I have this question and I believe the answer is 12.73 from the equation

V(N) = V(Rated) * sqrt(2).

Is this the correct formula and can someone show me how we arrive at it if it is. I am unsure about the derivation so I am unsure about whether the a change in the voltage across the diode will affect the no load voltage output. For example say if the forward voltage of the diodes was 0.9V or 1.1V?

Question 1


Your formula is ignoring the voltage drop from the diodes (given as 0.7V each). Otherwise it it is correct. To derive the sqrt(2) part you need to know that transformer output voltages are usually given as Vrms (volts root-mean-square). The problem doesn't explicitly say that but it is implied. You should be able to search that term and find a good description of the math involved.

Once you have the correct equation you will see that changing the diode voltage will change the output voltage. If the voltage drop was increased to 1.1V you would lose a total of 2.2V in the bridge rectifier, since the current always has to flow through two diodes.

  • \$\begingroup\$ So if I get what you are saying it should be (V(rated) - 2 * V(doide)) * sqrt(2). In this case that will give you 10.74 --> Answer C \$\endgroup\$ Apr 13 '15 at 20:12
  • \$\begingroup\$ Not quite, the transformer produces a peak voltage of V(rated)*sqrt(2). Then that is dropped 2)V(diode). \$\endgroup\$
    – Austin
    Apr 13 '15 at 20:13

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