Parallel RLC, calculating phasors

[James889]

Hi,

I have the following circuit:
[PLAIN]http://img252.imageshack.us/img252/9470/phasor.png

I need to find the current i(t).

I have come to a point where i don't know how to proceed.

I have the inductor and resistor in series.

40 +j120 in phasor form $$126.5\angle 71.56$$

which in turn is parallel with the capacitor.

$$\frac{1}{\frac{1}{126.5} +\frac{1}{-133.33}} = 2469$$

But how do i get the phase angle ?

 

[tiny-tim]

Hi James889!

1/(40 +j120) = (40 - j120)/(40² +120²).

It's some time since I've done this, but doesn't the https://www.physicsforums.com/library.php?do=view_item&itemid=303" depend on the frequency … 1/jωC ?

 

[James889]

Hi James889!

1/(40 +j120) = (40 - j120)/(40² +120²).

It's some time since I've done this, but doesn't the https://www.physicsforums.com/library.php?do=view_item&itemid=303" depend on the frequency … 1/jωC ?

Hi Tim.

Sure, but as you can see the complex impedance is already given.

And the actual numbers don't really matter.

I'm more interested in how to find the phase angle when you have two(or more) parallel phasors.

 

[tiny-tim]

Hi Tim.

Sure, but as you can see the complex impedance is already given.

And the actual numbers don't really matter.

I'm more interested in how to find the phase angle when you have two(or more) parallel phasors.

Hi James889!

(ah I misread that! )

I meant that you should keep the 40 +j120 as it is, do 1/ it, and add it to 1/-j133.33, and then 1/ that, and only reduce it to phasor from at the end