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The applied input ac power to a half wav...

The applied input ac power to a half wave rectifier is 100 W. The dc output power obtained is 40W. Find the rectifier efficiency.

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ac input power = 100 W, de output
power = 40 W.
Rectifier efficiency =`("dc power output ")/("Input ac power ") = (P_(dc ))/(P_(ac )) = ( 40 )/( 100 )`
`= 0.4 `
` therefore ` Percentage efficiency of the half wave rectifier = 40%
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