Parallel RLC circuit phasor diagram

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SUMMARY

The forum discussion centers on the calculation and interpretation of phasor diagrams for a parallel RLC circuit. The impedance (Z) is calculated at an angle of +65°, with the supply current (Isupply) at +30° and the supply voltage (Vsupply) derived from Z and Is, resulting in Vs at an angle of 95°. The community emphasizes that in parallel circuits, voltage is the reference, suggesting that Vs should be at 0°, while the current's angle needs clarification, particularly the significance of the 30° angle associated with Isupply.

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greg997
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I am having difficulties with phasor diagram for Is and Vs of an parallel RLC circuit. I have calculated all values but not sure how to use phasors.
Impedance Z has angle +65°. Current Isupply +30° and Vsupply is Z*Is= Vs at angle 95°
But with parrallel network V is the reference so I guess it could be horizontal line.
Is Isupply 30° from virtual horizonatl reference line, and Vs 95 degrees from virtual horizonatl line, therefore the difference between these two would be 65 degrees with voltage leading?
Any help is welcome
 
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greg997 said:
Impedance Z has angle +65. Current Isupply +30 and Vsupply is Z*Is= Vs at angle 95
It's a bit of a mystery how you got those three angles. Normally, I or V is taken as the reference zero angle, but you seem to have used some other undisclosed voltage or current as your zero degree reference.

But with parrallel network V is the reference so I guess it could be horizontal line.
Yes. So you'd have V at 0 degrees. You say Z has been calculated inductive at +65 degrees. So this means I=V/Z giving I an angle of -65 degrees.

In summary, it's V at 0. Z at +65. I at -65. You of course need the degree symbols here, but this browser is broken and doesn't allow me to insert symbols.

So check what you have done, and try and figure out what that 30 degrees associated with the current is. I can't explain it. Perhaps there are other elements in your circuit?
 

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