Smoothing of fullwave rectifying circuits

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In fullwave rectifying circuits, placing a capacitor in parallel with the load is crucial for effective smoothing of the output voltage. A capacitor in series would block DC current, preventing any power dissipation across the resistor. When the rectifier produces a perfect DC output, a series capacitor would result in no power being dissipated by the resistor, as it would be fully charged. Understanding the configuration of capacitors in relation to the load is essential for achieving desired circuit behavior. Proper placement of the capacitor significantly impacts the performance of the rectifying circuit.
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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 :)
 
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