another small breakthrough on our NERD technology.

[Rosemary Ainslie]

In which case you have the flow of current from a battery supply that is consistent with the applied voltage from the BATTERY SUPPLY SOURCE and the Ohm's value of the resistor.  I can't remember if you specified a resistive value.   but IF R = 10 and Vbatt = 12 - then you've got the measure of the current flow as v/r=1.2 amps or thereby.

If you do have a point - please explain it.  I suspect it's to do with the fact that the voltage over the resistor is established in anti phase to the battery supply voltage.  But since the resistor is NOT delivering any current from that applied potential difference - then the amount of energy that it's delivering back to the supply - is ZERO.  Until, obviously, that current from the battery is interrupted.

Rosemary

Added and changed.

Sorry.  I used the wrong term.  I amended it in this post.  NOT zero DISSIPATION which is nonsense.  It now reads - AS IT SHOULD - Zero discharge of current.  But I'll get back to your question.  I'm having difficulties getting a 'nested' quote number.

Be right back.
R

[Rosemary Ainslie]

The values are as follows:
VBAT = 50VDC
RLOAD = 10 Ohms

Please explain in greater detail why the resistor is dissipating 0W?

Reference my previous post.  This is explained. It most certainly is dissipating energy.

My question once again was this:

Please provide the actual value in Watts and polarity for:

a) the battery power and
b) the load power.

a) Given that the battery voltage is 50 V and the resistor 10 - then amperage = 50/10 = 5 amps.  Therefore 50 volts x 5 amps = 250 Watts.
b) The dissipated energy at the load = i^2r.  Therefore the amount of energy dissipated at the load is 5 x 5 x 10 = 250 Watts.
c) or the dissipated energy can be calculated as V^2/r.  Therefore 50 x 50 / 10 = 250 Watts.

Where does polarity come into this? 

Rosemary 

added emphasis - and included ^'2' - which was a small oversight
And added another option because I'm getting bored waiting for a reply

[Rosemary Ainslie]

Poynty Point

This is here for when you wake up.  I think I know what you're trying to point to.  Your argument is that there's a voltage over the load resistor is in antiphase to the applied voltage from the battery supply.  This is unarguable. But when a battery delivers it's current flow then that flow returns to that supply.  And in so doing it diminishes the potential difference.  When an element resistor or any load in series with that supply - dissipates energy in the form of heat and as a result of an 'exchange of energy' between it's components and that current flow - then this heat is irradiated outwards - from a source.  It has no polarity.

When and IF one interrupts the current flow from a battery supply - then - and ONLY THEN - can the potential difference across the load resistor - generate any current flow at all.  And then it DOES have a polarity.  And this is in terms of inductive laws where the induced magnetic fields then 'collapse' thereby generating counter electromotive force.  Which we all know so well.  But in the discharge of that energy - we have a more complex problem  Because in terms of that discharge - some energy is returned to the battery to 'recharge' it - which must be factored into that power analysis.  And some of that energy is ALSO dissipated at the resistor element, load.

So.  Polarity of that voltage across the element resistor - load - whatever - has NO BEARING on the rate of current flow from the supply NOR the rate at which energy is dissipated - in our example - as heat.  Voltage across the resistor is NOT responsible for the energy dissipated at that workstation.  The current from the supply source IS.

Rosemary

[poynt99]

You are getting close, however you're still struggling with the polarity.

Your own clue was that something is in anti-phase when comparing the battery and load, TRUE.

Explained in words, the power dissipated or supplied by any component (resistor OR battery) is the product of the voltage across it and the current through it.

Now, have a close look again at the diagram. The current is clockwise. Convention is that voltage "drops" across a load in the direction of the current (i.e. + to -).

Therefore both the current and voltage are "in-phase" when considering the load resistor. So we have:

P_RLOAD = +V x +I = W (a POSITIVE polarity)

The battery however is a different story. By observation, one can see that the current and voltage are NOT "in-phase", therefore ONE of them MUST have a negative sign associated with it. Since the current has not changed direction, the negative sign must be assigned to the battery voltage, therefore:

P_VBAT = -V x +I = -W (a NEGATIVE polarity)

So the answers to the question are:

a) Battery Power = -250W
b) R_LOAD Power = 250W

Understood? Agreed?

[Rosemary Ainslie]

Poynty. 

This last post of yours is the most astonishing piece of nonsense that I have ever had the misfortune to read.  What is alarming is that you are patently serious.  And my dilemma is that IF I argue it at all there will be the general impression that there's an OUNCE of argument worth arguing.  Which there isn't.  But what is REALLY worrisome - is this.  Not one of our members is actually saying anything.  Which means that they're 'undecided' - at best.  Or they're entirely bored with the argument in the first instance.  If you are seriously proposing to REPLACE standard protocols with this illogical nonsense - I assure you that you will FAIL.  No amount of 'academics' protesting in that paper - will endorse this Poynty.  I saw this as your argument.  I referred to it repeatedly in my own posts.  I then thought that - just maybe - I'd misunderstood you.  And I now see that you are actually EXPECTING us - not only to buy into this - BUT TO SERIOUSLY PROPOSE TO REPLACE OUR STANDARD MEASUREMENT PROTOCOLS with this?

You are getting close, however you're still struggling with the polarity.

I am NOT struggling with polarity.  I'm struggling to believe that you're actually serious.

Your own clue was that something is in anti-phase when comparing the battery and load, TRUE.

WHAT?  What do you mean by 'comparing the battery and load?  There's NO MAGIC THERE.  It's a FACT.  The voltage from a supply generates a current flow that is consistent with the voltage measured at the supply.  What it does do is generate an opposing VOLTAGE across the circuit components.

Explained in words, the power dissipated or supplied by any component (resistor OR battery) is the product of the voltage across it and the current through it.

Absolutely NOT.  The power delivered anywhere at all be it from a resistor or circuit component or battery supply - or even a GRID PLUG - is the product of the APPLIED VOLTAGE AND the rate of current flow. And current ALWAYS FLOWS relative to it's supply.  Therefore a negative voltage induces a current flow that is less than zero.  And a positive voltage induces a positive current that is greater than zero.

And since you've got a resistor there and NOT a motor - then your load will DISSIPATE HEAT which has absolutely NO POLARITY AT ALL.  it will be seen to irradiate outwards away from its source.  THAT'S IT.

R