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When an A.C voltage source in series with the DC voltage source are applied to a capacitor in series with a resistor they say that capacitor will block Dc and will let AC pass to the resistor. I am not understating it. By using super position theorem the statement can be proved but i am not getting the concept. Since the voltage/current across the capacitor is now pulsating Dc in this case then how the capacitor is blocking DC while letting AC passing through the resistor? How and why a capacitor makes a pulsating DC input to an AC output across the resistor?

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  • $\begingroup$ You said: Since the voltage/current across the capacitor is now pulsating DC. That's a confusing way to think about it, better to treat them separately. The voltage across the cap is alternating with a 2 V offset. But the current through the CAP is strictly alternating (after an initial charging occurs.) $\endgroup$ – Burt_Harris Apr 27 '18 at 23:17
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I never have cared for the phrase a capacitor blocks DC (it doesn't) but that's beside the point here.

DC is most often used to mean a constant voltage or current rather than the original meaning of unidirectional current.

If one takes a time varying voltage and finds the time average value, this value is often called the DC component of the voltage. That is, if you subtract off the time average value (DC component) from the time varying voltage, you get the AC component(s).

This is essentially what the capacitor does here - it subtracts 2V from the 'pulsating DC' leaving only an AC voltage across the resistor. That is, if you zeroed the AC source, you would find a constant 2V - of opposite polarity of the 2V source - across the capacitor and 0V across the resistor.

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The key is that it takes time to charge a capacitor; the time required is about equal to the time constant $RC$.

If only a DC source is connected, the capacitor will allow charge to flow at first, but as charge flows to the capacitor, voltage builds up across the capacitor. This voltage opposes the flow of additional charge, and so the charge eventually stops flowing (when the capacitor voltage matches the source voltage). If the DC source is then reversed, the capacitor helps the source send charge back the other way. If you keep reversing the source repeatedly, then you have an AC source. Here's the trick: if the reversals happen often enough (specifically, if the time between reversals is much less than the RC time constant), then the capacitor voltage never rises enough to match the source voltage; in this situation it's almost like the capacitor isn't there.

In the circuit you describe, the capacitor voltage will oscillate above and below the DC source voltage, so that the capacitor more or less cancels out the DC source. However, the AC source reverses too quickly to make a significant change in the capacitor's voltage, and so the capacitor never affects the AC source that much.

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  • $\begingroup$ The time required for the capacitor to be essentially charged is actually about 5 time constants. $\endgroup$ – Alfred Centauri Jun 24 '17 at 18:47
  • $\begingroup$ If the pulsating is fast enough, the capacitor would charge and discharge as if it was AC. Remember, the change in voltage is what is required for current to flow trough the capacitor, not the reversal of polarity acording to the ground voltage. $\endgroup$ – MaDrung Jan 19 '18 at 7:20
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Well, what do we know:

$$\text{V}_\text{in}\left(t\right)=1\cdot\cos\left(20000\pi t\right)+2=2+\cos\left(20000\pi t\right)=\text{V}_\text{C}\left(t\right)+\text{V}_\text{R}\left(t\right)\tag1$$

We also know:

  • $$\text{I}_\text{in}\left(t\right)=\text{I}_\text{C}\left(t\right)=\text{I}_\text{R}\left(t\right)\tag2$$
  • $$\text{V}_\text{R}\left(t\right)=\text{I}_\text{R}\left(t\right)\cdot\text{R}=\text{I}_\text{in}\left(t\right)\cdot22000=22000\cdot\text{I}_\text{in}\left(t\right)\tag3$$
  • $$\text{I}_\text{C}\left(t\right)=\text{I}_\text{in}\left(t\right)=\text{V}_\text{C}'\left(t\right)\cdot\text{C}\space\Longleftrightarrow\space\text{V}_\text{C}\left(t\right)=\int\frac{\text{I}_\text{in}\left(t\right)}{0.1\cdot10^{-6}}\space\text{d}t\tag4$$

So, we get:

$$2+\cos\left(20000\pi t\right)=\int\frac{\text{I}_\text{in}\left(t\right)}{0.1\cdot10^{-6}}\space\text{d}t+22000\cdot\text{I}_\text{in}\left(t\right)\tag5$$

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It seems that you understand why the capacitor blocks DC and passes AC, but you don't quite understand the case when DC and AC are both present.

Well, as you said, we need to use the principle of superposition, which, says that the total current through the capacitor will be equal to the sum of the currents produced by the DC source alone (AC is shorted) and by the AC source alone (DC is shorted).

If the DC source, alone, does not produce any current, the total current will be equal to the current produced by the AC source alone, i.e., the DC source would not produce any current whether it is alone or in series with the AC source.

A couple of clarifications here.

1) The current contributed by the AC source will be flowing through the DC source and will be changing direction every cycle. Here, we assume that the DC source is an ideal source and does not present any resistance to the AC current in either direction, i.e, it acts as a wire. This is not necessarily true for any DC source, but is a reasonable assumption for many practical cases.

2) As mentioned in other answers, before the capacitor blocks DC current, it will have to be charged to the voltage equal to the voltage of the DC source. So, there will be a short current pulse contributed by the DC source right after the circuit is connected together, but, in this exercise, we are focusing on the steady state mode, which reflects the operation of the circuit after things get settled.

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