Rosemary Ainslie circuit demonstration on Saturday March 12th 2011

[Rosemary Ainslie]

P1, P2, P3, and P4 are measurement points or nodes. These nodes are where the scope probes are placed in reference to the two equations.

P_R1 and P_Vbat means:

Power of R1, and Power of Vbat, respectively.

Well then?  How in HEAVEN'S NAME do you manage to show the battery voltage at a negative?  And I strongly recommend that you vary those references.  P = power and Probe positions - and potential difference?  What?  in the words of Nanny McPhee - 'BIG P - small fee'.  Very confusing.

Rosemary 

[poynt99]

How in HEAVEN'S NAME do you manage to show the battery voltage at a negative?

Start at node P1 and follow the + and - signs in a clockwise direction. By the time you arrive back at node P1, you will have noticed that the voltage drop across the battery is reversed wrt the voltage drop across the 3 resistors.

.99

[Rosemary Ainslie]

Start at node P1 and follow the + and - signs in a clockwise direction. By the time you arrive back at node P1, you will have noticed that the voltage drop across the battery is reversed wrt the voltage drop across the 3 resistors.

.99

Golly Pointy.  I've tried to take all this seriously and my best efforts are now sorely taxed.  Where you SHOULD start reading is from the power supply SOURCE.  THEN.  In the unlikely event that that the battery reads a NEGATIVE potential or voltage - then - if you also read a POSITIVE potential over the resistive components - you've found the biggest ANOMALY recorded in science.  It is the SOURCE that applies a consistent potential difference across the circuit - to begin with.  IF AND WHEN the source is disconnected and the circuit is OPEN - then that APPLIED POTENTIAL DIFFERENCE REVERSES - as required by INDUCTIVE LAWS and widely referred to as BACK OR COUNTER ELECTROMOTIVE FORCE.

Rosemary

added
Let me try this again.  The voltage drop - as you put it - across the battery has NEVER reversed.  It remains positive.  What happens is that it's potential difference is REDUCED or PARTIALLY DEPLETED.     

[WilbyInebriated]

tk...  you're back, good. we have some unfinished business, you and i... and you owe me a mea culpa...

who was the one to use the worst possible scientific method and yet still called it a "replication"? it was a pathetic hack, nothing more...
who was the one that said using a different transistor wouldn't make one whit of a difference? and then you lied and squirmed about what 'difference' meant when your error was made obvious by experiment.
who was the one who left in a fit of  profanity declaring they were "done"?

YOU are the one who should think hard tk... YOU are the one who embarrassed themselves in this thread previously. that's why you left in a fit of profanity remember? because you couldn't answer to your contradictions i had pointed out. you knew you were checkmated, so you swore up and down like a spoiled child and ran home to mommy... and now you're back... imagine that.

the ONLY measurement you should be concerned with tk is the one bruce so kindly pointed out to you. that is, your measure of character. which at the moment is at the bottom of the barrel due to your lies, misrepresentations, your hoaxes of the community ala the magnet motor hoax of yours...  etc, etc. etc. ad nauseam, ad infintum...

[Rosemary Ainslie]

I'm still trying to get my head around your argument Poynty.  Are you proposing that all measurements of energy now require that we take the positive flow of current from the shunt and multiply it with a negative voltage from the battery because it presents at the negative terminal?  Power measurements are based on vi dt.  Just ask yourself 'what is the actual voltage across the battery?'.  The actual voltage is STILL POSITIVE.  Therefore if the amperage is positive and the voltage across the battery is positive then you will INDEED get a positive product. That positive product will represent the amount of energy discharged by the battery.  And this, in turn should equate to the amount of energy dissipated on the circuit.  It's a net loss to the potential difference at the supply.  But it is still a positive value.  I'm not sure if all this is smoke and mirrors or if you are genuinely confused. 

Rosemary
Added.

And by the same token - if the amperage is negative then it is still a product of the battery voltage and that amperage flow - but it will now result in a gain to the potential difference at the supply as that product will be negative.  Golly.   It's NOT rocket science.  It's only elementary power analysis.