Help understanding phasor solution

AI Thread Summary
The discussion focuses on understanding the phasor representation and calculations related to inductance in a given circuit. Participants express confusion over the drawing of the phasor, particularly the placement of the current phasor \( I_c \) and its relationship to the current \( I \). It is clarified that if the inductor is ideal, \( I_c \) would lead \( I \) by 180°, while in a parallel RC arrangement, it would lead by 90°. Concerns are raised about the conditions provided in the problem, suggesting that the angular frequency may be too high, which could affect the current magnitudes. The conversation highlights the complexities of phasor analysis in circuits with both resistance and inductance.
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Homework Statement


From the given circuit and provided information draw a phasor and calculate the inductance.
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Homework Equations


3. The Attempt at a Solution
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The paper was given to me at class as a solution to our homework and i can't understand how they drew the phasor and drew the conclusion out of it.
I see that they took ##U_{12}## as their phasor axis since its the same for both sides but why do they draw the ##I_c## the way they did? How did they know that it goes up and how did they get the ##I_c/2## part?
The continuation of the left side of the third equation i do get but not the right side. Why is the right side like that?
 
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If the inductor were ideal (with R=0) then IC would lead I by 180°. If the inductance were zero, making it a simple parallel RC arrangement, then IC would lead I by 90°. So with both R and L then IC will lead I by some angle between those 2 extremes. That constrains the shape of triangle that you draw.

You also need to bring into the geometry the condition stated in your data of 2 currents having equal magnitude, viz., I = Ig
 
NascentOxygen said:
If the inductor were ideal (with R=0) then IC would lead I by 180°. g
Do you mean 90° ?
 
Numbskull said:
Do you mean 90° ?
No. I is the inductor current. Ic would lead U12 by 90° and if the inductor were ideal, inductor current I would lag behind U12 by 90°. This means Ic would lead I by 180°.
 
I believe that there is a flaw in the question as posed. Perhaps the angular frequency given is to high? If it is truly ##10^7 rad/sec## then the capacitor reactance is just 10 Ω. The real resistance of 100 Ω in the other branch will make it impossible for it to conduct a current with the same magnitude as the source current, even taking into account resonance effects. If I'm not mistaken the maximum magnitude for the current ##I## would be just 1 mA.
 

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