# RLC circuit impedance

1. Apr 7, 2016

### ReidMerrill

1. The problem statement, all variables and given/known data
A series RLC circuit consists of a 60.0 Ω resistor, a 2.30 mH inductor, and a 690 nF capacitor. It is connected to an oscillator with a peak voltage of5.80 V .

Part A
Determine the impedance at frequency 3000 Hz.
Part B
Determine the peak current at frequency 3000 Hz.
Part C
Determine phase angle at frequency 3000 Hz

2. Relevant equations
Z=sqrt[R2+(Xl-Xc)2]
Xc1/wC
Xl=wL

3. The attempt at a solution

I found Xc=1/(2pi*3000Hz) =0.76886
and XL2pi*3000=43.354

so Z=73.5765

As you might have guessed, this is not the correct answer.

What am I doing wrong? I suspect it has something to do with how I converted Hz to Rad/s

How would I go about parts B and C too?

2. Apr 7, 2016

### Staff: Mentor

Did you involve the value of C in your calculation for XC?

3. Apr 7, 2016

### ReidMerrill

That was a typo on my part. I used the correct equation when I worked it out.

4. Apr 7, 2016

5. Apr 7, 2016

### ReidMerrill

I converted nF to F wrong.... Thanks you!

6. Apr 7, 2016

### ReidMerrill

Now I'm working on a later part that involves finding impedance at 5000Hz
Xc=1/(2pi*5000z) =46.13
and XL=2pi*5000*L=72.26

so Z=54.013 but this is somehow wrong even though the exact same process worked for 3000 and 4000 Hz

7. Apr 8, 2016

### Staff: Mentor

Your individual reactance values look fine. Must be a calculator/finger interface issue

Can you try again? If it still doesn't look right, give us a breakdown of the calculation step by step.

8. Apr 8, 2016

### ReidMerrill

XL=2pi(5000)(0.0023)=72.2566
XC=1/(2pi5000*(6.9*10^-7))=46.1318

Z=sqrt[(60^2)-(72.2566-46.1318)^2]=54.0138

9. Apr 8, 2016

### Staff: Mentor

Ah. Why have you used a minus sign between the two terms within the square root?

10. Apr 8, 2016

### ReidMerrill

I keep doing that. I don't know why I keep doing that