# Overdamped parallel RLC

1. May 12, 2006

### sevens

For some reason i cant get my A1 and A2 in the correct spots. I get the correct value of these two, but if someone could point out where i've gone wrong, i would apreciate it.

V(0) = 125v
i(0) = 0A

value of the inductor is 1.25h
value of the resistor is 2 Ohms
value of the capacitor is 50mf ( yes i know huge... but :grumpy: its theory, not practice)

the final answer is to find the voltage accross the capacitor

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2. May 12, 2006

### doodle

3. May 12, 2006

### Staff: Mentor

Why do you say $$\omega_c = \frac{1}{2} Cv^2$$ ?

What are the units of this $$\omega$$ ?

Last edited: May 12, 2006
4. May 12, 2006

### sevens

The units of W are Jouls, its energy stored in the capacitor, and i use that formula to find the initial voltage in the capacitor. The answer i got is V(t) and then under that i wrote what the answer should be "actual answer" i keep getting A1 and A2 in the wrong place for several of my circuit analysis. Am i missing some tiny detail.?

5. May 12, 2006

### Staff: Mentor

Oh, you meant W like work or energy, not $$\omega$$ like angular frequency. I get that now.

But you're given v(0) = 250V as part of the problem statement. And the "final" voltage across the capacitor will be zero. What exactly does the question ask for? It wants the capacitor voltage when?

6. May 12, 2006

### sevens

constants in wrong spot

Yes its true that final voltage across the capacitor will be zero given enough time. The question wants an equation that can be defined in terms of t. So if you take the found equation for example t =2 sec the capacitor will be zero. however if i take t=.0001s there will be some voltage left in the capacitor.

Originaly i was told how much energy was stored in the capacitor which is why i used Wc to find initial Voltage.

overal the problem is that my A1 and my A2 are with the wrong positions. they are the correct values themselves but if you see where i wrote "actual answer" thats the position they should be in.

Thanks for the help
this problem will crack sooner or later

Last edited: May 12, 2006
7. May 15, 2006

### SGT

How did you get $$i_R(0) = 0$$?
$$i_R(0) = \frac{V_C(0)}{R}$$.

8. May 15, 2006

### Staff: Mentor

He just said i(0)=0. I think it's just one of the initial conditions, like if you were closing a switch at t=0 to connect the cap into the rest of the circuit.

9. May 16, 2006

### SGT

Since the initial energy in the inductor is 0, it follows $$i_L(0) = 0$$.
The resistor is in parallel with the capacitor, wich has an initial voltage of 125 V. So, the initial current in the resistor is $$\frac{125}{R}A$$.

10. May 16, 2006

### Staff: Mentor

You've worked with initial conditions in differential equations before, right? As usually stated, the current in the resistor would be zero at t=0-, and the value you state at t=0+. Standard stuff.

11. May 16, 2006

### doodle

Hmm... SGT is right. If Vc(0+) = 125V, then Ir(0+) = 125/2 = 62.5A since current across the resistor can change suddenly. This means that Ic(0+) = -Ir(0+) = -62.5A which then gives the actual answer.

12. May 17, 2006

### SGT

When you replace initial conditions in the general solution, you use the conditions at t = 0+. The conditions at t = 0- are irrelevant, except in the case of energy storage elements, like capacitors and inductors (or masses and sprigs). For those elements the conditions at t = 0- and t = 0+ are the same, except in the case of an impulsional excitation.

13. May 17, 2006

### acceler8

i hate it when experiments dont work the way they should

soooooooooo frustrating!!!!!!!!!!!!!!!!!!

xxxx Gareth