1.

An A.C. supply of 350 V, 60 Hz is applied to a full wave rectifier. If internal resistance of each diode is 200Omega and load resistance R_(1)=5k Omega, then the maximum value of output current is …….

Answer»

0.065 A
0.092 A
0.095 A
0.07 A

SOLUTION :0.092 A
`I_(m)=I_("rms")XX sqrt(2)`
`=(V_("rms")xx sqrt(2))/(R_(2)+2r_(f))`
where `r_(f )=` resistance of DIODE in forward bias
`=(350xx1.414)/(5000+2xx200)=0.0916A`
`therefore I_(m)~~0.092A`


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