# NCEES 515 vs. 518



## Rei (Apr 14, 2010)

Both questions involve with diodes but one used square root of 2 and the other used 2*square root of 2? Why?


----------



## Flyer_PE (Apr 14, 2010)

For #515, they are solving for the potential voltage across the diode. The capacitor will keep the voltage on the load side at the peak positive value for the voltage when the source voltage is at the negative peak. Potential difference is then 2*VPeak.

For #518, they are solving for the voltage on the load side of the rectifier.


----------



## tv20 (Jun 17, 2010)

Dear Flyer

Could you please a detailed explanation of these two problems? I always failed to understand the voltages across capacitors with a diode. Thank you in advance


----------



## ucla_7 (Aug 14, 2010)

For #518

Should we Convert Vrms (208V) to Vdc

and then subtract Vdc by the DC battery voltage ( 60V) ?

what is the formular for Vdc in term of Vrms after the fullwave rectifier ?

thanks


----------



## Flyer_PE (Aug 14, 2010)

^The voltage of concern for the rectifier is essentially the peak voltage of the AC waveform. With a 208Vrms voltage, Vp=sqrt(2)*Vrms = sqrt(2)*208 = 294 Volts Although it isn't a DC voltage, it can be treated as such for the purposes of this problem since the idea is to determine the required resistance to limit the peak charging current.


----------



## ucla_7 (Aug 14, 2010)

Flyer_PE said:


> ^The voltage of concern for the rectifier is essentially the peak voltage of the AC waveform. With a 208Vrms voltage, Vp=sqrt(2)*Vrms = sqrt(2)*208 = 294 Volts Although it isn't a DC voltage, it can be treated as such for the purposes of this problem since the idea is to determine the required resistance to limit the peak charging current.


Thank You for the information


----------

