Recent content by dfs730

ωo = 1/(LC)^1/2 -> L = ((1/ωo)^2)/C Q = (1/R)(L/C)^(1/2) -> L = C(QR)^2 -> C=1/QRωo = 7.96x10^(-9) Farads -> L=Q^2(R^2)(C) = 3.184 I think this looks right! Thanks a bunch, really appreciate it!

oh, so Qo= 100

1. You have a series circuit consisting of a ac power supply, a 600 ohm resistor and a 73 nF capacitor. If the circuit is configured as a low pass filter, what frequency will cause the gain to be 0.25? 2. If the circuit is configured as a high pass filter, what frequency will cause the gain...

Q= 2pif x (energy stored / energy dissipated per cycle) = ωo(energy stored / energy dissipated per cycle) = ωo(0.01) ?

So ωo = 2pif = (2pi)1.0 khz and the resistance would have something to do with Q?

w=1/(LC)^(1/2) f=w/2pi = 1/(2pi(LC)^(1/2)) = 1.00kHz some how this is supposed to relate to the equation for a damped object on a spring. L(d^2Q/dt^2) + R(dQ/dt) + Q/C = 0 <---> m(d^2x/dt^2) + b(dx/dt) + kx = 0 Other than this I really have no idea...

The energy of an RLC circuit decreases by 1.00% during each oscillation when R=2.00 ohms. If this resistance is removed, the resulting LC circuit oscillates at a frequency of 1.00 kHz. Find the values of inductance and capacitance.

Ahh...yes. I had put in Vr-Vd. Vd + Vr worked, I got 8.99V/ Thanks!

I'm not sure how to get the second part: What is the voltage across the diode-resistor combination? Any ideas?

You are doing it right, I had the same question but different values: 120 ohm resistor, current in the circuit (I) = 73mA, saturation current (Io) = 0.010mA, and T=300K. My answer came out to be 0.230V which was accepted as correct. Just try it again and make sure all your calculations are correct.