why capacitance depends upon frequency physically irrespective of mathematical fact?

Sir,I know that when f=0 capacitive reactance will be infinity according to mathematical formula and thats why capacitor blocks dc.But what is its physical significance?I mean why capacitor blocks dc when frequency is null?I really dont get its answer to clearly understand it.Thank you.

• Do you know how a capacitor is constructed? Nov 28, 2014 at 19:50
• Capacitance doesn't depend on frequency. Reactance does. Nov 29, 2014 at 9:13

There is a barrier...

Capacitors block DC because there is a physical barrier (non-conductor) that prevents current flow. AC can pass simply because the charges on one side push (repel) and pull (attract) the charges on the other side during each half-cycle of the AC waveform.

The phantom current

This causes the charges to appear to circulate even though they are not actually circulating. This concept is called a phantom current because it appears that there is a current crossing the non-conductor at the center of the capacitor even thought there really isn't. What is crossing over is the electric field lines. A more intuitive way to grasp this is that the forces exerted by a concentration of like charges on one side applies across the gap on the charges of opposite likeness on the far side.

Reactance

Reactance is the opposition of a circuit element to a change of voltage. When at DC, there is no effort made by the signal to "change the voltage" and therefore a perfect ability of the capacitor to resist the change (ergo infinite reactance). Physically, the inability of the static charges to generate a phantom current means that the capacitor can perfectly resist the transfer of energy in the ideal case.

• "it appears that there is a current crossing the non-conductor at the center of the capacitor even thought there really isn't." So if I connect a capacitor in series with a load and apply AC voltage, the load won't draw any current? Please explain because this is contrary to what I know.
– user59273
Nov 29, 2014 at 3:44
• The load will draw current. It's just that the electrons will not cross over the dielectric in the capacitor, they will build up one one side and deplete on the other. If you just look at the pins leading in to the capacitor, it looks like there is current flowing though the capacitor. Nov 29, 2014 at 3:52

The basic formula for a capacitor is: -

Charge on the capacitor = capacitance * Voltage or Q=CV

The rate that Q changes equals current and if capacitance is constant then: -

I = C$\dfrac{dV}{dt}$

This basically means the current thru a capacitor is related to rate of change of voltage. Static DC voltages produce no current, slowly changing voltages produce a small current and rapidly changing voltages produce a big current.

A capacitor is built of two metal plates with an insulator in between. Electrons cannot jump across this gap (well, they can if the voltage is too high, but this is not an issue within the specified operating range of the cap). When a voltage is applied across the plates, electrons will flow IN to one plate and OUT of the other plate. As the electrons move, an electric field will be built up in between the plates due to the charge imbalance. The current from the displaced electrons is called the displacement current. Once enough electrons have moved that the internal electric field equals the applied potential, the displacement current will stop flowing. Changing the applied potential will displace more electrons until the internal field again equals the applied voltage. If the applied voltage does not change, no current will flow - hence at DC, a capacitor looks like an open circuit. As for the reactance, voltage and current are related by Ohm's law, V = I * R. If I is zero, then R = V/0 and hence is undefined. Since passive linear components always have positive resistance, the capacitor's DC resistance is considered to be infinite.