Recent content by Aigik

YS = 0.01 - j0.01 or: YL = 0.01 + j0.01 So I have two equations for the admittance of the load: YL = 0.01 + j0.01 YL = 1/RL + j377C If I set these equal to each other, though, I still have two unknowns. Wait: could I set 0.01 = 1/RL and 0.01 = 377C?

I'm not quite sure I'm following. I know that to achieve maximum power transfer, we need to set the impedance of the load to the complex conjugate of the source (or vice versa). I have the admittance for the load: YL = 1/RL + j377C ...whose complex conjugate is YS = 1/RL - j377C I also...

The net admittance of the load would be: YL = 1/RL + jωC Or, since we know ω: YL = 1/RL + j377C

For the capacitor: Ycapacitor = jωC (since Y = 1/Z) For the resistor: Yresistor = 1/R Is this what you were looking for?

Thanks for replying. Well, I hadn't considered trying that before. Mainly because I'm more familiar with impedance. So going down that route, the admittance formula is: Y = (R - jX) / (R2 + X2) After plugging in the given resistance and inductance values, I can determine the...

Homework Statement For the following circuit, find the values of RL and CL that maximize the power extracted from the generator. https://dl.dropboxusercontent.com/u/15390227/circuit.jpg Homework Equations Z = R + jX XL = jωL XC = 1/jωC The Attempt at a Solution Immediately...

The angle of 75 degrees is with respect to the horizontal.

Do you by any chance have professor Brown at NIU as your instructor? Working on this problem set too, and I'm stumped by the last problem (the one where it asks for the smallest angle she can push before the block falls), and stumbled upon this. That's definitely troubling that the problem is...