Engineering RC parallel circuit, find the value of R and C

[Agent47]

Homework Statement

The circuit is a Parallel RC

Z= 105KΩ∠-27 and Frequency is 50 Hz.

I was given the following information and hints.

1/z = 1/R + jwc

I need to find the common denominator of 1/R + jwc, Invert it to find z and simplify Z into one real and one imaginary part.

Homework Equations

Need to find the value of R and C

The Attempt at a Solution

I started with finding the common denominator,

= 1/R +(jwc*R)/R

1/z = (1+jwRC)/R

To find z i simplified the equation above,

R = z(1+jwCR)

z = R/(1+jwCR)

Im not sure if i did this right as i can't find R and C. Any help will be much appreciated.

 

[gneill]

Why not start by calculating 1/Z, since you're given a numerical (complex) value for Z?

 

[Agent47]

If i do 1/z i would get 8.485x10-6 +j 4.323x10-4. I am not sure on how i would get and RC from 1/z = (1+jwRC)/R.

 

[gneill]

If i do 1/z i would get 8.485x10-6 +j 4.323x10-4. I am not sure on how i would get and RC from 1/z = (1+jwRC)/R.

You don't need RC. You can pick out the admittances of the two components directly from the complex value of 1/z.

1/z is the admittance of the parallel circuit. How do admittances in parallel add?

(The resistor analogy is two conductances in parallel).

 

[Agent47]

I worked out R= 118K ohm and C = 1.37x10-8. I checked by putting the values back into the equation and got the correct z. I guess i don't understand what my lecturer is going on about finding the common denominator etc.

 

[gneill]

I worked out R= 118K ohm and C = 1.37x10-8. I checked by putting the values back into the equation and got the correct z. I guess i don't understand what my lecturer is going on about finding the common denominator etc.

I suppose your lecturer had envisioned a different path to solution. But this one is pretty straightforward, no?

 

[Agent47]

I agree that the way you have explained is much simpler, ill ask my lecturer on what his method is. Thanks for your help.