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:a.c. power and d.c. power output

A crystal diode having internal resistance r_f =20Omega is used for ha

1.	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:a.c. power and d.c. power output
Answer» SOLUTION :v=50 sin `OMEGAT`therefore` maximum voltage, `V_m=5V`a.c.power input`=(I_(RMS))^2XX(r_f+R_L)=(30.5/1000)xx(20+800)=0.763`watt d.c.power OUTPUT `I_(dc)^2xxR_L=(19.4/1000)^2xx800=0.301"watt"`

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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:a.c. power and d.c. power output

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