# Solving circuit using complex numbers

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

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

No just $R=X_L$.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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
Mentor
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:

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.

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
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.