1.

A `p-n` diode is used in a half wave rectifier with a load resistance of `1000 Omega`. If the forward resistance `(r_(f))` of diode is `10 Omega`, calculate the efficiency of this half wave rectifier.

Answer» Load resistance `R_(L) = 1000 Omega`
Forward resistance of the diode `= r_(f) = 10 Omega`
Efficiency of half wave rectifier
`[(0.406 R_(L))/(r_(f)+R_(L))] =(0.406 xx 1000)/(1010) = 0.4019`
The percentage efficiency of the half wave rectifier `eta = 40.19 %`.


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