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.
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0
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%
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