another small breakthrough on our NERD technology.

[Rosemary Ainslie]

Here is the simple DC power problem I gave you before:

Please provide the calculations for power delivered by the battery, and power dissipated by RLOAD. BE SURE TO INCLUDE THE POLARITY OF EACH RESULT.

Nothing wrong with what you're showing there.  Now.  What is your point? 

Here's what I see.  The battery is able to release the flow of current through the load resistor - in this example in a clockwise justification.  This current flows back to the battery negative terminal thereby reducing the potential difference at that battery supply.  No problem.  BECAUSE there's a flow of current - then the amount of energy delivered by that battery is based on an analysis of wattage, which, in turn is the product of the battery volts and the amperage flow over time.  And all things being equal - because the wattage is from the battery - that number is represented as a positive number.

THEN. ONLY WHEN THAT CURRENT FLOW IS INTERRUPTED - the induced potential difference over the circuit components - including the LOAD resistor that you show, is able to discharge its 'stored' potential energy.  BUT.  That voltage that was INDUCED across that Load resistor - is in ANTI PHASE to the applied voltage from the supply.  Therefore - in relation to the supply that voltage is NEGATIVE.  IT THEN DISCHARGES that potential difference and this induces a current flow that is in anti phase to the previous flow of current.  Therefore the current flow is negative. Therefore a product of a negative voltage and the negative current flow would result in a POSITIVE SUM? Is that your argument?  I have no problem.

BUT.  You cannot argue that the power that is dissipated as a result of the collapsing fields in that element has been DISCHARGED BY THE BATTERY SUPPLY.   It MAY have been first delivered by the battery  But it is now moving in a counter clock-wise direction.  It does not DISCHARGE the battery.  If anything it RECHARGES the battery.

So.  What is your point?

Rosemary
added

[poynt99]

Nothing wrong with what you're showing there.  Now.  What is your point? 

Here's what I see.  The battery is able to release the flow of current through the load resistor - in this example in a clockwise justification.  This current flows back to the batter negative terminal thereby reducing the potential difference at that battery supply.  No problem.  BECAUSE there's a flow of current - then the amount of energy delivered by that battery is based on an analysis of wattage, which, in turn is the product of the battery volts and the amperage flow over time.  And all things being equal - because the wattage is from the battery - that number is represented as a positive number.

THEN. ONLY WHEN THAT CURRENT FLOW IS INTERRUPTED - the induced potential difference over the circuit components - including the LOAD resistor that you show, is able to discharge its 'stored' potential energy.  BUT.  That voltage that was INDUCED across that Load resistor - is in ANTI PHASE to the applied voltage from the supply.  Therefore - in relation to the supply that voltage is NEGATIVE.  IT THEN DISCHARGES that
potential difference and this induces a current flow that is in anti phase to the previous flow of current.  Therefore the current flow is negative.  Therefore a product of a negative voltage and the negative current flow would result in a POSITIVE SUM? Is that your argument?  I have no problem.

BUT.  You cannot argue that the power that is dissipated as a result of the collapsing fields in that element has been DISCHARGED BY THE BATTERY SUPPLY.   It MAY have been first delivered by the battery  But it is now moving in a counter clock-wise direction.

So.  What is your point?

Rosemary

First of all, you seem to think that the circuit I posted is related to YOUR circuit somehow, am I correct?

That is NOT my intention, nor is that the case when you look at the circuit. There are no switches, there are no transistors or MOSFETs, there is no inductance. What we have there is simply a pure DC voltage supplied by the battery, connected to a pure resistive load RLOAD.

Care to try again?

[Rosemary Ainslie]

First of all, you seem to think that the circuit I posted is related to YOUR circuit somehow, am I correct?

That is NOT my intention, nor is that the case when you look at the circuit. There are no switches, there are no transistors or MOSFETs, there is no inductance. What we have there is simply a pure DC voltage supplied by the battery, connected to a pure resistive load RLOAD.

Care to try again?

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.

[poynt99]

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.

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

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.

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.

[poynt99]

Until, obviously, that current from the battery is interrupted.

What 'obviously' happens if the current from the battery is interrupted in my circuit?