A 240V (rms) 60 Hz mains supply is fed to a trans which has 20:1 ratio. The output v from the 2nd winding of the transformer is connected to a full-wave rectifier circuit and a 10kΩ R. Assuming ideal diodes, what is the peak voltage and the peak diode I


240 x sq root of 2 = 339

339 x 1/20 = 17 for peak voltage

Peak voltage / resistor, will give diode current

  • 1
    \$\begingroup\$ You haven't calculated the rectified voltage or diode current at all, much less correctly. If you want help with a homework problem we expect you to show a significant amount of effort and ask a specific question. In this case you should also use the built-in schematic editor to draw your entire circuit. \$\endgroup\$ – Elliot Alderson Nov 6 '19 at 13:26
  • \$\begingroup\$ @ElliotAlderson Sorry but i did not mean to come across as low effort attempt, because I legitimately think this is the way to do this question right. rms by sq of 2 to get primary voltage peak voltage, then for secondary voltage its 339 x n2/n1 and then divide that value by the resistor to get the secondary current. \$\endgroup\$ – Nexus_Valentine Nov 6 '19 at 13:38

Assuming you mean this:


simulate this circuit – Schematic created using CircuitLab

Then you would first calculate the AC output of the transformer.

Since you have a 20 to 1 turns ratio, that'd be the input voltage divided by 20. Use the RMS value of the input voltage. In your example, that's 240VAC. Dividing by the turns ratio gives you the transformer output voltage as RMS.

Once you have the RMS voltage on the low voltage side of the transformer, you can figure out the peak voltage from the rectifier:

  1. AC peak to peak: \$V_{pp} = V_{RMS} \times 2\sqrt{2}\$
  2. Rectified peak: \$V_R = V_{pp}/2 - 2V_{f}\$
  3. Peak current: \$I_P = V_R / R\$

Where \$V_f\$ is the forward voltage of one diode in your bridge rectifier, and \$I_P\$ is the peak current through \$R1\$.

Since you say "assume ideal diodes," \$V_f\$ would be zero.

Pretty sure that's right. If not, someone please let me know and I'll correct it and Nexus_Valentine and I will both have learned something.

| improve this answer | |
  • \$\begingroup\$ Thanks! So the input voltage is 240 x sq root of 2? \$\endgroup\$ – Nexus_Valentine Nov 6 '19 at 14:05
  • \$\begingroup\$ The input voltage is just the RMS voltage. \$\endgroup\$ – JRE Nov 6 '19 at 14:11
  • \$\begingroup\$ @Nexus_Valentine if you mean the input peak voltage (at the primary), yes it is correct \$\endgroup\$ – Wheatley Nov 6 '19 at 14:12
  • \$\begingroup\$ @JRE , thanks for your help , I am getting the same values from your method and from original post method, so it must be right! \$\endgroup\$ – Nexus_Valentine Nov 6 '19 at 14:16

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