Capacitor with pulsating dc voltage source/current

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 ?

• Do you understand what people mean when they say a capacitor blocks DC? – user253751 Feb 28 '18 at 3:42

Maybe this image helps you understand:

From the time domain point of view you might think of this "pulsating DC" concept and think that it should pass over the capactior.

From the frequency domain point of view, there are two different components, the DC (0Hz) and AC. If you put a capacitor in series, you are using a High Pass Filter, meaning that the DC component will be filtered and won't pass.

simulate this circuit – Schematic created using CircuitLab

Figure 1. A simple audio amplifier.

Let's take a simple example. Figure 1 shows a simple audio buffer amplifier fed from a single ended 9 V supply.

Input

• The IN signal will alternate about 0 V and its average voltage will be zero.
• Because of the single-ended supply we can't drive the amplifier output negative. It just can't go below the ground voltage (0 V). Since we are only interested in the audio AC signal we can do a trick. By adding C1, R1 and R2 we can bias the voltage at A to half the supply voltage = 4.5 V. It should be clear that the average voltage at IN is zero and $V_A$ = 4.5 V.
• Now when the input voltage rises above 0 V point A will rise above 4.5 V by the same amount. Similarly when IN goes negative point A will go below 4.5 V. You can think of the capacitor as maintaining 4.5 V difference between IN and A.
• Note that this only works if the frequency of the audio in is sufficiently high that the capacitor doesn't have enough time to discharge back to 4.5 V. This is a high-pass filter. Low frequencies will be attenuated.

Output

For the output we now need to remove the 4.5 V DC bias to restore our AC signal about 0 V. C2 serves this purpose.

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?

The voltage at IN is pure AC. The voltage at A is DC with the IN signal superimposed on it. If you measured it at A with your multimeter on DC range you would read 4.5 V. If you measured it on AC range it would read the same as IN. They're both there simultaneously.

How a capacitor is makes a pulsating DC input to an AC output across the resistor?

Let me know if I've not answered this question.

• Nice explaination "How a capacitor makes a pulsating DC input to an AC output across the resistor? Let me know if I've not answered this question." i guess you havent answer this question. i have come to know about the differentiator circuits that reshapes the input wave forms and this circuit is a type of differentiator circuit and i am seeking its expalnation. i shall be very grateful to you if you help me in this regard.@transistor – Alex Jun 24 '17 at 12:50