1.

In the diode-based rectifier circuit given below, if V_(s) =V_(m)sin omega t and the diode is ideal, then the average value of V_(L) is

Answer»

`(R_(L))/((R_(L) + R_(S))) (V_(m))/(PI)`
`R _(L) V_(m) SIN omega t `
`(R_(L))/((R_(L) + R_(S)))V _(m)`
`(R_(L))/((R_(L) + R_(S)))V _(m)sin omegat`

Solution :Given, AC voltage, `V_(s) =V_(m)sin omega t and V_(m)=` maximum value of voltage In the given circuit diode will only conduct current in forward bias, so output across `R_(I),` will be

Average voltage,` V _(av) = (V_(m))/(pi)`
average current `I _(av) = (V_(av))/(R)`
Toal resistance, `R = R_(s) + R_(L)`
`L _(av )=(V_(m))/((R_(S) + R_(L))pi)`
Now, voltage across `R_(L),`
`V_(L) =I_(av) . R_(L)`
`V_(L) = (R_(L))/(R_(S) + R _(L)). (V_(m))/(pi)`


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