Multisim and Ultiboard

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About AC Power current Phase

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))"

1.PNG2.PNG

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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

 

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Thanks to G. Goodwin

I got the result like this:

4.PNG

 

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!

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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.

 

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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.

 

=============

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.

6.PNG7.PNG

 

With sincere appreciation.

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