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Thank you Sir Case1 :for the postive cycle and then the diode open and i =0A vi=v-=v+ So vi-vi/finite=i10k=0A Thus vo+(10k*0)=vi ...answerCase2 : Negativ cicyle ,diode is short and v-=0v So i10k=vi-0/10k=vi/10k, vo+(vi/10k)*10k=0,vo=-vi....answer
Thank you Sir
Case1 :for the postive cycle and then the diode open and i =0A
vi=v-=v+
So vi-vi/finite=i10k=0A
Thus vo+(10k*0)=vi ...answer
Case2 : Negativ cicyle ,diode is short and v-=0v
So i10k=vi-0/10k=vi/10k, vo+(vi/10k)*10k=0,vo=-vi....answer