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A crystal diode having internal resistance r_f =20Omega is used for half-wave rectification. if the applied voltage v = 50 sin omegat and load resistance R_L = 800Omega find: I_m,I_(dc), I_(rms)

Answer»

SOLUTION :`v=50 SINOMEGAT therefore` maximum VOLTAGE, `V_m=5V I=V_m/(r_f+R_L)=50/(20+800)=0.061mA I_(dc)=I_m//pi=61//pi=19.4mAI_(rms)=I_m//2=61//2=30.5mA`


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