RLC Circuit w/ switch

1. Apr 14, 2010

phufool

1. The problem statement, all variables and given/known data
Provided in the picture below.

2. Relevant equations

3. The attempt at a solution
So what I've done so far is use KVL to obtain Vc(t) +Vr(t) + Vo(t) = 1/2Vin(t).
Could someone tell me if I started this problem correctly? Would I just take the derivative of the equation next?

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2. Apr 14, 2010

Staff: Mentor

Welcome to the PF. I think the input Vi is meant to have the value of 1/2V, but it's hard to tell for sure from the drawing.

And yes, writing the KVL around the loop is a good approach, but you need to use the differential equations that relate I and V for the capacitor and inductor. That's where you end up with one differential and one integral. And yes, you then differentiate that equation to get a 2nd order DE, which you then solve and apply the initial conditions to.

3. Apr 14, 2010

phufool

Thanks for the reply! And yes I think you're right, it is probably 1/2 V. So then the differential equation would look like: Vc(t) + VL(t) + Vo(t) = 1/2 .
Vo(t) = Ri(t)
VL(t) = Ldi(t)/dt
Vc(t) = 1/C integral from -infinity to t of i(a)da
So if I plug these into the equation and take the derivative, would I get:
1/C*i(t) + Ld^2i(t)/dt^t + Rdi(t)/dt = 0? So the 1/2 would just be irrelevant?

So for these C, L, and R variables, would I just substitute these for the numbers given in the problem? Say C = 1/2, L= 1/4, and R = 1?

PS: Is there a program/website I can use to make these equations look nicer? I'm sure what I'm typing must be hard to read lol.

4. Apr 15, 2010

phufool

So I was hoping if someone could still help me with this equation?
From the above equation: 1/C*i(t) + Ld^2i(t)/dt^t + Rdi(t)/dt = 0?
Since i(t) = 1/R * Vo(t), I would substitue i(t) and the values of C, L and R and get:
1/4*d^2Vo(t)/dt + dVo(t)/dt + 2Vo(t) = 0
So I assume Vo(t) = Ae^st and get the roots: s = -2+-2i
Can anyone tell me what to do next or if I'm doing this correctly? Thanks

5. Apr 16, 2010

Staff: Mentor

The response will be a damped sinusoid, so your exponential term should have both sigma and j*omege in it. Something like:

$$V_o(t) = A e^{B(\sigma + j\omega)}$$

Use that, differentiate, substitute back, and apply initial conditions to solve for the constants.

6. Apr 16, 2010

phufool

Thanks so much! Now the trouble I'm having is knowing what are the initial conditions. I can't seem to understand it based on the circuit given. Do you think you could help explain it to me?

7. Apr 16, 2010

Staff: Mentor

Glad that helped. BTW, the coefficient B may just be 1, but I'm not sure. You should be able to tell as you apply the ICs.

At time t=0-, the switch is open, so there is zero current, and zero voltage across the elements to the right of the switch. At t=0, the switch is closed, so all of a sudden you have the supply voltage across the series RLC combination. That will let the current start to build, and the voltage division between those 3 components will start to change with time...