Complex current calculation in Parallel R L Circuit

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Discussion Overview

The discussion revolves around calculating the complex current in a parallel R-L circuit, specifically focusing on the contributions from a resistor, inductor, and capacitor. Participants are exploring the application of Ohm's law and impedance in the context of AC circuit analysis, with an emphasis on expressing the current in terms of complex numbers.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants express uncertainty about the correct approach to find the complex current, with references to using Ohm's law and total impedance calculations.
  • There is a discussion about the configuration of the circuit, with some clarifying that the resistor and inductor are in series, while the capacitor is in parallel with this combination.
  • Participants propose that the total impedance should be calculated symbolically before substituting numerical values, suggesting that the impedance of the inductor is represented as jωL.
  • Some participants correct earlier claims about impedance addition, emphasizing that the series impedance of the resistor and inductor should be treated as R + jωL.
  • There are attempts to derive expressions for the total current, with some participants suggesting that the total current can be calculated as the sum of the currents through the capacitor and the series R-L branch.
  • Discussions include the need to handle complex numbers carefully, particularly when combining real and imaginary components in impedance calculations.
  • Some participants express confusion about how to incorporate the capacitor's impedance into the overall calculations.

Areas of Agreement / Disagreement

Participants generally agree on the need to calculate the total impedance and current in the circuit, but there are multiple competing views on the correct approach and the role of the capacitor in the calculations. The discussion remains unresolved regarding the best method to express the total current and how to handle the capacitor's contribution.

Contextual Notes

Limitations include the potential for algebraic errors and the challenge of managing complex numbers in impedance calculations. Some participants note that the circuit diagram is not visible to all, which may hinder understanding.

  • #31
I think I will give up for now .. I looked at it too much that I cannot come up with anything that makes sense ... Thanks a lot for all the effort you made to explain me :)
my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL) or
(R^2+(wL)^2)%20).gif
 
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  • #32
gl0ck said:
my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL)

This is right.. keep going when you're ready.

I = E (jwC + 1/(R+jwL)) = E ((jwC)(R+jwL)+1)/(R+jwL)

(I found a common denominator and added)

Multiply that out and see if you can find a C that will minimize |I|. The j*j you see in the numerator above will introduce a negative quantity that will let you make |I| smaller.
 
  • #33
Thanks a lot man but I have to go to sleep.. I hope tomorrow to find the right solution .. For now I give up..
Thanks for the efforts both of you
 
  • #34
gl0ck said:
I think I will give up for now .. I looked at it too much that I cannot come up with anything that makes sense ... Thanks a lot for all the effort you made to explain me :)
my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL) or
(R^2+(wL)^2)%20).gif

Good. Separate the total real and imaginary components and factor out the common bits:
$$I = E \left[ \frac{R}{R^2 + (ω L)^2} + j ω \left(C - \frac{L}{R^2 + (ω L)^2}\right)\right]$$
What value of C will minimize the magnitude of the whole?
 
  • #35
Thanks! So when I plugged in the numbers It gets really nasty. First can I use w = 2Pi and to treat jEwC as j220*2*Pi*C? also w*L to be 12.57 because jwL = j12.57 Ohms
so to the equation I = 220 [ R/(R^2+12.57^2) + j2*Pi(C - 0.4/(R^2+12.57^2)]
I= 220[8/(64+158.0049) + j220*6.2830 ( C - 0.4/(64+158.0049)]
I = 7.92+(220jwC - 220*0.0018jw)
I= 7.92 + j1382.26C - j2.4807C
I=7.92+j1379.78C

I hope to do the right calculations

Thanks
 
  • #36
The impedances of reactive components (inductors, capacitors) change with frequency. Furthermore, they change in opposite directions so that inductor impedance goes up with increasing frequency while capacitor impedance goes down. So no, in general you can't just set ω to a convenient value like ##2\pi##; you must use the value specified.

For part (a) you just need to derive the expression for the current, you don't have to plug in any numbers.

For part (b) you need to first determine an expression for C that will minimize the magnitude of the current. Use the given hint.
 

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