# Equivalent impedance

1. Apr 14, 2006

### thenewbosco

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

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

Last edited by a moderator: May 2, 2017
2. Apr 14, 2006

### Andrew Mason

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

Last edited by a moderator: May 2, 2017
3. Apr 14, 2006

### 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]

and make that in parallel with $$Z_c$$,

which gives you: $$Z_{eq} = (Z_c \backslash \backslash (Z_r + Z_L) )$$

Last edited: Apr 14, 2006
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