Question

Find ripple factor and efficiency of a half wave and full wave rectifier having load resistance of 2 kΩ from a 25 V r.m.s. supply. Diode forward bias resistance is 100Ω.​

Answer 1

Answer:

40

Explanation:

The efficiency η of half wave rectifier is given by, η=

r

f

+R

L

0.406R

L

η=

1+

R

L

r

f

0.406

Where, R

L

is the resistance of load resistor and r

f

is the resistance of diode in forward biased condition.Hence, the percent efficiency is

η=(

1+

R

L

r

f

0.406

)×100

η=(

1+

2000

20

0.406

)×100

η=(

1+0.01

0.406

)×100

η=(

1.01

0.406

)×100

η=(0.4019)×100=40.2%