GR9677 #45



Alternate Solutions 
casseverhart13 20190924 06:06:16  I must say, what a problem. Orlando Towing Partners  IDrive   flyboy621 20101024 16:07:24  We can solve this without knowing anything about AC circuits.
First, you can eliminate the circuits with diodes, since neither would behave differently at different AC frequencies (why would they?). That rules out (B) and (C).
To narrow it down further, analyze the lowfrequency limit, by assuming DC voltage (lowest frequency you can get, right?). We are given that at low frequency.
(A) wouldn't work because the voltage across the capacitor, the resistor, and must each equal .
(D) wouldn't work for a similar reason.
(E) is the only choice left. It looks like it would work because the capacitor would charge until the voltage drop across it equals , and then there would be no more voltage drop across the rest of the circuit.
We don't need to know the highfrequency behavior of any of these to know that (E) is right.   Blake7 20070923 05:50:32  Folks, whenever you see the symbol for a cap, two parallel, UNCONNECTED lines, you should immediately think "DC 'Block'"; a "short". And, conversely, with a cap being a DC block, it is also an AC, Hi Freq "Pass". Go back and study Max's eqns for caps if you don't get this, yet.
So, by inspection, the high freqs in circuit (E) just ripple along through the cap and show up "ACROSS" the (output) resistor.
In (A) and (D), the hi freqs "shunt" down through the cap; they are "low" "pass". (B) and (C) have no caps and therefore no frequency filtering performance compared to the rest.
Onceuponatime, all the EEs to be looked to us for this stuff! (They still DO in reality, so let's not let them down!)
  nitin 20061030 01:51:34  This question can be answered simply as follows:
In (E), the capacitor and resistor are connected in series. When driving frequency, f_d (equivalently angular frequence w_d), of V_{in} is high, the capacitive impedance X_C is small, so no voltage is dropped across the capacitance, and V_{out}=V_{in}. On the other hand, when f_d is low, X_C is high, and X_C>>R, so that almost all voltage is dropped across the capacitor. Hence, V_{out}\approx 0.
This is not the case for the other circuit arrangements.  

Comments 
casseverhart13 20190924 06:06:16  I must say, what a problem. Orlando Towing Partners  IDrive   flyboy621 20101024 16:07:24  We can solve this without knowing anything about AC circuits.
First, you can eliminate the circuits with diodes, since neither would behave differently at different AC frequencies (why would they?). That rules out (B) and (C).
To narrow it down further, analyze the lowfrequency limit, by assuming DC voltage (lowest frequency you can get, right?). We are given that at low frequency.
(A) wouldn't work because the voltage across the capacitor, the resistor, and must each equal .
(D) wouldn't work for a similar reason.
(E) is the only choice left. It looks like it would work because the capacitor would charge until the voltage drop across it equals , and then there would be no more voltage drop across the rest of the circuit.
We don't need to know the highfrequency behavior of any of these to know that (E) is right.   antithesis 20071002 22:46:44  A shortcut for this (and other filter problems) is to remember the following mnemonic: For RC, in a configuration such as that of figures D and E, the position of the capacitor determines the type of filter: If it it "up" (as in E), it is a high filter. If it is "low" (as in D), it is a low filter.
Then, remember that if you have LR, inductor is roughly opposite to a capacitor, so up means low filter, and vice versa.
I usually don't like to memorize useless stuff (I never took a real circuit course, with impudence), but this seems almost too trivial to remember.   Blake7 20070923 05:50:32  Folks, whenever you see the symbol for a cap, two parallel, UNCONNECTED lines, you should immediately think "DC 'Block'"; a "short". And, conversely, with a cap being a DC block, it is also an AC, Hi Freq "Pass". Go back and study Max's eqns for caps if you don't get this, yet.
So, by inspection, the high freqs in circuit (E) just ripple along through the cap and show up "ACROSS" the (output) resistor.
In (A) and (D), the hi freqs "shunt" down through the cap; they are "low" "pass". (B) and (C) have no caps and therefore no frequency filtering performance compared to the rest.
Onceuponatime, all the EEs to be looked to us for this stuff! (They still DO in reality, so let's not let them down!)
Blake7 20070924 02:42:57 
Ooooops!!!
A DC Block is an OPEN! A cap looks like an OPEN to DC! (not a short!) Pardon me!
At this point the physical intuition means more to me than the terms themselves.
Also, as far as the cap and the circuit are concerned, lower frequencies look closer to DC and therefore the GREATER their IMPEDANCE.
(Wow, Yosun, it might be nice if folks could edit their own comments.)

Blake7 20070924 02:44:21 
Ooooops!!!
A DC Block is an OPEN! A cap looks like an OPEN to DC! (not a short!) Pardon me!
(At this point the physical intuition means more to me than the terms themselves.)
Also, as far as the cap and the circuit are concerned, lower frequencies look closer to DC and therefore the GREATER their IMPEDANCE.
(Wow, Yosun, it might be nice if folks could edit their own comments.)

  nitin 20061030 01:51:34  This question can be answered simply as follows:
In (E), the capacitor and resistor are connected in series. When driving frequency, f_d (equivalently angular frequence w_d), of V_{in} is high, the capacitive impedance X_C is small, so no voltage is dropped across the capacitance, and V_{out}=V_{in}. On the other hand, when f_d is low, X_C is high, and X_C>>R, so that almost all voltage is dropped across the capacitor. Hence, V_{out}\approx 0.
This is not the case for the other circuit arrangements.   yosun 20051111 13:41:15  keflavich: the parsererror has been manually fixed, and i'd hopefully get to look at the code later this week to permanently fix it. (apparently, the symbol V_{in} confused it.)
high freq filters (as with low freq filters) are pretty easy. once u find the impedance, just take the right approximation in the particular regime you're interested in... and you're done. see above.   keflavich 20051111 12:02:11  Just a heads up: there appear to be a lot of HTML tags in your equations.... unless those 's are supposed to be there.
I don't know a thing about highpass filters, but since an 'empty' (uncharged) capacitor lets all the current through and a full one lets none, E is the option that allows highfrequency current through but not low frequency. That's probably what you said, give or take, but I couldn't read it too well.  

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This question can be answered simply as follows:
In (E), the capacitor and resistor are connected in series. When driving frequency, f_d (equivalently angular frequence w_d), of V_{in} is high, the capacitive impedance X_C is small, so no voltage is dropped across the capacitance, and V_{out}=V_{in}. On the other hand, when f_d is low, X_C is high, and X_C>>R, so that almost all voltage is dropped across the capacitor. Hence, V_{out}\approx 0.
This is not the case for the other circuit arrangements.

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