Oh! I see! Well in that case, let's see... There's 2 X pi radians in a cycle, so basically ω = 6283.19... Would that be what I have to use in those forumlas?
Q = 2pi * 100/1 = 628.319
and R = 2Ω
We have Q = (ω0L) / R
628.319 = (6283.19 X L) / 2
solve for L like I previously did
And...
Isn't that what I did? The formula for frequency is the one for the LC circuit, and the information for the frequency is given for the LC circuit.
Edit: Oh wait, is it on the first formula that lies the problem, since I'm finding L in an RLC circuit with the frequency for an LC circuit?
If...
Ok well if I do that, this is what I get (I left out the units on here because it gets confusing and I don't know how to do it like you guys:
Q = 2pi * 100/1 = 628.319
and R = 2Ω
We have Q = (ω0L) / R
628.319 = (1000 X L) / 2
1256.638 = 1000L
L = 1.2566
And then
ω0 = 1 / √(LC)...
I don't think I have the proper tools to answer that question just yet, all of this is Chinese to me... The teacher didn't really cover it, he just skimmed over, guess I'll have to read the book. Even the concepts you mentioned, or what's in wikipedia, have not been covered in class (and...
I copy/pasted the whole problem, so it's not mentioned if it's series or parallel
I could find Q with the formula you gave me, if the system loses 1%, then it would be:
Q = 2pi X 100/1 = 2pi X 100
Attenuation factor, that's not even something we covered in class... Do you simply mean...
Homework Statement
The energy of the RLC circuit decreases by 1% during each oscillation when R=2 Ohms. If this resistance is removed, the resulting LC circuit oscillates at a frequency of 1 kHz. Find the inductance and the capacitance. Hint: be aware of the difference between regular and...
Awesome thanks, I just want to make sure I understand correctly though.
On one end, we have the current that decay from 24A to 0A with a sudden drop from 24A to 12A, thus giving an inversely proportional graph
On the other end, according to my book (I don't have the reflex to picture the...
Ok well the current decaying over time, I'm assuming it's non-linear, otherwise it would have been specified, so it's an inverse non-linear graph (I don't know it that is the right term, it's been a while since I studied graphs)
The formula that I have posted, wouldn't also yield a non-linear...
Ok, then the next step would be to find the time constant to get from 24A to 12A, here's how I would do it:
I = ε/R X (1 - e-t/time constant, isolate for time constant:
IR / ε = 1 - e-t/time constant
1 - (IR / ε) = e-t/time constant
ln (1 - (IR / ε) = -t / time constant
time constant = -t...
Should I assume it's an RL circuit then? Because if I do then I would get a completely different set of equation...
For R, would (assuming it's an RL circuit) the equation be:
I = (ε/R) (I being the initial current at t = 0) ==> R = ε/I = 0.125Ω.. which I just noticed is what I previously...
Homework Statement
To measure the inductance and resistance of a real inductor, a physicist first connects the inductor to a 3V battery. At these conditions, the final, steady, current is 24 A. Then the physicist suddenly short-circuits the inductor with a thick (resistance-less) wire place...