A full wave rectifier uses two diodes, the internal resistance of each diode may be assumed constant at `30 Omega`. The transformer r.m.s. secondary voltage from centre tap to each end of the secondary is 50 V and load resistance is `970 Omega`. Find (i) the mean load current (ii) the rms value of load current.

Explanation

Here, `r=30 Omega, R=970 Omega, V_(rms)=50V`,
so `V_(0)=sqrt(2)V_(rms)=sqrt(2)xx50=70.7V`
Max. load current,
`I_(0)=V_(0)/(r+R)=70.7/((30+970))=70.7mA`
(i) Mean load current,
`I_(m)=(2I_(0))/pi=(2xx70.7)/(22//7)=45mA`
R.M.S. value of load current, `I_(rms)=I_(0)/sqrt(2)=70.7/sqrt(2)=50mA`