Solving circuit using complex numbers

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Homework Statement


Find the current in them in this circuit, if we know [itex]R=X_L, X_C[/itex] and [itex]u=5sin(314t)[/itex]


The Attempt at a Solution


First , [itex]5=U_0, 314=\omega[/itex] and voltage we can write as [itex]u=U_0cos(\omega t + \frac{\pi}{2})[/itex] and [itex]u=U_0 e^{i\frac{\pi}{2}}=iU_0[/itex]. [itex]U[/itex] is the voltage at the source [itex]U_1[/itex] in the branch and [itex]U_2[/itex] at the resistor. Now [itex]U=U_1+U_2[/itex] or [itex]U=IR+IZ[/itex] where [itex]\frac{1}{Z}=\frac{1}{iL\omega}-\frac{C\omega}{i}[/itex] or [itex]Z=\frac{iL\omega}{1-CL\omega^2}[/itex]
Now the current is [itex]I=\frac{U}{R+Z}[/itex] or when it is arranged [itex]I=\frac{iU_0(1-CL\omega^2)}{iL\omega (2-CL\omega^2)}[/itex]. Now I don't know how to complete the calculation to reduce it to the form like this [itex]I_0sin(\omega t + \theta)[/itex].
 

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Answers and Replies

  • #2
gneill
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If it is true that R = XL,XC, then XL = XC. What does that tell you about the LC tank circuit?
 
  • #3
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No just [itex]R=X_L[/itex].
 
  • #4
SammyS
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Homework Statement


Find the current in them in this circuit, if we know [itex]R=X_L, X_C[/itex] and [itex]u=5sin(314t)[/itex]


The Attempt at a Solution


First , [itex]5=U_0, 314=\omega[/itex] and voltage we can write as [itex]u=U_0cos(\omega t + \frac{\pi}{2})[/itex] and [itex]u=U_0 e^{i\frac{\pi}{2}}=iU_0[/itex]. [itex]U[/itex] is the voltage at the source [itex]U_1[/itex] in the branch and [itex]U_2[/itex] at the resistor. Now [itex]U=U_1+U_2[/itex] or [itex]U=IR+IZ[/itex] where [itex]\frac{1}{Z}=\frac{1}{iL\omega}-\frac{C\omega}{i}[/itex] or [itex]Z=\frac{iL\omega}{1-CL\omega^2}[/itex]
Now the current is [itex]I=\frac{U}{R+Z}[/itex] or when it is arranged [itex]I=\frac{iU_0(1-CL\omega^2)}{iL\omega (2-CL\omega^2)}[/itex]. Now I don't know how to complete the calculation to reduce it to the form like this [itex]I_0sin(\omega t + \theta)[/itex].
Doesn't [itex]R=X_L, X_C[/itex], mean that [itex]R=X_L=X_C\ ?[/itex]
 
  • #5
gneill
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No just [itex]R=X_L[/itex].

Ah, my bad. I read it as both being R, rather than as part of a list of known values. Where's a semicolon when you need one? :smile:

Anyways, you can write the complex impedances directly:

ZR = R

ZL = 0 + jR

ZC = 0 - jXC

These can then be used in formulas in the usual way to find voltages and currents (complex). Convert the values to phasor form at the end.
 
  • #6
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These can then be used in formulas in the usual way to find voltages and currents (complex). Convert the values to phasor form at the end.

I have used them and I got that last equation for [itex]I[/itex].
 
  • #7
gneill
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I'm not sure why you've gone to the "raw" L and C versions of the impedances when you've been given XL and XC. You should be able to find the total current as a complex function of u, R, and XC.
 

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