Finding VL(t) and Vc(t) in a RLC circuit with sinusoidal excitation

In summary, the homework statement is to find VL(t) and Vc(t) in the circuit below. The Attempt at a Solution found VL(t) and Vc(t) using capacitance and impedance.
  • #1
hogrampage
108
1

Homework Statement


Find VL(t) and Vc(t) in the circuit below

rlc1.jpg


Where Vs = cos(200[itex]\pi[/itex]t) V

Homework Equations


[itex]\alpha[/itex] = [itex]\frac{1}{2RC}[/itex]
ω0 = [itex]\frac{1}{\sqrt{LC}}[/itex]
[itex]\omega[/itex]0 = [itex]\sqrt{\omega^{2}-\alpha^{2}}[/itex]

The Attempt at a Solution


Well, I found the total capacitance

C = 1.5 uF

Then, I found [itex]\alpha[/itex]:

[itex]\alpha[/itex] = [itex]\frac{1}{2(100Ω)(1.5uF)}[/itex] ≈ 3333 s-1

and

ω0 = [itex]\frac{1}{\sqrt{(20mH)(1.5uF)}}[/itex] ≈ 5774 s-1

So Vc = B1e-3333cos([itex]5774\pi[/itex]t) + B2e-3333sin([itex]5774\pi[/itex]t)

That's as far as I went. I have no idea what I'm doing :|. Also, please note that I have NOT had differential equations. This class does not require it, so please give an answer without using DQ's.
 
Physics news on Phys.org
  • #2
I looked through some articles/lessons on these types of circuits, and they used impedance to solve for voltage drops and all that. However, none of them showed a time-dependent function. So, can I solve this problem by using similar steps (i.e. find XL, XC, and then impedance Z)?

Here are the values I came up with (noting that XC is inductive reactance of both capacitors combined):

XL = 2[itex]\pi[/itex](200[itex]\pi[/itex])(20mH) = 78.96Ω
XC = 1/2(200[itex]\pi[/itex])(1.5uF) = 168.87Ω
Z = 1/(sqrt((1/R)2+(1/XL-1/XC)^2) = 81.56Ω
I = V/Z = 1/81.56 = 10mA
 
  • #3
No picture.
 
  • #4
Not sure why it isn't showing up in first post, but here it is.
 

Attachments

  • rlc1.jpg
    rlc1.jpg
    7 KB · Views: 391
Last edited:
  • #5
Have you learned to use phasors yet? That would make the problem look like it was all resistors with complex numbers (phasors) for voltage and current.
 
  • #6
Unfortunately, no. We are starting the chapter with those tomorrow. This assignment is supposed to use the form that looks something like this:

A-Ae^-t/tau
 
  • #7
hogrampage said:
I looked through some articles/lessons on these types of circuits, and they used impedance to solve for voltage drops and all that. However, none of them showed a time-dependent function. So, can I solve this problem by using similar steps (i.e. find XL, XC, and then impedance Z)?

Here are the values I came up with (noting that XC is inductive reactance of both capacitors combined):

XL = 2[itex]\pi[/itex](200[itex]\pi[/itex])(20mH) = 78.96Ω
XC = 1/2(200[itex]\pi[/itex])(1.5uF) = 168.87Ω
Z = 1/(sqrt((1/R)2+(1/XL-1/XC)^2) = 81.56Ω
I = V/Z = 1/81.56 = 10mA

No, because that does not account for the phasing of the voltages within your circuit. Like Antiphon says, you need to invoke phasors (comlex values for impedances, voiltages and currents). Pay particular attention to the fact that your voltage source is a cosine rather than the more usual sine generator.
 
  • #8
hogrampage said:
I looked through some articles/lessons on these types of circuits, and they used impedance to solve for voltage drops and all that. However, none of them showed a time-dependent function. So, can I solve this problem by using similar steps (i.e. find XL, XC, and then impedance Z)?

No, because that does not consider phasing of voltages and currents. Like Antiphon says, you need to invoke phasors (complex V, I and X). Also take note in this regard that your voltage source is a cosine rather than the more usual sine generator.
 
  • #9
Okay, that makes sense. However, at the time of this (it was a lab yesterday), we had not learned about phasors, so I'm not sure how the professor wants us to do it. I do know I can use this as the voltage source:

sin(200[itex]\pi[/itex]t+[itex]\frac{\pi}{2}[/itex])

So now that I got that, I'm not sure where to go. This problem was for the pre-lab, so I'm wondering how to solve it without phasors. It doesn't seem possible.
 
  • #10
hogrampage said:
Unfortunately, no. We are starting the chapter with those tomorrow. This assignment is supposed to use the form that looks something like this:

A-Ae^-t/tau

That's not what your answer will look like. You have a sinusoidal excitation plus you have BOTH inductor & capacitor. Either fact precludes a response like you state and in fact necssitates a sinusoidal one for both Vl and VC..

If this is doable without either phasors, diff. eq.'s or a transform method, I don't know what it would be but I do know I wouldn't want to find out, since for sure it would be a very awkward process. It reminds me of the ARRL Handbook, trying to teach this stuff without the necessary tools. I would sit on it until you get the phasor stuff given to you.
 

Related to Finding VL(t) and Vc(t) in a RLC circuit with sinusoidal excitation

What is an RLC circuit?

An RLC circuit is an electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C) connected in series or parallel. These components work together to create a resonant frequency that can be used to filter or amplify specific frequencies of an electrical signal.

What is total capacitance in an RLC circuit?

Total capacitance in an RLC circuit refers to the combined capacitance of all capacitors in the circuit. It is calculated by adding the individual capacitances of each capacitor in series, or by using a formula to calculate the equivalent capacitance of capacitors in parallel.

How do I calculate total capacitance in an RLC circuit?

The formula for calculating total capacitance in a series RLC circuit is: 1/Ct = 1/C1 + 1/C2 + 1/C3 + ... where Ct is the total capacitance and C1, C2, C3, etc. are the individual capacitances. In a parallel RLC circuit, the formula is: Ct = C1 + C2 + C3 + ...

Why is total capacitance important in an RLC circuit?

Total capacitance affects the resonant frequency of an RLC circuit, which is the frequency at which the circuit will oscillate most efficiently. It also determines the impedance of the circuit, which can affect the flow of current and voltage through the circuit.

How can I adjust the total capacitance in an RLC circuit?

Total capacitance can be adjusted by changing the values of the individual capacitors in the circuit, or by adding or removing capacitors. This can be useful for tuning the circuit to a specific resonant frequency or for adjusting the impedance of the circuit.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
291
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
13
Views
12K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
5K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
12K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
22K
  • Engineering and Comp Sci Homework Help
Replies
13
Views
2K
Back
Top