chiques
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Just to be sure, is the voltage source, on the circuit at the top side, a real one and has, therefore, an internal resistance/impedance close to zero ohm?I went through this tutorial which seems fairly straight forward: Parallel Resonant Circuit Math
Something like this? I’m not sure about the way I calculated the voltage coming from the signal generators 50 Ohm impedance though...Use complex notation or phasor diagram.
The same unit type as the source voltage, whatever it is. What do you prefer?Are the current magnitudes the Irms currents or are they peak currents?
This is reassuring.When you use complex/phasor convention, you find the right value.
So your math is correct.
View attachment 182427
View attachment 182428
No, it was a weirdness of AWR. Voltage Source has a particular state. There is 90 degree lagging anyway..This is reassuring.
I'm curious as to why the phases are so off though??
I also had some typos on my imaginary signs for the resistor and inductor so all those match your simulation results.No, it was a weirdness of AWR. Voltage Source has a particular state. There is 90 degree lagging anyway..
Keysight ADS gives the right results. There is single error in your MathCad space on The Current of the Capacitor. If you correct that error, you'll find the same results.
View attachment 182460
Z=-0.33+j*0.055 Rect-->Polar = 0.0641405 \[ \angle \]120.96376
according to my hand calculator.
I used the polar solver in my TI and got the same results you did.View attachment 182465
Here is my hand calculator version in radians and degrees. Still scratching my head
Z=-0.33+j*0.055 Rect-->Polar = 0.0641405 \[ \angle \]120.96376
according to my hand calculator.
The right answer is:
Z=-0.33+j*0.055 Rect --> Polar = 0.334551939∠170.5376778
The magnitude should be close to the real number because its imaginary one is much smaller (this can be seen by drawing the 2 vectors).
Also, it is clear that the vector is in the second quadrant and close to its end (close to 180 deg).
--- Updated ---
And
Z=-0.033+j*0.055 Rect --> Polar = 0.064140471∠120.9637565
--- Updated ---
I was confused because of the mistyping in the real number which, I noticed now, is -0.033. Sorry
BigBoss said:
Z=-0.33+j*0.055 Rect-->Polar = 0.0641405 \[ \angle \]120.96376
according to my hand calculator.
I was confused because of the mistyping in the real number which, I noticed now, is -0.033. Sorry
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