RLC Circuit: Calculating Current I(ω)

[VictorWutang]

Homework Statement

Write the current I(ω) for the RLC circuit above

Homework Equations

I = V/Z

The Attempt at a Solution

I believe Z should be Z = ( 1/R + 1/X_L + 1/X_C)^-1

which would give Z = ( 1/R + 1/(ωL) + ωC )^-1

and I is simply emf / Z


but I don't think you can add R L and C like this since they have different phase angles.


Help?

 

[gneill]

Homework Statement

Write the current I(ω) for the RLC circuit above

Homework Equations

I = V/Z

The Attempt at a Solution

I believe Z should be Z = ( 1/R + 1/X_L + 1/X_C)^-1

which would give Z = ( 1/R + 1/(ωL) + ωC )^-1

and I is simply emf / Z


but I don't think you can add R L and C like this since they have different phase angles.


Help?

You are correct that phase angle must be taken into account. Rather than use reactances (X_L, X_C), have you considered using complex impedances (Z_L, Z_C) which take into account the phase angles automatically?

 

[VictorWutang]

Oh I see... in that case

I know for RL and RC circuits in series

Z_L = √ R² + ω²L²

and

Z_C = √ R² + 1 / (ω²C²)

but I don't understand how I would use this for RLC in parallel. Help?

 

[gneill]

Are you familiar with complex numbers?

 

[VictorWutang]

I think I'm familiar enough that i'll understand whatever you explain using them.EDIT* sorry, I realized that was a little vague. Yes, I understand basic use of complex numbers but have not taken a college level class on them yet.

 

[gneill]

I think I'm familiar enough that i'll understand whatever you explain using them.


EDIT* sorry, I realized that was a little vague. Yes, I understand basic use of complex numbers but have not taken a college level class on them yet.

Okay, well you already know that there is a phase shift of 90 degrees between the voltage and current for reactive components; For the capacitor the current leads the voltage by 90 degrees, while for the inductor it lags the voltage by 90 degrees. Complex numbers are convenient to use here because one can consider the imaginary portion to be "at right angles to" the real portion of a complex number. This is made obvious when you plot the numbers on a complex plane, since the imaginary axis is at right angles to the real axis.

So, if you write the impedances as complex values, then you can use all the usual formulas that you use for resistances but with the complex impedances. Obviously you need to use number arithmetic in the manipulations, but otherwise it's just plug and chug with the formulas.

For capacitors and inductors the impedances have the same magnitude as their reactances (That is, reactance is the magnitude of the complex impedance). To find the impedance of these components just replace ω with jω in the reactance expressions. Here "j" is the imaginary value $\sqrt{-1}$.