I need an explanation for voltage drop cancellation

1. There should not be any 'voltage drop cancellation'. If everything (source voltage, current draw, connection resistances, etc.) is steady, the measurements you have taken should all jive the way you seem to be expecting them to.
2. The voltage drop of 2V along the neutral sounds a little severe for 8A on a 25' cord. If there is actually a poor connection resistance contributing significantly to the 2V measurement, the 2V may change a lot if the cord is moved / touched.
3. The source voltage may vary by a few volts. Perhaps when you read 121V at the motor, the source was around 125V? Perhaps when you read 122V at the source, the motor voltage was down around 118V?
4. Redo the measurements for confirmation? Make sure the motor is running steady state when you measure. Seeing something funny going on, I would go ahead and measure the voltage drop of the hot conductor, 'just because'.

I note in passing that some of your (4 V) drop in the cord will be inductive, and not contribute to power (Watts) loss. Your calculation of

35.36 W will be a little high (and shows too many sig figs). With the conductors it seems you are using, the resistive component will dominate, so your calculation is probably correct within one or several percent.

BTW, I recommend a larger AWG extension cord.

j

Construct a phasor diagram of the system and you will find the reason. You are forgetting the existance of Watts, VAR and VA. Gene

The motor is a complex load (i.e. you use imaginary numbers). You have to use complex math to solve this problem. e.g. if all power going into the motor produced useful work, it would appear as a resistive load. If zero power went into the load and in addition there were no losses in the motor, then it would appear as an inductive load. In your case the motor is between these two extremes. The resistive part of the load is a real number and the inductive part of the load is an imaginary number. You combine the resistive drop in the motor with the resistive drop in the wiring, then combine that with the inductive drop by doing a 'square root of the sum of the squares' to get the total voltage drop.

You don't say what you're measuring with, but it sounds pretty likely that you are using an rms meter and are not accounting for the phase shift of an inductive load such as the motor. Voltage and current go out of phase and thus you cannot get accurate readings the way you're doing it. For another, you need to use only ONE reference point for ALL your readings. The way your'e doing it you don't have a valid reference for the measurements, making them oranges and apples. If phasors, VI lead/lag, eli the iceman, and all that are meaningful, that's what you've forgotten. If that's not meaningful, then you have a little homework to do yet.

HTH,

Pop

2.0 x 2 : =4V. I calculated the resistive loss in the extension cord to be : 8.84A x 4.0V = 35.36 watts : : Strangely, when I measure the voltage at the motor end and the source : end of the extension cord, the difference is only 1V. Taking power : readings at both ends, the difference is 37watts, which is consistent : with my previous math. If the power lost in the wiring is 37W and the : current is 8.84A, then the voltage drop should be around 4V, but : somehow, the voltage drop is only 1V and 3V is getting cancelled out. : The inconsistency does not happen with a resistive load. : : Explanation? :

Your measurements are fine except for the number of decimal places used. You do have a voltage drop along the wire as measured-near 4V for the pair for a total wire resistance of 0.45 ohms which corresponds to that from the difference in power between the sending end and the motor (0.47 ohms on the basis of 37 watts). However, your calculations are wrong. The voltage drop measured is not in phase with the terminal voltage as the current is not in phase. As you have been told, take into account the power factor. If your power readings were taken with a decent wattmeter, you pf at the load is about

0.23 (lousy -no load?) as V*I =121*8.84=1070 implying that current lags voltage by about 77 degrees. Assuming that the cord is purely resistive at 0.47 ohms the input voltage magnitude will be about 122V so your drop of 1V is also correct. The difference between this and 122.4 may be due to measurement errors or due to some inductance in the extension cord.

Measuring input and output voltages will give the correct "drop" at any power factor. Measuring the voltage between ends of one side of the cord will not unless the motor is at unity pf - not likely- because current and voltage will not be in phase.