I am trying to measure the real power of an inductor, and in the manual it is said that I have to connect a resistor with low resistance to the inductor and then see the voltage difference between the current and voltage.
Question: Why do I need a resistor? Why don't we just connect the oscilloscope across the inductor and see the phase shift?
To see phase shift you need to compare two separate wave forms. Looking at one wave form won't tell you anything.
The two wave forms to compare are current and voltage. Seeing the voltage is easy, like you said, just connect the probe across the inductor. Seeing the current is more difficult because the oscilloscope can't "see" current. There are two options, use a current transducer to convert the current to voltage or use a resistor to do the same thing. Ohms law says that current through the resistor can be seen as a voltage drop across the resistor.
Be careful, don't connect the oscilloscope grounds in different places. That's a fast way to burn up your scope. The problem with that is the wave forms will have an additional 180 degrees phase shift. Just compensate for it when you do the calculations.
simulate this circuit – Schematic created using CircuitLab
I voted for vini_i's answer, but there were two questions.
simulate this circuit – Schematic created using CircuitLab
Connect both grounds in-between components (or use one ground and leave the other disconnected) to serve as reference. Connect Channel 1 (or A) on resistor, 2 (or B) on inductor. Set \$ V_R \$ as trigger. Scope is measuring \$ V_R \$ vs \$ V_L \$.
Measure phase angle \$ \theta \$ on scope. It will be less than 90° because it is a non-ideal inductor. You may have to invert channel 2 (Pull Invert).
If it was an ideal inductor, the phase angle would be 90°. Real inductor is made of wire and wire has resistance \$ R_{DC} \$, plus a \$ R_{AC} \$ due to ac effects (hysteresis, eddy currents, skin effect, radiation), so phase angle will be less than 90°.
Measure \$ V_{R_{PP}} \$ on scope. Convert to RMS. $$ V_{R_{RMS}} = \frac {V_{R_{PP}}} {2} \times 0.707 $$
Measure R. Use Ohm's Law to calculate current.
Measure \$ V_{L_{PP}} \$ on scope. Convert to RMS.
Calculate effective voltage using trig. $$ V_{Eff} = V_L\ cos\ \theta $$
Calculate average power due to effective resistance in inductor. $$ P_{Eff} = V_{Eff} \ I $$
Edit with reference to comment...
In the above circuit, you are looking into right triangle of the phasor diagram or \$ \theta_{Measured} \$ (Brown). You are measuring \$ V_{R_{Measured}} \$ (Blue) and \$ V_{L_{Measured}} \$ (Red). You cannot see the ideal inductor voltage \$ V_{L_{Actual}} \$ (Teal) or \$ V_{R_{Eff_{Actual}}} \$ (Purple). You cannot see I on the scope, but \$ V_R \$ is in phase with I, so seeing \$ V_R \$ effectively sees I.
So if you connect it like so, scope is measuring \$ V_S \$ vs \$ V_L \$. You are seeing the angle at the top of the phasor diagram.
You could probably use sine instead of cosine, but I am NOT confident of that answer. By connecting it like my original answer or vini_i's, it will work.
The problem with making this measurement has to do with the number of "not shown" reference points (like the signal generator's output being tied to ground through power cord, and from there back to the oscilloscope). Optimum measurement reference would be your point V_L because it is common to both the voltage and the "current" measurement. Probably, there is a problem with this since it basically shorts out the signal generator.
Second best alternative is to measure from Ref to V_L and Ref to V_S, and use Chan A + CHan B (with one of them inverted) to pull out the current signal.
Third best alternative (only because it is more costly) is to acquire a DC (or maybe only AC iff) current probe. They generally measure the magnetic field around the wire, which is a linear function of current.
FYI... DO NOT CLIP THE THIRD WIRE SAFETY GROUND ON THE SCOPE. It doesn't solve anything if you have two channels.
