# Solving circuit using complex numbers

1. Mar 4, 2012

### masterjoda

1. The problem statement, all variables and given/known data
Find the current in them in this circuit, if we know $R=X_L, X_C$ and $u=5sin(314t)$

3. 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 $I_0sin(\omega t + \theta)$.

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2. Mar 4, 2012

### Staff: Mentor

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

3. Mar 4, 2012

### masterjoda

No just $R=X_L$.

4. Mar 4, 2012

### SammyS

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

5. Mar 4, 2012

### Staff: Mentor

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:

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. Mar 4, 2012

### masterjoda

I have used them and I got that last equation for $I$.

7. Mar 4, 2012

### Staff: Mentor

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.