RLC circuit with variable capacitance problem

[megansanford]

Consider a series RLC circuit. The applied voltage has a maximum value of 120 V and oscillates at a frequency of 81 Hz. The circuit contains a variable capacitor, an 820 $$\Omega$$ and a 3.7 H inductor. Determine the value of the capacitor such that the voltage across the capacitor is out of phase with the applied voltage by 48 degrees, with V_max leading V_c. Answer in units of H.
$$\omega$$ = 2$$\pi$$f; Z = sqrt(R² + ($$\omega$$L - 1/$$\omega$$C)²)
So i figured I needed to use that last equation, but I don't know Z or C. What does it mean "out of phase" and what kind of equations use that value?Thank you so much for any help, in advance

 

[gneill]

Why does the circuit contain two capacitors, a fixed one of 1.5 μF and a variable one of unspecified value? How can the problem be solved if both the value of the inductor and variable capacitor are unknown? Perhaps you could review the problem statement?

 

[megansanford]

Oops, I fixed the problem.

 

[gneill]

In an RLC circuit the voltage and current waveforms are not necessarily in phase (their peaks and zeros don't happen at the same time). So the voltages expressed by the three components in the circuit will also not have coincident peaks and zeros.

If you were to treat the circuit as a voltage divider and write the expression for the voltage across the capacitor (use complex impedances / phasors to do so), then you can determine the phase angle of that voltage with respect to the supply voltage waveform.

 

[ideasrule]

So i figured I needed to use that last equation, but I don't know Z or C. What does it mean "out of phase" and what kind of equations use that value?

The last equation is the equation for total impedance, which is not what you want.

The phase angle is how much the voltage leads the current. If you plot voltage vs. time and current vs. time on the same graph, the phase angle is how many radians it takes for the current to peak after voltage peaks, or for the current to drop to a minimum when voltage drops to a minimum.

Inductors try to increase the phase angle because they oppose changes in current, causing voltage to lead current even more. Inductors decrease the phase angle because current is highest when they have no charge (and hence generate no back voltage), and lowest when they're fully charged. The phase angle is just tan(phi)=(XL-XC)/R, where XL and XC are the reactances of the inductor and capacitor. If all of this is new, I think you should review your textbook; I'm very sure that it talks about phase factors in reasonable detail.

 

[megansanford]

Thank you. My textbook was rather confusing, as was the lecture, which makes sense given that my professor wrote the textbook. I just have trouble grasping the concepts.

 

[megansanford]

Is the maximum voltage just extraneous information given to confuse?

 

[ideasrule]

Yes, it is. I also don't understand what "H" is; it's certainly not the SI unit for capacitance, which is F.

 

[megansanford]

So i did the tan(phi) = (XL - XC)/R and still got the incorrect answer:

tan (48) = (omega*L - 1/omega*C)/R

1/[omega(omega*L - tan(48)*R)] = C

C = 2.020711 microFarad

My calculator is in degrees and I changed frequency from Hz to radians. Can you guys see where I might have gone wrong?

 

[megansanford]

Yes, it is. I also don't understand what "H" is; it's certainly not the SI unit for capacitance, which is F.

H is supposed to be microfarad. I typed the problem incorrectly. Sorry for the confusion

 

[ideasrule]

48 itself is not the phase angle. Take a look at these lecture notes, especially Figure 34.9: http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter34/chapter34.html

Phase angle is considered positive if the resultant (supply) voltage is counterclockwise from the resistor voltage, which is pointing in the same direction as current. In other words, the phase angle is how much supply voltage leads Vr. For your case, the supply voltage leads Vc by 48 degrees, meaning that it lags Vr by 90-48=42 degrees.

 

[megansanford]

48 itself is not the phase angle. Take a look at these lecture notes, especially Figure 34.9: http://teacher.pas.rochester.edu/phy122/Lecture_Notes/Chapter34/chapter34.html

Thank you. Those notes helped me with my other homework problems as well.