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We know that for a half wave rectifier \$V_{rms}=\frac{V_m}{2}\$ and \$I_{rms}=\frac{I_m}{2}\$

But the Transformer utilization factor uses \$V_{rms}=\frac{V_m}{\sqrt{2}}\$ (But uses \$I_{rms}=\frac{I_m}{2}\$) to calculate the "ac rated power" as \$\frac{V_m}{\sqrt{2}}*\frac{I_m}{2}\$. Why?

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    \$\begingroup\$ A diagram would be nice. \$\endgroup\$ – Transistor Jan 6 at 9:35
  • \$\begingroup\$ @Transistor For half wave rectifier? \$\endgroup\$ – paulplusx Jan 6 at 10:19
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I think I got it and it is quite silly. There are two \$V_{rms}\$ voltages, it can be considered as:

\$V_{rms(input)}\$ : The voltage that is measured before the diode i.e. at the input of the diode and it's the direct output of the transformer.

\$V_{rms(output)}\$ : The voltage that is measured after the diode.

Also, since it's a diode in series, the current before the diode is the same current that is after the diode hence there is only one \$I_{rms}\$

The AC rated power uses the \$V_{rms(input)}\$ which is quite obvious because that's the voltage across the secondary winding of the transformer.

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