A Little Bit of Guidance on this RLC Circuit (Please )

In summary, the problem is that V_o(0) and \frac{dV_o}{dt}(0) are not specifically defined, and it would be helpful to have a hint as to what they should be.
  • #1
mushiman
10
0
I've been working on this problem to review (see the attachments), and my circuit analysis skills are quite rusty. I've determined that the circuit should be underdamped, and it will have a step response. Based on this, I'm using the equation V0(t)=Vf+B1e^(αt)cos(ωdt)+B2e^(ω0t)sin(ωdt). As you can see from the third attachment, I'm not sure as to what the values of B1 and B2 should be. Vf should be 20v as I interpreted, correct?

Also, it would seem that 50mA source would loop between the resistor while the switch is open (since the inductor would effectively be a short and the capacitor would be open), and when the switch is closed, the current would pass through the switch and loop in that manner, correct?

As I stated earlier, assuming that all of my former work is correct, I'm not sure what values I should assign to B1 and B2. A push in the right direction from here would be great -- I have terrible handwriting but hopefully it is clear enough to read.
 

Attachments

  • 4P1(s).jpg
    4P1(s).jpg
    24.1 KB · Views: 369
  • 4P2(s).jpg
    4P2(s).jpg
    20.2 KB · Views: 376
  • 4P3(s).jpg
    4P3(s).jpg
    14.2 KB · Views: 417
Engineering news on Phys.org
  • #2
mushiman said:
I've been working on this problem to review (see the attachments), and my circuit analysis skills are quite rusty. I've determined that the circuit should be underdamped, and it will have a step response. Based on this, I'm using the equation V0(t)=Vf+B1e^(αt)cos(ωdt)+B2e^(ω0t)sin(ωdt). As you can see from the third attachment, I'm not sure as to what the values of B1 and B2 should be. Vf should be 20v as I interpreted, correct?

Also, it would seem that 50mA source would loop between the resistor while the switch is open (since the inductor would effectively be a short and the capacitor would be open), and when the switch is closed, the current would pass through the switch and loop in that manner, correct?

As I stated earlier, assuming that all of my former work is correct, I'm not sure what values I should assign to B1 and B2. A push in the right direction from here would be great -- I have terrible handwriting but hopefully it is clear enough to read.

You mean [tex]V_0(t) = V_f + B_1e^{\alpha t}cos(\omega_dt) + B_2e^{\alpha t}sin(\omega_dt) [/tex] with [tex]\alpha[/tex] in both exponents.
The integration constants [tex]B_1[/tex] and [tex]B_2[/tex] are determined from the initial conditions [tex]V_o(0) = V_c(0)[/tex] and [tex]\frac{dV_o}{dt}(0)[/tex]. Those initial conditions can be determined from your diagram for t<0.
 
  • #3
Right (forgot about special characters on this forum).

The problem is that I'm not exactly sure what [tex]V_o(0)[/tex] and [tex]\frac{dV_o}{dt}(0)[/tex] are.

Isn't it true that the 50mA source will loop through the resistor and then through the switch when it is closed? If so, then I believe my work up till the portion where [tex]B_1[/tex] and [tex]B_2[/tex] must be determined is correct. This is where I was hoping for a hint -- I'm not exactly sure what either of them ([tex]V_o(0)[/tex] and [tex]\frac{dV_o}{dt}(0)[/tex]) should be interpreted as.
 
  • #4
mushiman said:
Right (forgot about special characters on this forum).

The problem is that I'm not exactly sure what [tex]V_o(0)[/tex] and [tex]\frac{dV_o}{dt}(0)[/tex] are.

Isn't it true that the 50mA source will loop through the resistor and then through the switch when it is closed? If so, then I believe my work up till the portion where [tex]B_1[/tex] and [tex]B_2[/tex] must be determined is correct. This is where I was hoping for a hint -- I'm not exactly sure what either of them ([tex]V_o(0)[/tex] and [tex]\frac{dV_o}{dt}(0)[/tex]) should be interpreted as.

[tex]V_o(0)[/tex] is the voltage in the capacitor at the closing of the switch. It is the difference between the voltages on the resistors.
[tex]\frac{dV_o}{dt}(0)[/tex] is the derivative of the voltage in the capacitor at the same instant.
Remember that [tex]i_c(t)=C\frac{dV_C}{dt}[/tex]
 
  • #5
Great, that's what I was thinking it would be.

Thanks for the help.
 

Related to A Little Bit of Guidance on this RLC Circuit (Please )

1. What is an RLC circuit?

An RLC circuit is an electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C). These components are connected in series or parallel to create a circuit that allows for the flow of alternating current (AC) electricity.

2. How does an RLC circuit work?

An RLC circuit works by using the properties of each component to regulate the flow of electricity. The resistor limits the current, the inductor stores energy in a magnetic field, and the capacitor stores energy in an electric field. Together, they create a circuit that can resonate at a specific frequency.

3. What is the purpose of an RLC circuit?

RLC circuits have many practical applications, such as in filters, oscillators, and amplifiers. They can also be used to tune radio frequencies and improve power factor correction in electrical systems.

4. How do I calculate the values of an RLC circuit?

To calculate the values of an RLC circuit, you will need to know the resistance, inductance, and capacitance of the components. From there, you can use Ohm's Law and other formulas to determine the current, voltage, and frequency of the circuit.

5. What are some common problems with RLC circuits?

Some common problems with RLC circuits include resonance, where the circuit can become unstable and produce excessive current, and damping, where the circuit dissipates energy and does not reach steady-state. Other issues may include component failures, incorrect component values, and incorrect circuit configurations.

Similar threads

  • Electrical Engineering
Replies
26
Views
5K
  • Introductory Physics Homework Help
Replies
10
Views
1K
Replies
12
Views
2K
  • Electrical Engineering
Replies
6
Views
6K
  • Electrical Engineering
Replies
4
Views
4K
  • Electrical Engineering
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
305
  • Electrical Engineering
Replies
5
Views
1K
  • Electrical Engineering
Replies
8
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
34
Views
5K
Back
Top