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MIT 6.002 Circuits and Electronics, Spring 2007 17 Online
OpenStudy (anonymous):

what is the effect of load resistor on ripple factor

OpenStudy (shayaan_mustafa):

Hi aaaaaa:) Nice name :D How are you? Well in order to reduce ripples you have to increase the value of your load. Load value is inversely to ripple factor.

OpenStudy (anonymous):

Shayan_Mustafa, a small point here, take into into account that most texts consider that increasing the load must be understood as "reducing the value of the load resistor". The complete answer would be: ripple factor can be reduced by increasing the value of the load resistor (which means reducing the load of the circuit)

OpenStudy (shayaan_mustafa):

Yes bro. You are right.

OpenStudy (anonymous):

@KIRANP Some points here: 1.-Ripple factor is normally measured over a load resistor, never over an impedance 2.-Should there be capacitances or inductances or both we assume they have been added between the PS (after the rectifier) and the resistive load in order to filter and reduce the ripple factor. If nothing is said, it is assumed a filter made with a capacitor after the rectifier and in parallel with the load 3.-In the case of a capacitor (C) in parallel with the load (R), its voltage will be impacted by the time constant T=RC. The higher the capacitance, the higher is T, therefore the lower the ripple factor. Same for R, the higher R (the lower the load), the higher T and the smaller the ripple factor 4.-In case of an inductor (L) in series with the load (R), the current in the load will be impacted by T=L/R. The higher L, the higher T and the lower the ripple factor. The higher R (the lower the load), the lower T and the higher the ripple factor. Obviously we have referred to case in point 3 but one thing is clear, the load (R) surely impacts the ripple factor whatever the case you take

OpenStudy (anonymous):

yeah carlos i agree,u r correct

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