1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

AC to DC

  1. Jul 20, 2014 #1

    Zondrina

    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data

    A single phase half wave rectifier produces a DC signal ##V_{DC} = (4.6 \pm 0.78) V## and AC signal ##V_{AC} = (0.22 \pm 0.036) V##. What is the error on the efficiency?

    2. Relevant equations

    Functional dependence, propagation of error:

    ##\sigma_f = \sqrt{ (\frac{\partial f}{\partial x})^2 \sigma_x^2 + (\frac{\partial f}{\partial y})^2 \sigma_y^2 + ... }##

    Efficiency:

    ##\eta = \frac{0.405}{1 + \frac{R_{Gen}}{R_{Load}}}##

    3. The attempt at a solution

    I'm actually not quite sure where to start this. Does the efficiency have a functional dependence on the AC and DC voltage somehow?

    ##V_{DC} = I_{DC}R_{Load}##
    ##V_{AC} = I_{AC}R_{Gen}##
     
  2. jcsd
  3. Jul 22, 2014 #2

    Zondrina

    User Avatar
    Homework Helper

    Hey guys, turns out there was a typo in the question. The error on the quality factor is what was required, not the efficiency.

    The quality factor is given by: ##Q = \frac{V_{DC} - V_{Ripple}}{V_{DC}}##

    Applying error propagation:

    ##\sigma_Q = \sqrt{ (\frac{\partial Q}{\partial V_{DC}})^2 \sigma_{V_{DC}}^2 + (\frac{\partial Q}{\partial V_{Ripple}})^2 \sigma_{V_{Ripple}}^2 }##

    This gives:

    ##\sigma_Q = \sqrt{ (\frac{V_{AC}}{V_{DC}^2})^2 \sigma_{V_{DC}}^2 + (\frac{1}{V_{DC}})^2 \sigma_{V_{AC}}^2 }##

    Which yields the correct answer after subbing everything in.

    Sorry for the confusion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: AC to DC
Loading...