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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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Categories: 0877 Tags: , , , ,
1. 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. 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. 22.10.2015 at 00:39

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

4. 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. 24.10.2016 at 16:35

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