06-06-2021 02:36 AM
I tried to get AC power current phase and the I(R1) phase...
Could anyone help me solve the question "Why they are different ( I(V1)&I(R1))"
06-06-2021 06:50 AM
Hi dowithme,
Try these: delete the two wires that connect to the resistor, horizontally mirror/flip the resistor, put the wires back, run the simulation.
Best regards,
G. Goodwin
06-06-2021 07:38 AM
Thanks to G. Goodwin
I got the result like this:
But...If the phase of AC power V(1) was 0 degree,then the both phases of I(R1) and I(V1) should be 0 degree ,too.
Am I wrong? I got confused.
I appreciate your answer!
06-06-2021 12:16 PM
Right, it's confusing. In fact I gave a wrong suggestion in my first reply because I got confused too. I recommended to reverse the orientation of the resistor while the phase of the current through it is correct (0°) in your first table. It's I(V1), which has a phase of 180° that should be analyzed.
I changed my point of view about this problem. I think we just have to interpret the results in your first table. The values in that table are correct if I(V1) is defined as the current through voltage source V1, entering its positive terminal and exiting its negative terminal. Since the conventional current in the circuit is from the positive terminal of V1 to the resistor and to the negative terminal of V1, I(V1) got a phase of 180°.
The salient idea here is I(V1) is not exactly the same as the total current or "loop current" in this simple circuit. I believe it will further help if you experiment with the other end of the resistor not returned to ground but connected to a second AC voltage source V2. For simplicity V2 should have a phase of 0° but its magnitude should not be equal to that of V1. I speculate that the tabulated current for the smaller voltage source will be in phase with the current through the resistor while the tabulated current for the larger voltage source will be 180° out of phase with the current through the resistor.
06-07-2021 04:09 AM
Thanks to G. Goodwin
============
I changed my point of view about this problem. I think we just have to interpret the results in your first table. The values in that table are correct if I(V1) is defined as the current through voltage source V1, entering its positive terminal and exiting its negative terminal. Since the conventional current in the circuit is from the positive terminal of V1 to the resistor and to the negative terminal of V1, I(V1) got a phase of 180°.
You've made a good point.
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I tried to connect to a second AC voltage source V2.
Here is resoult.
The I(V1) and I(V2) are the same ,but different form the I(R1)..
Maybe I have to try another way.
With sincere appreciation.
06-18-2021 03:21 PM
My speculations are wrong and the tabulated values are now more confusing. Aside from the phases of the currents through the voltage sources, the magnitude of the current through the resistor becomes 0 and the magnitudes of the currents through the voltage sources dropped to very small values which are practically 0 as well.
Nevertheless, I highly appreciate that you tried my recommendation and posted the results in your informative reply. If ever you obtain a convincing explanation about this matter you may share it to us by posting the information here.