Smoothing of fullwave rectifying circuits

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SUMMARY

The discussion focuses on the configuration of capacitors in fullwave rectifying circuits, specifically addressing the placement of a capacitor in parallel versus series with the load. It is established that placing a capacitor in series with the load blocks DC current, preventing power dissipation in the resistor. In contrast, a parallel configuration allows for smoothing of the output voltage, ensuring steady power delivery to the load. The conclusion emphasizes the importance of capacitor placement for effective circuit performance.

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  • Understanding of fullwave rectifying circuits
  • Knowledge of capacitor functions in electrical circuits
  • Basic principles of DC voltage and current
  • Familiarity with power dissipation concepts in resistors
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Homework Statement


http://s13.postimg.org/l780tdg1z/screenshot_33.png
it's b ii)

Homework Equations


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The Attempt at a Solution


well, I just drew a capacitor in series with the load, but when I looked for the answer it was written that it has to be in parallel, I just wanted to know why?how would it make any difference
thanks in advance :)
 
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Imagine that the rectifier produces perfect DC output, and you had the capacitor in series with the resistor. What would be the steady state power dissipated by the resistor?
 
if you mean by perfect DC output that it doesn't need any smoothing(like the voltage is constant)
I think that there will be no power dissipated by the resistor,because the capacitor is fully charged.is that correct?
 
Yes, that it what I meant. The capacitor in series essentially blocks the DC current.
 
MisterX said:
Yes, that it what I meant. The capacitor in series essentially blocks the DC current.

ahh, ok. thank you very much :)
 
Last edited:

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