InterviewSolution
Saved Bookmarks
| 1. |
A fullwave P.N diode rectifier used load resistor of `1500 Omega`. No filter is used. Assume each diode to have idealized characteristic with `R_(f) = 10 Omega` and `R_(r) = ac`. Since wave voltage applied to each diode has amplitude of 30 volts and frequency 50 Hz. Calculate. (i) Peak, d.c rms load current (ii) d.c power input (iii) A.C power input (iv) Rectifier efficiency. |
|
Answer» (i) Peak current `I_(m)=(V_(m))/(R_(f)+R_(L)):. I_(m)=(30"volts")/(10+1500)=19.9mA` d.c load current `I_(dc)=(2I_(m))/(pi)=0.636 I_(m)=0.636xx19.9mA=12.66mA` `I_("rms")=(I_(m))/(sqrt(2))=(19.9)/(sqrt(2))=14 mA` (ii) DC Power output `P_(dc)=I_(dc)^(2)xxR_(L)=(12.66xx10^(-3))^(2)xx1500 "Watt" = 240.41 mW` (iii) AC power input `P_("in")=I_("rms")^(2)(R_(f)+R_(2))=(14xx10^(-3))^(2)(10+1500)"Walt" = 295.96mW` |
|