Equivalent Impedance Calculation for Circuit - Help Needed

[thenewbosco]

see diagram http://img.photobucket.com/albums/v11/biggm/z.jpg

The question is to find the equivalent impedance for the circuit.

I am not sure how to solve this, it is simple in the case of all three elements in series but with the capacitor in parallel i am not sure how to solve it.
thanks for the help

 

[Andrew Mason]

see diagram http://img.photobucket.com/albums/v11/biggm/z.jpg

The question is to find the equivalent impedance for the circuit.

I am not sure how to solve this, it is simple in the case of all three elements in series but with the capacitor in parallel i am not sure how to solve it.
thanks for the help

Find the impedance of the inductor and resistor in series. Then find the impedance (pure reactance) of the capacitor in parallel. Then add the impedances using:

$$\frac{1}{Z_{total}} = \frac{1}{Z_1} + \frac{1}{Z_2}$$

It gets a little difficult because the impedances have phase differences.

$$\frac{1}{Z_{total}} = \frac{1}{\sqrt{R^2 + \omega^2 L^2}} + \omega C$$

AM

 

[Cyrus]

An equivalent alternative is to say,

$$Z_r = R$$

$$Z_c = - \frac{j}{ \omega C} = \frac{1}{j \omega C } = \frac {1}{\omega C } \angle {-90^o}$$

$$Z_L = j \omega L = \omega L \angle {90^o}$$

Then treat the R-L in series, and call it [itex]Z_{RL} [/tex]<br /> <br /> and make that in parallel with $$Z_c$$, <br /> <br /> which gives you: $$Z_{eq} = (Z_c \backslash \backslash (Z_r + Z_L) )$$[/itex]