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

1. Nov 25, 2006

### mushiman

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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2. Nov 26, 2006

### SGT

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.

3. Nov 26, 2006

### mushiman

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.

4. Nov 26, 2006

### SGT

$$V_o(0)$$ 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 $$i_c(t)=C\frac{dV_C}{dt}$$

5. Nov 26, 2006

### mushiman

Great, that's what I was thinking it would be.

Thanks for the help.