I think that trying to use RMS values of voltage and current for this is not going to work. Imagine shifting the current waveform 4ms later; neither the RMS voltage nor the RMS current would change at all, but the power drawn would change by an order of magnitude.
The instantaneous power drawn by your circuit is V * I. In any small time dt, the energy consumed will be V * I * dt. The energy consumed in 1s, the power drawn, will be the integral of V * I * dt from T=0 to T=1s. You could compute this directly from the sample values in your excel spreadsheet. At each time sample, multiply the instantaneous voltage by the instantaneous current, and that gives the instantaneous power drawn. Multiply that by the sample interval, and that is the energy consumed in that sample interval. Add all those up over an AC cycle, and multiply by the number of cycles per second and that is the energy drawn per second, otherwise known as the power.
Looking at the scope traces, the current drawn by the circuit is usually 0. Once per AC half-cycle, the current increases to about 90mA very quickly, then drops linearly to 0 over about 820us. It's a 60Hz circuit so it does this every 8.3ms. When the circuit is drawing current, the voltage is more-or-less constant at 170V. That's an average current of 45mA over the 820us at 170V = 7.65 W, but it only takes this power 1/10 of the total time, so the final power consumption is 0.76 W.
In my experience, the probability of wiring up a current probe backwards is exactly 0.5!
If you know the input power and the output power then the difference is the power lost.... in the feed wires.
Do you know how to calculate the power dissipated in a resistor based on what current is flowing through it? If yes, then you reverse this formula to uncover what current must be flowing to dissipate the power lost in the feed wire. When you have done this you'll find that your prof is right and he also made the assumption that the feed wire between generator and load was purely resistive.
Let's see if you can figure this out by your comments. PF is 0.72 by the way so let's see if you can figure this out too.
Best Answer
This kind of wattmeter does not directly work from the power factor. It averages the instantaneous voltage times current. The force on the needle is proportional to the voltage times the current, which is the instantaneous power. A spring causes the needle to deflect linearly with applied force. The inertia of the needle averages the instantaneous product of the current and the voltage, so shows averge real power.
If the voltage and current are out of phase with each other by 90° (power factor = 0), then the power would be positive for half the cycle and negative for half the cycle. If the meter is intended for normal power line frequencies, then the mechanical mechanism will not respond to the individual 50 or 60 Hz half-cycles but show you the low pass filtered power with a significantly longer time constant than individual power cycles. At a power factor of 0, this average will be 0, and the meter will show 0.
If you separately measure the RMS voltage and current then multiply them, you get the VA value (magnitude of power including both real and imaginary parts). You can take the real power shown by the meter and divide it by the VA value to get the power factor.
So again, this meter doesn't directly deal with power factor. It measures real power going back to first principles and whether the voltage and current signals are sines or not.
By the way, the electric meter on your house works on the same basic principle. Instead of deflecting a needle against a spring, the force between the voltage and current coils cause a small motor to spin. The total number of rotations is the time-integral of the power, or the total energy delivered. A series of gears rotate various labeled dials so that this integral can be accumulated and displayed so the meter guy can read the total every month. Newer meters send the accumlated reading to the power company automatically via various communication means.