SevenUP

Yes. In circuit A, If the amp meter is removed and a volt meter is connected in series (the same as in circuit B) then Vaa' will equal Vab which is the “unloaded” source voltage.

_7UP_

An ideal amp meter is a conductor and has zero impedance, so there is no potential difference hence no voltage drop, whether current is passing through or not.

V_ab will always have voltage (if the supply is working).

_7UP_

Here’s why:

The Circuit in 25.18b is effectively an open circuit. Modern volt meters can have input impedances in the billions of ohms (even the cheap ones are in the tens of millions of ohms). Because this circuit is effectively open, there is no current flow, so there is no voltage drop on the resister. Without a voltage drop, the source voltage is “read” by the meter.

There are many voltage sources that require a “load” to be accurately measured. Without a load, the source voltage may be several times higher than the specified rating.

_7UP_

This is how I interpret it. (Elucidating comments from Alan are always appreciated)

This is called radial current flow. It is calculated via an infinite number of “shells” radiating out from the point of origin. As the current flows out from the center, each shell has a progressively greater surface area and a lower corresponding resistance. The shells’ surface areas, and its corresponding resistances, are summed to calculate the total resistance. However, because the area, and the rate of area, increases as the current moves outward, the cross-sectional area formula is integrated to effectively make the area uniform over predefined distances.

For DC circuits this phenomena seems trivial. For AC circuits it becomes important because the current flow moves closer to the surface of a conductor as the frequency increases. Where it really becomes interesting is in the conduction of signals in biology, specifically nerve cells that are mediated by sodium, potassium, and calcium channels. Understanding this is a brain strain (and a nerve pinch), and how anyone ever figured it out in the first place is mind boggling. 

_7UP_

I believe the answer is C, they quadruple. I’m not absolutely certain, but it is logical because each charge increases 100% and there are 4 charges. I know that the potential at any point on the square is the algebraic sums of the individual charges. So a change of magnitude of a charge is directly proportional.

Alan can verify the accuracy or error of my statements.

Your question was lost in the sea, even after Melody bumped it.

_7UP_

Thank you, Alan. This is like finally seeing the three dimensional picture in a blended color dot image. The two velocities relate to the solution via the harmonic mean. Not unusual considering they are rates. It sure is easy to see this when they are the right values.

2/((38.2)^-1+(6.8)^-1) = 11.54

I think the most interesting thing is how the lower energy auto dramatically slows the higher one. I saw similar energy-shedding patterns in high-mass low-velocity vs low-mass high-velocity comparisons.

All of this is very cool. Thank you again.

_7UP_

Thank you Alan. I usually understand these intuitively, first, then learn the maths. In this case, it is the other way around. I need to develop a mental picture from the scalar and vector dimensions via the maths. 

_7UP_

A question for Alan or Heureka

Why does this equation set fail? Would this work if this were an elastic collision?

Kinetic energy autoN = (0.5(1300)*(30²)) = 585,000J

Kinetic energy autoS = (0.5(900)*(15²)) = 101,250J

585,000J + (- 101,250J) = 483,750J

Total mass now as a unit = (1300kg) + (900kg) = 2200kg

Solve for velocity: Sqr(483750J / 0.5(2200kg)) = 20.97m/s in north direction.

_7UP_

Sorry Alan, I didn't see your post until I made mine.

The answer to your question is yes.

A positively charged rod will pull electrons away from the ball until it reaches an equilibrium. This will leave the ball with a net positive charge.

_7UP_