Phase difference of voltages at one side of a transformer

Consider the circuit in picture simulate this circuit – Schematic created using CircuitLab

Suppose that I measure VA and VB with respect to GND using an oscilloscope.

Will VA and VB be in phase or out of phase with respect to each other?

Edit : I tried to make measurement on a real circuit with the values of parameters as in picture (V1 is 220V). I measured VA and VB with an oscilloscope and I got

VA=6 V

VB=17.5 V

I used the MATH function on oscilloscope to get the difference VA-VB

VA-VB= 20 V

I did not measure directly the phase shift between VA and VB signals, but, from the amplitude I can deduce that pase shift d was such that

sqrt(VA^2+VB^2-2*VAVBcos(d))=20

Solving I find d around 106°.

Is this normal? I expected a 180° shift while I get a value which is much lower

• Even after the question has already been accepted: Could you show us the waveforms for VA and VB measured in your circuit (simulated or mounted)? – Dirceu Rodrigues Jr Jan 31 '17 at 15:05
• @DirceuRodriguesJr I do not have the waveforms but I edited the question adding the measured amplitudes – Sørën Feb 1 '17 at 0:03
• What's the turns ratio for each transformer? – Dirceu Rodrigues Jr Feb 1 '17 at 3:33

If the transformers shown are ideal, VA will be 180 degrees out of phase with VB. If not ideal then VA and VB will tend to be 180 degrees out of phase plus or minus a small error.

• Thanks a lot for the answer! Therefore, if I want to measure VA-VB (voltage across the transformer), can I connect VA to CH1, VB to CH2, then use INVERT button for VB and finally use ADD button? Or is it wrong to use the INVERT button here? – Sørën Jan 30 '17 at 9:54
• You can do it like that for sure. Or, you can move the 100 ohm resistor into the VA line thus grounding VB and then you just measure VA. – Andy aka Jan 30 '17 at 9:56
• It's a s simple as one side being positive and the other side being negative and then of course that alternates. If both sides had the same polarity then there would be dc and ac together and that doesn't make sense. – Andy aka Jan 30 '17 at 10:39
• If there is zero current flowing in R1 (or R1 is zero ohms) then VA magnitude is the same as VB. – Andy aka Jan 30 '17 at 16:10
• This is due to non-idealities like saturation currents and leakage inductance. The distortions are probably due to saturation currents. When you performed "VA-VB= 20 V", was 20 volts what you expected? – Andy aka Feb 1 '17 at 8:17

In generic terms, for a given voltage difference between the $a$ and $b$ points, there will be an infinite number of possibilities for the voltages $v_a(t)$ and $v_b(t)$to assume, when measured relative to a reference point. The exact setup will be given by the conditions of the circuit. For example, in the case of AC (sinusoidal), the $v_{ab}(t)$ voltage can be given as the difference:

$$V_{ab}\cos\omega t = V_a\cos(\omega t+ \theta_a) -V_b\cos(\omega t+ \theta_b)$$

including $\theta_a$ and $\theta_b$ phase shifts. Assuming that $V_a$ and $\theta_a$ are known and developing the right side, leads to:

$$\theta_b = atan2(V_a\sin\theta_a,V_a\cos\theta_a-V_{ab})$$ $$V_b = \sqrt{ V_a^2-2V_aV_{ab}\cos\theta_a+V_{ab}^2}$$

Considering TR1 turn ratio as 11, I simulate your circuit (obtaining results consistent with the above expressions) for two values of R1: The phase difference is 111.6 deg ($\theta_a$=64.19 deg and ($\theta_b$=175.78 deg). Changing R1 to 10R: Here the phase difference is 108 deg ($\theta_a$=54.69 deg and ($\theta_b$=162.69 deg). That is, many possibilities for phase difference other than 180 deg.

• Thanks so much for this detailed answer!! May I ask which software you used for this simulation with digital oscilloscope? – Sørën Feb 1 '17 at 11:23
• It is the Proteus, widely used in educational establishments – Dirceu Rodrigues Jr Feb 1 '17 at 14:40