Electronic Devices - Voltage Regulators - Discussion
Discussion Forum : Voltage Regulators - General Questions (Q.No. 5)
5.
Calculate the ripple voltage of a full-wave rectifier with a 75-
F filter capacitor connected to a load drawing 40 mA.
F filter capacitor connected to a load drawing 40 mA.Discussion:
6 comments Page 1 of 1.
Rathi said:
1 decade ago
@Kismat.
Ripple voltage of the full wave rectifier is Vr(rms) = I/(240 x sqrt(3) x C)
= (40 x 10^-3)/(240 x sqrt(3) x 75 x 10^-6)
= 1.28 V.
Ripple voltage of the full wave rectifier is Vr(rms) = I/(240 x sqrt(3) x C)
= (40 x 10^-3)/(240 x sqrt(3) x 75 x 10^-6)
= 1.28 V.
(1)
Maui said:
7 years ago
Vr = Edc / (2 * f * C * Rload).
Vr = I(load)/(2*f*C) = 40ma/(2 * (60hz) * (75uF)).
Vr = 4.44V.
Vr(rms) = Vr / (2*√(3)).
Vr(rms) = 1.28.
Vr = I(load)/(2*f*C) = 40ma/(2 * (60hz) * (75uF)).
Vr = 4.44V.
Vr(rms) = Vr / (2*√(3)).
Vr(rms) = 1.28.
(1)
Pranab samanta said:
1 decade ago
Ripple voltage of the full wave rectifier is Vr(rms) = (2.4*Idc)/c.
= (2.4*40*10^-3)/(75*10^-6).
= 1.28v.
= (2.4*40*10^-3)/(75*10^-6).
= 1.28v.
(1)
Kismat said:
1 decade ago
2.4*40/75 = 1.28.
(10^-3)/(10^-6) = 1000.
1280 V.
@Pranab Samanta explanation is wrong.
(10^-3)/(10^-6) = 1000.
1280 V.
@Pranab Samanta explanation is wrong.
Vijaya said:
8 years ago
You are right @Kismat.
Dilip said:
1 decade ago
how does
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