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

  1. Nov 25, 2006 #1
    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.
     

    Attached Files:

  2. jcsd
  3. Nov 26, 2006 #2

    SGT

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    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.
     
  4. Nov 26, 2006 #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.
     
  5. Nov 26, 2006 #4

    SGT

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    [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]
     
  6. Nov 26, 2006 #5
    Great, that's what I was thinking it would be.

    Thanks for the help.
     
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