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Suppose a capacitor of capacitance $C$ is being used in a rectifier to smooth out the p.d. across a resistor of resistance $R$. What is the effect on the smoothing curve when the $R$ is lowered?

The time constant $\tau = RC$ should be lowered and hence the time to discharge is decreased. Hence should the smoothing decrease? Am I right? Because my book suggests otherwise.

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    $\begingroup$ Can you give an example of what you mean by a "smoothing curve"? It isn't something that has a universally known meaning. $\endgroup$ – The Photon Apr 20 '15 at 14:41
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Yes, when you have a smaller resistance, your RC product decreases and so does the time constant. This means that the ripple will be greater, and the smoothing will be less if your capacitor is in parallel with the resistor.

Unfortunately, there are some words (like "resolution", "smoothing") for which "greater" can mean one of two things. If I say "greater resolution", does it mean it went from 1 mm to 2 mm (the number is greater, but the resolution is poorer) or does it mean "better" (it went from 2 mm to 1 mm)? It is possible that the confusion in your book came about from a similar unfortunate phrasing.

In general, if your resistance is less, the current is greater and you need a bigger "reservoir" (capacitance) in order to ride through the bumps of the supply.

Finally - you really need to look at the entire circuit to answer questions like this properly: there is usually more than one resistor value, and so there is the "charging" and "discharging" current of the capacitor that needs to be taken into account. If the resistor in question is in series with the capacitor, then the capacitor will be charged more quickly - and so as a percentage of the voltage, it might give you a smaller ripple.

This is why it's not easy to answer this question "definitively" without seeing the actual circuit diagram you are using.

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  • $\begingroup$ The diagram can be found at the first answer to the question electronics.stackexchange.com/questions/73863. Is not your answer in conflict with that answer? The author their suggests that the heavier the load (higher current), the faster the capacitor discharges, thus the more ripple there will be. $\endgroup$ – PdX Apr 20 '15 at 15:29
  • $\begingroup$ In the first paragraph I state "the ripple will be greater". So I agree with the answer you link. What did you think I meant? $\endgroup$ – Floris Apr 20 '15 at 15:32
  • $\begingroup$ Well apparently I got confused about HEAVIER load whereas the author of that post means heavier current. The load in that question must draw huge amounts of current unlike our simple case. And by the way, after seeing the diagram, do you think $RC$ will still fall if we connect another resistor parallel to the first resistor and capacitor? $\endgroup$ – PdX Apr 20 '15 at 15:58
  • $\begingroup$ As I said - this is one of those areas where it is very easy to get confused by the words used. "Heavy" load means small resistor, means larger ripple. When you say "parallel to the first resistor and capacitor" what exactly do you mean? In the circuit drawn, there is a resistor with LED in series parallel to a capacitor. Do you want to put a second resistor parallel with the capacitor? That would increase the ripple, yes. $\endgroup$ – Floris Apr 20 '15 at 16:00
  • $\begingroup$ Well, you understood it right. Thank you for your input. $\endgroup$ – PdX Apr 20 '15 at 16:13

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