This is either strawman at it`s best, or you`ve totaly missed the point I was trying to make with regards to PD.

 

"A Pool" without dimensions is impossible to calculate, However certain basic principals still apply with regards to PD (basic electronics).

and so, if 1 million volts (I`m being conservative here) is equaly distributed over a body of water, with so many volts per square metre end to end, then a 2 meter man will get his fair share of it, more so infact because we`re Ionic in nature as opposed to pure water.

 

edit: 2 x 100 ohm resistors in parralel 10 volts is passed across them, how many volts is dropped across any one resistor?

edit: 2 x 100 ohm resistors in parralel 10 volts is passed across them' date=' how many volts is dropped across any one resistor?[/quote']

Hmmmm, let's see if I got this straight.

2 * 100 ohm resistors is 2, 100 ohm resistors in series equals 200 ohm's

The branches of the resistors are in parallel, so we got 100 ohms.

We assume the 10 volts is DC .....no worry about inductive or capacitive reactance.

Ok, I = E ÷ R equals 100 mA, P = E * I equals 1W.

For conservation of power, 500mW each leg.

Ok, Any branch R = 200 Ohms E = 10V therefore, I = 50 mA

V Drop across any resister = 5V

Power each resistor is 250 mW, 250 mW * 4 = 1W

 

How'd I do?

 

One for you, what's the Q of an inductor and resistor in parallel? You pick the L,C and F ?

edit: 2 x 100 ohm resistors in parralel 10 volts is passed across them, how many volts is dropped across any one resistor?

 

Exactly! But what happens with 1 x 20k and 1 x 0.001 ohm in parallel? That’s the point I was trying to make. (talking about lightning rod not pool)

 

As for the pool I was taking the taller-on-ground-than-in-pool stance.

YT2095:

Ah, on second thought, the above example is too easy.

This one is slightly harder.

You have a frequency divider in a circuit following a local oscillator. The divider divides by two. It (the divider) has an internal flat white noise level of -130 dBm/hz. The oscillator output is +16 dBm, it has a single sideband noise level of -140 dBc/Hz at 200 KHz removed from the carrier. What is the resultant noise level out of the divider at Fo ± 200 KHz expressed in dBc/Hz ??

YT2095:

Ah' date=' on second thought, the above example is too easy.

This one is slightly harder.

You have a frequency divider in a circuit following a local oscillator. The divider divides by two. It has an internal flat white noise level of -130 dBm. The oscillator output is +16 dBm, it has a single sideband noise level of -140 dBc at 200 KHz removed. What is the resultant noise level out of the divider, given in dBc?[/quote']

 

ummm... I think maybe you or I is missing the point.

ummm... I think maybe you or I is missing the point.

Lance, no, I was making a point.

Anyway, getting back on topic. Your post was to YT2095. I'll answer, it's a little less than .001 ohms.

Lance' date=' no, I was making a point.

Anyway, getting back on topic. Your post was to YT2095. I'll answer, it's a little less than .001 ohms.[/quote']

 

I was about to say that’s an awful lot of work for such an obvious answer.

 

Anyway that was a rhetorical question. A lower resistance just makes my point more clear.