Second Order RLC source switching cct

In summary: Keep up the good work.In summary, the conversation is about a person asking for help with analyzing an electrical circuit. They have correctly set up the problem and determined the equations to use, with the only potential mistake being a decimal in their answer for the angular frequency. They are seeking confirmation that they are on the right track.
  • #1
jess_88
7
0
Hey guys

I'm just wondering if I'm on the right track for the problem. I have started analysing the circuit, but I'm really unsure of my method of analysis. Did I use the formulas correctly? :confused:

The circuit switches from position "a" to position "b" at t=0
and I'm trying to fined i(t) for t>0
[PLAIN]http://forum.allaboutcircuits.com/cache.php?url=http://i.imgur.com/Dpyep.png

from what I understand, I need to determine i(o), which is for the circuit at t=0 (containing an inductor and 4A supply).

Then determine the α and wo for the circuit at t>0
where α = 1/2RC and wo = √(1/LC).
which will indicate the formula to use for the step response, to allow for determining i(t).


my calculated values so far-
by KCL
i(0) = 4A(6/2+8)
= 3A

This is where I think I may have gone wrong-
at t>0
I used source transformation for the 4A supply (24V)
then added the total voltage in the loop (= +12V)
[PLAIN]http://forum.allaboutcircuits.com/cache.php?url=http://i.imgur.com/WjGl5.png

α = 1/(2 x 20Ω x 0.02F)
= 1.25

wo = √(1/ (2 x 0.02))
=5

therefor I understand I need to use
v(t) = (A1cos(wd)t + A2sin(wd)t)e^(-αt)

wd = √((wo^2) - (α^2))

= √(375/16)

... which isn't a very nice looking number

have I made a mistake somewhere? Is my whole understanding of the method in answering this question just plane wrong? :confused:
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Thanks in advance for any advice/help. :)Hi there,It looks like you have set up the problem correctly and you have determined the appropriate equations to use. Your calculations also look correct. The only thing I might suggest is that when you are calculating wd (angular frequency), you can simplify it to √(25/8) instead of having a decimal in the answer. Otherwise, it looks like you are on the right track!
 

1. What is a Second Order RLC source switching circuit?

A Second Order RLC source switching circuit is an electrical circuit that uses a resistor (R), an inductor (L), and a capacitor (C) to control the flow of current from a source. It is called "second order" because it involves two energy storage elements (inductor and capacitor) in addition to the resistor.

2. How does a Second Order RLC circuit work?

In a Second Order RLC circuit, the inductor and capacitor store energy in the form of an electric and magnetic field, respectively. The resistor dissipates this energy as heat. When the circuit is switched on, the inductor resists changes in current, while the capacitor resists changes in voltage. As a result, the circuit produces an oscillating current and voltage, which can be used for various applications.

3. What are the advantages of using a Second Order RLC circuit?

Second Order RLC circuits have several advantages over other types of circuits. They have a higher quality factor, which means they have a narrower bandwidth and can produce more precise oscillations. They also have a higher resonance frequency, making them useful for applications that require high-frequency signals. Additionally, they can be easily tuned by adjusting the values of the components.

4. What are the applications of Second Order RLC circuits?

Second Order RLC circuits have a wide range of applications in various fields, including electronics, telecommunications, and power systems. They are commonly used in filters, amplifiers, oscillators, and frequency-selective networks. They are also used in radio transmitters and receivers, as well as in audio systems to produce high-quality sound.

5. What are the limitations of Second Order RLC circuits?

While Second Order RLC circuits have many advantages, they also have some limitations. One limitation is that they can be sensitive to changes in temperature, humidity, and other environmental factors. This can affect the accuracy and stability of the circuit. Additionally, they can be more complex to design and analyze compared to simpler circuits. Finally, they are not suitable for high-power applications as the components may overheat and fail under high currents.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
2
Views
819
  • Engineering and Comp Sci Homework Help
Replies
7
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
2K
Back
Top