Complex Parallel Impedance of 20 Ω Resistor & 110 mH Inductor

[fonz]

Homework Statement

Find the complex parallel impedance of a 20 Ω resistor and 110 mH inductor from a 50 Hz supply.

Homework Equations

Z = R (X_L² / R²+X_L²) + j X_L (R² / X_L² + R²)

The Attempt at a Solution

[/B]
R = 20 Ω
X_L = j34.6 Ω

Z = 20 ( j34.6² / j34.6² + 20²) + j34.6 ( 20² / j34.6² + 20²) Ω

Z = 20 ( -1197 / 400 - 1197 ) + j34.6 ( 400 / 400 - 1197 ) Ω

Z = 30 - j17 Ω

Apparently this is wrong but I can't find where I have gone wrong.

Thanks

 

[BvU]

Hi,

Not clear where your relevant formula comes from.
Why not simply $${1\over Z} = {1\over R} + {1\over j\omega L}\ \ ?$$

 

[BvU]

Correction: I confirm your equation, so it's in a later step. Check out what X_L should be.

 

[fonz]

I calculated X_L at 50 Hz and got j34.6 Ω

Then I tried to calculate X_L² as follows:

X_L² = (j34.6)² = -1197

Are you suggesting -1197 is incorrect?

Thanks for your response.

 

[BvU]

http://mlg.eng.cam.ac.uk/mchutchon/ResonantCircuits.pdf. The complex impedance is ${\bf j}\omega L$ but the reactance X_L is a real number.

 

[fonz]

http://mlg.eng.cam.ac.uk/mchutchon/ResonantCircuits.pdf. The complex impedance is ${\bf j}\omega L$ but the reactance X_L is a real number.

I've been sat trying to interpret what you said and getting nowhere unfortunately. 34.6 is a real number? 34.6² = 1197

 

[BvU]

So X_L² is 1197, not -1197. Indeed.

And: be clearer with the brackets when you write down an expression.

 

[fonz]

So X_L² is 1197, not -1197. Indeed.

And: be clearer with the brackets when you write down an expression.

Right I understand what you are saying and thanks for the help.