NCEES Practice Problem 58

[giginubs]

Hi everyone,

I was wondering if anyone can explain the answer for question 58 from the official NCEES Practice Exam.

The problem and answer is:



Thank you!

 

[RedRaider2020]

You find your Max AC Voltage which is equal to the DC voltage for 3 ph full wave rectifiers.
Because of the battery polarity it's seen as a -60 volts.
Then you use I=V/R. You know you're looking for the resistor at 10 amps
you have to add the batteries 25 ohm resistance.
The equation becomes IR=V because you're trying to solve for R.
.25ohms * 10 AMPs = 25V which you can subtract from the voltage.
R=V/I so divide 209.2V/10A

 

[giginubs]

You find your Max AC Voltage which is equal to the DC voltage for 3 ph full wave rectifiers.
Because of the battery polarity it's seen as a -60 volts.
Then you use I=V/R. You know you're looking for the resistor at 10 amps
you have to add the batteries 25 ohm resistance.
The equation becomes IR=V because you're trying to solve for R.
.25ohms * 10 AMPs = 25V which you can subtract from the voltage.
R=V/I so divide 209.2V/10A

The Max AC voltage would be sqrt(2)* V_rms and that we need that value due to the diodes for the full-wave rectifier?

 

[RedRaider2020]

Yea, you have to represent the sine wave as a function of time so you'll have to convert to the Peak (Max) value for most of the power supplies and converter questions. For most problems you'll need to find the Average voltage Vo. For power you have to convert the average value to the RMS value which is the DC equivalent value. Intuitively if you have a lot of ripple voltage you'll need to do more voltage conversions to solve problems because the wave will be much different than a solid DC signal that's just a straight line.

 

[giginubs]

Thank you for the clarification.

 

[e152645]

Yea, you have to represent the sine wave as a function of time so you'll have to convert to the Peak (Max) value for most of the power supplies and converter questions. For most problems you'll need to find the Average voltage Vo. For power you have to convert the average value to the RMS value which is the DC equivalent value. Intuitively if you have a lot of ripple voltage you'll need to do more voltage conversions to solve problems because the wave will be much different than a solid DC signal that's just a straight line.
View attachment 20994

Why we are not using 3 phase full wave rectifier average output voltage formula? Which is 0.955*Vm,L-L as per handbook. I got output voltage of 0.955*208*sqrt(2)= 280.1 V. As a result resistance value should be 19.6 ohm. Is multiplying by sqrt(2) rms input just an estimation of formula provided by handbook?

 

[xnuke]

The question asks for the value of the current-limiting resistor to limit the peak charging current to 10 A, so you use the peak voltage that comes out of the rectifier, not the average voltage.

 

[Marcocast6]

Why we are not using 3 phase full wave rectifier average output voltage formula? Which is 0.955*Vm,L-L as per handbook. I got output voltage of 0.955*208*sqrt(2)= 280.1 V. As a result resistance value should be 19.6 ohm. Is multiplying by sqrt(2) rms input just an estimation of formula provided by handbook?

This is what I got, why wouldnt we be able to use that formula, seems like the correct approach