# Parallel LR impedance

1. May 4, 2017

### fonz

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Z = R (XL2 / R2+XL2) + j XL (R2 / XL2 + R2)

3. The attempt at a solution

R = 20 Ω
XL = 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

2. May 4, 2017

### BvU

Hi,

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

3. May 4, 2017

### BvU

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

4. May 4, 2017

### fonz

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

Then I tried to calculate XL2 as follows:

XL2 = (j34.6)2 = -1197

Are you suggesting -1197 is incorrect?

5. May 4, 2017

### BvU

I am indeed. The complex impedance is ${\bf j}\omega L$ but the reactance XL is a real number.

6. May 4, 2017

### fonz

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

7. May 5, 2017

### BvU

So XL2 is 1197, not -1197. Indeed.

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

8. May 5, 2017

### fonz

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