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

AI Thread Summary
The discussion focuses on analyzing an underdamped RLC circuit and determining the values of B1 and B2 in the voltage equation V0(t). The user has identified that the final voltage Vf should be 20V and is seeking clarification on the initial conditions V0(0) and dV0/dt(0) to calculate B1 and B2. There is a consensus that the 50mA current source will loop through the resistor when the switch is open and will pass through the switch when closed. The user is advised that V0(0) represents the capacitor voltage at the switch's closure, while dV0/dt(0) is the rate of change of that voltage. The discussion concludes with the user expressing gratitude for the guidance received.
mushiman
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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.
 

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

The problem is that I'm not exactly sure what V_o(0) and \frac{dV_o}{dt}(0) 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 B_1 and B_2 must be determined is correct. This is where I was hoping for a hint -- I'm not exactly sure what either of them (V_o(0) and \frac{dV_o}{dt}(0)) should be interpreted as.
 
mushiman said:
Right (forgot about special characters on this forum).

The problem is that I'm not exactly sure what V_o(0) and \frac{dV_o}{dt}(0) 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 B_1 and B_2 must be determined is correct. This is where I was hoping for a hint -- I'm not exactly sure what either of them (V_o(0) and \frac{dV_o}{dt}(0)) should be interpreted as.

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

Thanks for the help.
 
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