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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: efficiency of rectification

Answer»

SOLUTION :v=50 sin `omegat`THEREFORE` MAXIMUM VOLTAGE, `V_m=5V` efficiency of RECTIFICATION `0.301/0.763xx100=39.5%`


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