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GR0877 Problem 69

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) The impedance of an ideal resistor is Z_R = R, and the impedance of an ideal capacitor is \displaystyle Z_C = \frac{1}{i \omega C}, where i is imaginary unit. The total impedance of the circuit is \displaystyle Z = Z_R+Z_C = R + \frac{1}{i \omega C}. The total current is \displaystyle I = V_i/Z = \frac{V_i}{R + \frac{1}{i \omega C}}. The amplitude V_o of the output voltage is \displaystyle V_o = IZ_C = \frac{V_i}{\left(R+ \frac{1}{i \omega C} \right) i \omega C} = \frac{V_i}{i RC \omega +1}. From this one has \displaystyle G = \frac{V_o}{V_i} = \frac{1}{i RC \omega +1}. Therefore, when \omega \to \infty, G \to 0 and when \omega \to 0, G \to 1. The only graph with such a behavior of G is (D).

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  1. Steve
    10.11.2011 at 03:40

    Quick way to get this one: the circuit is a low-pass filter, so for low w, Vout=Vin, and Vout goes to zero as w goes to infinity. Only (D) shows this behavior

  2. Stevie
    16.10.2013 at 17:37

    Exactly. From doing all the practice tests, I can tell you that high and low pass filters comes up a lot. Look at question 39 on the 2001 test for another problem and a great explanation here: http://grephysics.net/ans/0177/39 . In particular, look at the comments for that problem on that site. They helped me a ton.

  3. Kevin Tah
    22.10.2015 at 00:39

    Exactly!! After reading it I just figured it is low pass or high pass lol

  4. Xing
    23.10.2016 at 16:52

    Didn’t any one notice that the equivalent resistance of tandem inductance and capacitor is \omega L – 1/(\omega C)? so the magnitude of V_0 would be 1/(\omega^2 LC-1), with the increase of \omega, V_0 would be infinite when \omega equals \sqrt(1/LC), but the figures of the problem did not show this feature.

  5. Xing
    24.10.2016 at 16:35

    OK, I was wrong, I mistook resistor for inductance.

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