# LRC circuit and phase angle

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1. Nov 27, 2016

### ooohffff

1. The problem statement, all variables and given/known data
A 35 mH inductor with 1.0 resistance is connected in series to a 20 µF capacitor and a 60 Hz, 40-V (rms) source. Calculate the phase angle.

2. Relevant equations
tan φ = (XL - XC) / R

3. The attempt at a solution
Solving for φ:

φ = tan -1 [(XL - XC) / R]

XL = 2πfL = 13.194 Ω
XC = 1/(2πfC) = 132.63 Ω
R=1Ω

Plugging those values in I get
Φ = -89.5° = 270.48°

I submitted the 270.48 but it's wrong?

2. Nov 27, 2016

### Staff: Mentor

What you've found is the phase angle of the impedance. You want the phase angle of the current. What law relates the current to the voltage and impedance?

3. Nov 27, 2016

### ooohffff

Ohm's law I=V/Z ?

4. Nov 27, 2016

### Staff: Mentor

Yup. What will be the current's angle if your Z's angle is -89.5° ?

5. Nov 27, 2016

### ooohffff

Would it be tan φ = (VL - VC) / Z ? If not then I'm thoroughly confused.

6. Nov 27, 2016

### Staff: Mentor

No. The only voltage of consequence here is the supply voltage. If this was a DC circuit with resistors you'd write I = V/R. In this AC circuit you write I = V/Z.

V is the source voltage that also serves as the reference for the phase angle. As such its phase angle is 0°. Your Z has a phase angle of -89.5°. So what's the phase angle of V/Z? (How do you handle angles when you do a division)?

7. Nov 27, 2016

### ooohffff

Would you break it into components like:
I = (40/(Zcosφ)) i + (40/(Zsinφ)) j

and then find the angle of I?

8. Nov 27, 2016

### Staff: Mentor

No need. You have the angles already. There's a simple rule for dividing two complex numbers when you know the angles. What's the rule?

9. Nov 27, 2016

### ooohffff

I just derived this because I don't think I know what you're talking about or I'm misguided, but I got tan-1 (cotφ)

10. Nov 27, 2016

### Staff: Mentor

No, you're getting way too complicated. When you divide two numbers in complex polar form you simply subtract the angle of the denominator from the angle of the numerator:

$\frac{a ∠ θ}{b ∠ φ} = \left(\frac{a}{b}\right) ∠ (θ - φ)$

11. Nov 27, 2016

### ooohffff

But then the resulting angle would be at 89.5°?

12. Nov 27, 2016

### ooohffff

I've tried that before initially but that angle was also incorrect.

13. Nov 27, 2016

### Staff: Mentor

Yes. That should be the phase angle of the current with respect to the voltage.

You tried +89.5° and it was considered incorrect? Perhaps they're being picky about significant figures?

14. Nov 27, 2016

### ooohffff

Yup. That was the first one I tried with the equation: φ = cos -1 (R/Z) = 89.52028539°

Yes, maybe I should try with more sig figs.

15. Nov 27, 2016

### Staff: Mentor

How many significant figures does the given data suggest?

16. Nov 28, 2016

### ooohffff

Ah I figured out the problem! My original answer in this post -89.5° was correct. The mistake I made was that I should not have converted it to 270.48°, since technically φ should be negative since XC > XL, and you should generally take the smaller angle of the angles between two vectors.

17. Nov 28, 2016

### Staff: Mentor

What was the exact phrasing of the question as you received it? Generally "phase angle", unless qualified, refers to the phase angle of the current with respect to the voltage. Were they only looking for the phase angle of the impedance?

18. Nov 28, 2016

### ooohffff

That is the exact phrasing of the question. That phase angle, according to the formula that I used, should be the angle between voltage and current, not impedance and current.

tanΦ = (VL-VC )/ VR = (XL - XC)/ R

19. Nov 28, 2016

### Staff: Mentor

Yes, it yields the angle associate with the impedance which is also the phase angle of the current with respect to the voltage.

But as I mentioned previously, unless otherwise specified, generally when one talks about phase angle one is referring to the phase angle of the current with respect to that of the voltage, not the voltage with respect to the current. Basically, I'm not very happy with the problem as it is presented. But if you reached the answer that they're looking for, not much more can be said