I'm an electronics hobbyist, just getting in to the basics. I bought a
copy of "Make: Electronics" and I'm teaching myself from that.

I understand how Ohm's Law works for calculating real-world voltage/
amperage/resistance, but I am intrigued by the theoretical
implications of it. Given that V = R x I, if there was zero resistance
in a circuit (i.e. R = 0), then that implies that there would be no
voltage (because the voltage would calculate out to 0). I want to make
sure I understand this conceptually.

Using the water-pressure analogy, if there are two tanks connected to
each other with one tank full and the other empty, and a pipe at the
base of the tanks connecting them, then I understand that the voltage
is equivalent to the pressure of the water in the filled tank, and the
amperage the speed with which this pressure is relieved as current
flows from the filled tank to the empty one until equilibrium is
achieved. If the pipe between the tanks offers zero resistance, then
the speed with which the water moves between the tanks to equalize the
pressure should be extremely high (i.e., infinite amperage). In other
words, in a truly zero-resistance system, equilibrium would be
achieved instantly.

Is the fact that equilibrium is achieved instantly the reason why the
voltage would be zero? I.e., the potential between the tanks is
instantly resolved by the lack of resistance, with a theoretical
amperage of infinity?

Thanks,

- max

On Sun, 4 Apr 2010 08:07:50 -0700 (PDT), "maxim.porges"
wrote:

I'm an electronics hobbyist, just getting in to the basics. I bought a
copy of "Make: Electronics" and I'm teaching myself from that.

I understand how Ohm's Law works for calculating real-world voltage/
amperage/resistance, but I am intrigued by the theoretical
implications of it. Given that V = R x I, if there was zero resistance
in a circuit (i.e. R = 0), then that implies that there would be no
voltage (because the voltage would calculate out to 0). I want to make
sure I understand this conceptually.

Using the water-pressure analogy, if there are two tanks connected to
each other with one tank full and the other empty, and a pipe at the
base of the tanks connecting them, then I understand that the voltage
is equivalent to the pressure of the water in the filled tank, and the
amperage the speed with which this pressure is relieved as current
flows from the filled tank to the empty one until equilibrium is
achieved. If the pipe between the tanks offers zero resistance, then
the speed with which the water moves between the tanks to equalize the
pressure should be extremely high (i.e., infinite amperage). In other
words, in a truly zero-resistance system, equilibrium would be
achieved instantly.

---
Well, no.

There is the time required for the equilibrium to occur.
---

Is the fact that equilibrium is achieved instantly the reason why the
voltage would be zero? I.e., the potential between the tanks is
instantly resolved by the lack of resistance, with a theoretical
amperage of infinity?

---
No.

Equilibrium will be reached when the level of water in both tanks is
equal, which will take some time to achieve.

Also, equilibrium won't happen in one pass since the water has mass,
which will cause the water columns in the tanks to oscillate for a while
before everything quiets down.


JF

Thanks John - I think you are focusing on the water analogy a bit too
much, though. This is how voltage/amperage/resistance is typically
explained to newbies like me, but I'm trying to understand the
conceptual side of things as it relates to electronics.

My boss was an electrical engineering student for a while and I also
sent this question to him. Here was his reply for anybody who is
interested.

"Yes, no resistance, aka hard short, means no voltage. There are some
very clever ways you can work with just a light bulb to find and debug
shorts and opens. Remember that one over zero is not "infinity" but
rather "something different". In practice, hard shorts are not long
represented by unrestricted flow, because the I-squared in the power
law overwhelms the zero resulting in destructive heat typically. The
water analogy does break down some here. Check the m3m.com pages on
debugging pinball machines for much insight on theory and practice of
shorts."

On Mon, 5 Apr 2010 09:08:57 -0700 (PDT), "maxim.porges"
wrote:

On Mon, 05 Apr 2010 09:27:50 -0500, John Fields
wrote:

On Sun, 4 Apr 2010 08:07:50 -0700 (PDT), "maxim.porges"
wrote:

I'm an electronics hobbyist, just getting in to the basics. I bought a
copy of "Make: Electronics" and I'm teaching myself from that.

I understand how Ohm's Law works for calculating real-world voltage/
amperage/resistance, but I am intrigued by the theoretical
implications of it. Given that V = R x I, if there was zero resistance
in a circuit (i.e. R = 0), then that implies that there would be no
voltage (because the voltage would calculate out to 0). I want to make
sure I understand this conceptually.

