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I was given a homework assignment shown below and I have a few quick questions that I appear to be stuck on. Any help would be appreciated.


enter image description here

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I can get Part i) and the answer is 168.64 degrees.

Part ii) I can't seem to get correct. My professor gave us the equation:

$$\displaystyle V_{L_{rms}} = \frac{1}{\pi} \int_{Q_{1}}^{Q_{2}} (V_p\sin(\theta)-1.4)^2 d\theta$$

with Q1&2 being the two points at the which diode conducts (5.68 and 174.32 degrees respectively). When I integrate this is, I get huge numbers like 249Vrms which doesn't seem be correct. Does anyone have any guidance on what I could possibly be doing wrong?

Thanks

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    \$\begingroup\$ Are you correctly converting θ to radians where required? \$\endgroup\$
    – Dave Tweed
    Dec 9, 2019 at 16:04
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    \$\begingroup\$ I was just about to comment on the same thing Dave just said. You seem to have \$\pi\$ in your integral formula but you're also talking about degrees and not radians. \$\endgroup\$
    – user103380
    Dec 9, 2019 at 16:06
  • \$\begingroup\$ Yes I tried with both radians and degrees, I edited the post to add my workings, Thanks to both of ye in advance. \$\endgroup\$
    – S_G
    Dec 9, 2019 at 16:36

1 Answer 1

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Calculations I have managed to get a bit closer to the correct answer. Is VL,rms in the equation meant to be squared? If so that would then mean VL,rms would be 8.49V based on my calculations

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  • \$\begingroup\$ it is meant to be squared yeah, but you take the square root after it is basically the absolute value of the mean of the voltage. Your final answer(8.49V) does imply you are doing that so I think you are fine \$\endgroup\$
    – Juan
    Dec 10, 2019 at 5:39

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