I'm trying to do a problem for class and I'm having trouble on clarification and where to start.
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I understand how the turns ratio works, but I don't know how to find it here. I can't find anything in my book or in handouts; I only find them when there are resistors on both sides.

I'm also confused on what VL and Vp are. I think Vp is the peak of the whole wave but I don't know what VL is showing here. Vrpp is also confusing me. I just need some clarification and guidance.


  • \$\begingroup\$ \$V_L\$ is the load voltage. \$V_p\$ and \$V_s\$ are the primary and secondary voltages. \$V_{r\ PP}\$ is the peak-to-peak voltage on \$V_L\$. \$\endgroup\$ Dec 12, 2017 at 18:53
  • \$\begingroup\$ Ohhh, that makes more sense. So since the voltage across the load is 10, the turns ratio would then be 1:11? Or would I have to take into account the .7 per diode used then use that voltage? \$\endgroup\$
    – John
    Dec 12, 2017 at 18:59
  • \$\begingroup\$ You need to take diode drops and ripple into account. They made it a bit more difficult by specifying RMS output voltage rather than average. \$\endgroup\$ Dec 12, 2017 at 20:03

1 Answer 1


Since Vp is given as 110 Vrms, it must be the RMS voltage of the power source, not the peak of anything. VL is given as 10 V rms, so it must be the RMS value of the voltage at that point including Vr(pp), the peak to peak ripple voltage. You mostly need to work backwards from VL.

  • \$\begingroup\$ So how would I go about working backwards, would I have to take the load value and bring it back through the two diodes and take that value to find the secondary peak and turns ratio? \$\endgroup\$
    – John
    Dec 12, 2017 at 19:15
  • \$\begingroup\$ First find the rectified but unfiltered peak value and filter capacitor value that will give the filtered VL voltage with the specified ripple. \$\endgroup\$
    – user80875
    Dec 12, 2017 at 19:35

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