Not for me....
If km/h equal voltage, as long the water has more resistance it should
create a higher voltage drop at the same current...
So, if air is 10Km/h, water should be more than 10km/h and not 5km/h...

Let's try my version. Current (I) is represented by one ant. The copper
wire is an empty water pipe. The ants walks through the pipe. Resistance
(R) is the pipe diameter. Voltage (V) is the difference between the number
of ants at the pipe ends. As lower the pipe diameter (meaning as higher the
resistance), as many ants will stay on the pipe ends and only a few with
guts (a small current) will pass through the pipe.
:)


Vasile

On Mon, Mar 16, 2015 at 6:22 PM, <jim@jpes.com> wrote:

>
>  It sounds reasonable to me.
>
> Regards,
>
> Jim
>
> > -------- Original Message --------
> > Subject: [OT] Volts, amps, resistance (again)
> > From: Lindy Mayfield <Lindy.Mayfield@sas.com>
> > Date: Mon, March 16, 2015 11:13 am
> > To: "Microcontroller discussion list - Public." <piclist@mit.edu>
> >
> >
> > Again this stupid question.  I've learned that I can work with volts,
> amps and resistance pretty well without really understanding.  I have the
> water analogy but that's not the best.  So I thought I would make my own
> analogy, and ask if it is correct.
> >
> >
> > I am running at 10 km/h through the air.  It takes my muscles so much
> energy to keep me running at that speed. If I switch to running through
> water, for example, in order to run 10 km/h I need to be stronger. If I a=
m
> running through oatmeal I need to be stronger still to keep my speed at 1=
0
> km/h.
> >
> >
> > So 10 km/h equals voltage, amps is how strong I am, and what I am
> running through is resistance. If the resistance is more than I am strong
> then my speed will decrease and I will only be able to run 5 km/h.
> >
> >
> > I can now see how all of those three pieces, V, I and R, balance out.
> >
> >
> > Is this a good analogy?
> >
> > Thanks!
> > Lindy
> > --
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