I have been having some trouble understanding the functioning of a capacitor and a resister attached in parallel to a rectifier to smooth out the time-variant dc output. From what I have read, a smoothing capacitor would essentially convert a rectified dc voltage \$V_{in}=V_0|\sin(\omega t)| \$ to a sort of triangular waveform with an offset.a parallel RC circuit with a graph of voltage across it This doesn't make sense to me since we take it as an axiom that the voltage across components in parallel remains the same, so how can a capacitor attached in parallel to the input voltage change it? The explanations I read talk about the capacitor discharging once the voltage reaches quarter(half?)-cycle, which I guess makes sense but I still don't see how this doesn't contradict the 'axiom'.

Further, why doesn't this smoothing happen with ac voltage? If the principle behind smoothing is the capacitor getting fully charged every quarter-cycle and then proceeding to discharge, why doesn't this apply to ac voltage?

When considering a parallel RC circuit with an AC current source, we get the Kirchhoff equations $$i=i_R+i_C$$ $$V=V_0\sin(\omega t)=i_RR=\frac1C\int_0^t{i_Cdt}$$ which we solve to find out the respective currents, but this too assumes that voltage is the same in the capacitor and the resistor and equal to the input voltage. So why doesn't the parallel RC circuit used after a rectifier behave in the same manner, with similar equations $$i=i_R+i_C$$ $$V=V_0|\sin(\omega t)|=i_RR=\frac1C\int_0^t{i_Cdt}$$

Why does using \$V=V_0|\sin(\omega t)|\$ instead of \$V=V_0\sin(\omega t)\$ make such a huge difference in the way circuits usually behave? It seems to my I am making a very elementary conceptual mistake. What am I missing here?

  • \$\begingroup\$ What's the question really? Without load the capacitor charges to peak AC voltage. But usually you want to power a load that is approximated by the resistor, so the capacitor discharges when it is not charging. \$\endgroup\$
    – Justme
    Commented Jul 9, 2022 at 16:04

2 Answers 2


Your drawing omits the rectifier, which contains diodes (it also can be built using transistors). The diodes do not conduct all the time, which means that the input voltage (a sinusoid, say) is not always connected to the parallel combination of resistor and capacitor. When it is connected, current flows from the source through some source impedance (dropping some voltage) and then through the diode(s) (dropping some more voltage) and finally flows onto the plates of the capacitor. Because of the source resistance, the capacitor does not charge instantaneously. The load is always drawing current from the capacitor. It's true that this somewhat inhibits or slows the charging operation, but if the source provides enough current, it will still permit the capacitor to charge. So in your sketch there are two phases. In one, the capacitor charges, and in the other it discharges through the load resistance.


Your concern is perfectly legitimate, you are just missing an important detail:

The diode!

What's not true is that you system is driven by

$$ V_{in}=V_0|\sin(\omega t)|$$ ideal voltage source, in this case you'd be on spot, the ideal voltage generator rules, everything connected just sees this voltage.

Instead a condition on the current shall be added to account for the diode

$$ V_{in}=V_0|\sin(\omega t)|\quad , \quad i>0$$


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