[Noisebridge-discuss] Homework Help?

[Mitch Altman]

Hi Daniel,

My email mangled your circuit (which must have taken awhile for you to draw in ASCII -- well done!).  Can you send it as an attachment?

Since I can't see the diagram, I can guess what it is from your math.  It seems right (from my guess of your diagram).  But one thing you might not have taken into account is that the diode has a voltage drop across it, usually 0.7V.

Best,
Mitch.


----------------------------------------
> Date: Mon, 28 Jan 2013 09:01:13 -0800
> From: Daniel at coloraura.com
> To: noisebridge-discuss at lists.noisebridge.net
> Subject: [Noisebridge-discuss] Homework Help?
>
> Hey Circuit Hackers.
>
> So, I'm taking a Linear Circuits class (at Diablo Valley College in
> Pleasant Hill), but due to scheduling constraints I skipped the "AC"
> electronics course. I have one particular homework problem that I think
> I've gotten correct, but I'd like to have someone with more experience
> confirm. Unfortunately most of my classmates finish the homework on the
> due-date, so I can't compare notes.
>
> The problem as stated:
>
> Find the peak and average power delivered to Rl in the figure:(view
> fixed-width font)
>
>
> .--o^-._-^-* -. 2:1 .---->|---o--.
> | Fuse ^ | | Diode^ |
> o S* || *S \
> S || S/
> 120V rms S || S\ 220 ohm
> S || S/
> o | |\
> | | | |
> `-------------. .------.---o--.
> |
> -----
> ---
> -
>
>
> My approach to solving this is:
>
> 120v rms becomes 60v rms on the secondary of the transformer. Assuming a
> sine waveform, the peak voltage is 60v*sqrt(2), approx 85v.
>
> Using the "ideal" model, the diode only conducts one direction, which
> cuts the Vrms in half.
> Pavg = 30v^2/220ohm ~= 4watts
> Ppeak = 85v^2/220ohm ~= 33watts
>
> Does this look correct?
>
> Thanks,
> Daniel.
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