Using the water-pressure analogy, if there are two tanks connected to
each other with one tank full and the other empty, and a pipe at the
base of the tanks connecting them, then I understand that the voltage
is equivalent to the pressure of the water in the filled tank, and the
amperage the speed with which this pressure is relieved as current
flows from the filled tank to the empty one until equilibrium is
achieved. If the pipe between the tanks offers zero resistance, then
the speed with which the water moves between the tanks to equalize the
pressure should be extremely high (i.e., infinite amperage). In other
words, in a truly zero-resistance system, equilibrium would be
achieved instantly.

---
Well, no.

There is the time required for the equilibrium to occur.
---

Is the fact that equilibrium is achieved instantly the reason why the
voltage would be zero? I.e., the potential between the tanks is
instantly resolved by the lack of resistance, with a theoretical
amperage of infinity?

---
No.

Equilibrium will be reached when the level of water in both tanks is
equal, which will take some time to achieve.

Also, equilibrium won't happen in one pass since the water has mass,
which will cause the water columns in the tanks to oscillate for a while
before everything quiets down.

JF


Thanks John - I think you are focusing on the water analogy a bit too
much, though. This is how voltage/amperage/resistance is typically
explained to newbies like me, but I'm trying to understand the
conceptual side of things as it relates to electronics.

---
The water analogy was your idea, and my intent was to show you why it
broke down when it was assumed that equilibrium occurred
instantaneously.

In _fact_ it does not, and cannot occur instantaneously, whether the
system is viewed from either a mechanical or electrical perspective.

The system you posit, electrically, would be superconducting, with 2
capacitors connected like this: (View in Courier)

S3 ---
+-----+-----+-O O-+-----+
| | | | |
O| O| | | O|
O|S1 O|S2 | | O|S4
| | | | |
[BAT] | [C1] [C2] |
| | | | |
+-----+-----+-----+-----+
|
GND


In practice, at the beginning of a run S1 and S3 would be open while S2
and S4 would be closed, disconnecting the battery from the circuit and
the caps from each other, while shorting the caps to ground.

Once the caps were discharged, S2 and S4 would be opened then S1 would
be closed, charging C1.

Next, when C1 was fully charged, S1 would be opened, disconnecting the
battery from C1 and, finally, C3 would be closed, allowing charge to
flow from C1 into C2 until it equalized between the two caps.

Now, even if there was no resistance at all in the circuit the energy
transfer wouldn't be instantaneous because the inductance in the circuit
would cause the circuit to "ring" at a frequency of:

1
f = --------------
2pi sqrt(LC)

---

My boss was an electrical engineering student for a while and I also
sent this question to him. Here was his reply for anybody who is
interested.

"Yes, no resistance, aka hard short, means no voltage. There are some
very clever ways you can work with just a light bulb to find and debug
shorts and opens. Remember that one over zero is not "infinity" but
rather "something different".

---
If we look at:

1
y = ---
n

and make n smaller and smaller, what'll happen to y when n gets to zero?
---

In practice, hard shorts are not long
represented by unrestricted flow, because the I-squared in the power
law overwhelms the zero resulting in destructive heat typically. The
water analogy does break down some here. Check the m3m.com pages on
debugging pinball machines for much insight on theory and practice of
shorts."

---
BTW, since you're a newbie, here are a couple of guidelines for you:

1. This is alt.electronics, which is a newsgroup that is part of USENET,
which is not part of Google or Google groups.

2. On USENET it's considered good practice to leave part of the
article (called "context") you're responding to intact and to post
your reply below it (bottom-posting) or to intersperse your reply
(in-line posting) within the article being replied to if that'll make
the exchange clearer.

The reason for including context is that it makes the thread readable by
someone accessing the thread late, and bottom posting creates an
easy-to-follow chronological for readers who may not have followed the
thread from the beginning.

Here's Google's take on it, from:

http://groups.google.com/support/bin...2348&topic=250

"Summarize what you're following up.

When you click "Reply" under "show options" to follow up an existing
article, Google Groups includes the full article in quotes, with the
cursor at the top of the article. Tempting though it is to just start
typing your message, please STOP and do two things first.
Look at the quoted text and remove parts that are irrelevant.
Then, go to the BOTTOM of the article and start typing there.
Doing this makes it much easier for your readers to get through your
post. They'll have a reminder of the relevant text before your
comment, but won't have to re-read the entire article.
And if your reply appears on a site before the original article does,
they'll get the gist of what you're talking about."

JF