire tracking feature to their millermatic 135 nd 175 . I hated it.
there is nothing like having control over heat/ pudle (molten metal ) in a welding application ..... that is why I am hooked on TIG.
sorry for swaying away from the topic.
Reply to
acrobat-ants
Hi John, As usual for this group, some excellent answers so far. In your post you also mention work and energy.
In simple electrical circuits, power is the voltage multiplied by the current, measured in watts.
Power is the rate at which work is done.
A watt is equivalent to expending 1 joule of energy per second.
Hope this helps
regards,
John
Reply to
john johnson
So in your average stick welder, if you change the amps, does the voltage change too? Is it different with AC and DC?
Eide
Reply to
Eide
Not really. The voltage varies according to how long the arc is.
Reply to
Ernie Leimkuhler
The way I remember it is to first think of voltage as a 'potential difference'. Basically you have two charges at two different locations at differnt charge levels. The more drastic the differnce the more potential for current.
Therefore: I = E/R or I is proportional to current
At the same time anything in the path will absorb potential energy and convert it to heat.
Therefore: I = E/R or R is inversly related to current
BTW: This eplains why doing powerline repair via helicopter can kill you if you don't connect a grounding line first. It's not the power line, it's the helicopter generating a higher charge (potential difference) than the wire.
Dave
Reply to
Dave

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