Solving circuit using complex numbers

[masterjoda]

Homework Statement

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

The Attempt at a Solution

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

 

[gneill]

If it is true that R = X_L,X_C, then X_L = X_C. What does that tell you about the LC tank circuit?

 

[masterjoda]

No just $R=X_L$.

 

[SammyS]

Homework Statement

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

The Attempt at a Solution

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

Doesn't $R=X_L, X_C$, mean that $R=X_L=X_C\ ?$

 

[gneill]

No just $R=X_L$.

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?

Anyways, you can write the complex impedances directly:

Z_R = R

Z_L = 0 + jR

Z_C = 0 - jX_C

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.

 

[masterjoda]

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 $I$.

 

[gneill]

I'm not sure why you've gone to the "raw" L and C versions of the impedances when you've been given X_L and X_C. You should be able to find the total current as a complex function of u, R, and X_C.