Electronic Devices - Voltage Regulators - Discussion
Discussion Forum : Voltage Regulators - General Questions (Q.No. 43)
43.
Calculate the ripple of a capacitor filter for a peak rectified voltage of 40 V, a capacitor value C = 75 
F, and a load current of 40 mA.
F, and a load current of 40 mA.Discussion:
8 comments Page 1 of 1.
Vins said:
3 years ago
I think 2.4 is the division of 1/4(1.73)(60) but it should be 2.4m not 2.4.
Parikiki said:
4 years ago
@Uthaya.
Where did you get the 2.4 value?
@Tankiso.
What is that 1.73 value?
Where did you get the 2.4 value?
@Tankiso.
What is that 1.73 value?
Kimi said:
4 years ago
@Bhavani.
The default value of frequency is 60Hz. If it isn't stated in the problem then use that value for the frequency.
The default value of frequency is 60Hz. If it isn't stated in the problem then use that value for the frequency.
Bhavani said:
5 years ago
Why you using 60 in terms of frequency.
Tankiso said:
7 years ago
Vr(rms)=Idc/4 * 1.73 * f * C.
Vr(rms)=40mA/4 * 1.73 * 60 * 75uf.
Vr(rms)=1.28V.
Ripple of filter=(Vr(rms)/Vdc) * 100.
= (1.28/40) * 100,
= 3.2V.
Vr(rms)=40mA/4 * 1.73 * 60 * 75uf.
Vr(rms)=1.28V.
Ripple of filter=(Vr(rms)/Vdc) * 100.
= (1.28/40) * 100,
= 3.2V.
(2)
Napawnu kadi said:
8 years ago
I agree @Uthaya.
Using that formula, I (current load) must be in mA and C (capacitance) must be in uF.
Using that formula, I (current load) must be in mA and C (capacitance) must be in uF.
Nobody said:
9 years ago
@Uthaya.
Vrms = (2.4 * 40mA)/75uF.
=1280 not 1.28V.
Vrms = (2.4 * 40mA)/75uF.
=1280 not 1.28V.
Uthaya said:
1 decade ago
Vrms = (2.4*Idc/C);
Vrms = (2.4*4*10^-3)/75*10^-6;
Vrms = 1.28V;
Ripple of a filter = (Vrms/Vdc)*100;
= (1.28/40)*100;
= 3.2%.
Vrms = (2.4*4*10^-3)/75*10^-6;
Vrms = 1.28V;
Ripple of a filter = (Vrms/Vdc)*100;
= (1.28/40)*100;
= 3.2%.
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