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Quick quiz

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allanw:
Just did poorly on a technical phone interview for a certain graphics card company. The recruiter told me that a lot of people also have trouble with this seemingly simple problem:

Qualitatively, what does the voltage at the resistor look like for this input assuming no initial charge on the cap?



I was pretty caught off guard because I thought they'd ask me questions about embedded systems, which is what I was applying for... also the fact that input went back down got me a little bit confused and I couldn't do it. Of course, I know the answer now... I'd like to see what some other people's reasonings are for this circuit though.

And no cheating with spice! :)

DJPhil:
After all the book studying I've been doing the last six months, I feel like I should be able to nail this, but I'm not positive about my answer.

After cheating with spice because I couldn't stand it: Ok, wow. I see what you mean.  :o

I was close, but for mostly the wrong reasons. I had the charge waveform right, but the discharge reversed in time because symmetry in my head became somehow more important than reality. Seeing it in spice makes what I was just guessing look like gibberish, which it was. I'm humbled.  :-[

EEVblog:
Ah yes, an old trap for young players.
Remember that the cap holds charge that doesn't go away instantly (unless you short it).

It might help to think of it like this. Once the cap is charged, picture taking it out of the circuit (still with it's charge), and putting it into another circuit with the new conditions (i.e. the battery replaced by a link).

Dave.

Zero999:
I think the thing that will confuse most nubes is where the negative voltage comes from. Dave is right, here's another clue: when the capacitor is charged and its positive is connected to negative, what will be the voltage across the resistor?

Of course it's a trick question anyway: the circuit never reaches steady state because the capacitor never charges to 100%.

EEVblog:

--- Quote from: Hero999 on September 24, 2010, 07:36:05 am ---Of course it's a trick question anyway: the circuit never reaches steady state because the capacitor never charges to 100%.

--- End quote ---

That's the smart-arse theoretical answer, so = instant FAIL!  :P

Dave.

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