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The short answer to the question is the capacitive reactance of a capacitor in a DC circuit is infinite.

In a DC circuit, disregarding transient behavior and any leakage effects, a capacitor is effectively an open circuit, and so its reactance is essentially infinite.

Capacitive reactance is calculated as Xc =1/(jwC) where w is the angular frequency in radians per second, w = 2*pi*f, C is in Farads, and f is in Hertz.

With DC, both f and w are zero, and, theoretically, the formula,

Xc =1/(jwC) = limw-->0 [1/(jwC)] becomes infinitely large. In any practical circuit, however, there is always some leakage, so the impedance of the a capacitor will be quite large, on the order of megohms, but still finite.

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A major use is for timing. When you have a capacitor and a resistor, you have an RC circuit, which has a time constant associated with it based on the values of the cap and resistor.

(V final) = (V initial)*e^(-t/RC) where t = time.

A: As frequency increases the capacitor internal inductance will eliminate some of the desired capacitance. A very hi speed PWM will have expenses capacitors with four leads to eliminate some of the internal inductance

In a dc circuit a capacitor acts like a battery, storing charge and releasing it whenever the dc load demands more charge than the dc voltage source can give.

Capacitors are used to give a steady dc output from an unregulated ac rectifier: whilst the voltage increasesthe the capacitor gets charged to the peak of the voltage applied. As the rectified ac voltage then reduces towards zero, the capacitor discharges and hence helps give a more steady dc output from the rectifier.

Changing the capacitance of a capacitor has no effect on the frequency of an AC

applied across it. Likewise, changing the frequency of an AC applied to a capacitor

has no effect on the value of the capacitor. In that sense, there is no relationship

between capacitance and frequency.

But . . .

If the capacitor happens to be one of the frequency-determining elements in the

circuit where the AC is being produced, only then, increasing the capacitance

generally decreases the frequency of the AC signal being produced.

Simple relationship: Capacitance is *defined* to be the ratio of charge to voltage: C=Q/V

Capacitive reactance is inversely proportional to frequency.

go and refer books .that wiil provide enough information.books are better than this...................................by murugapandi subbiah (EEE)

Capacitive reactance is a term that is applicable to AC, not DC.

Q: What is the Relationship beteen Capacitance and frequency?

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The element law of a capacitor in frequency domain is based on Ohm's Law, which is capacitance times voltage is equal to current. The higher frequency, the lower the capacitance and vice versa.

It can take a lot of capacitance to present a low impedance to a low frequency. Electrolytics offer lots of capacitance for a low price.

The relationship between resistance and capacitance in a clc circuit is the capacitive reactance given by XC.

You seem to be mixing up your terminology. There is no such thing as 'self-capacitance of an inductor'! If you know the frequency and equivalent capacitance for two capacitors, then you can find the equivalent capacitive reactance of the capacitors, but that's not what you seem to be asking! I suggest you rephrase the question.

the circuit will pass waves of a lower frequency

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Answer: Capacitance is unaffected by frequency; it does not change. Details: Capacitance is unaffected by frequency. In a capacitor, what increases with Frequency is Admittance (analogus to Conductance) . The capacitive Reactance is inversely proportional to Frequency. Therefore, when Frequency is increased, current flow may increase.

This is a very broad generalization, but in general, increasing the value of one or more capacitors in an electronic circuit will decrease the resonant frequency of one or more sections of the circuit.

Reactance (in ohms) = 1/(2 pi * capacitance * frequency). Capacitance is in farads. Frequency is in Hertz (cycles/second). So increasing capacitance or increasing frequency will decrease reactance.

The element law of a capacitor in frequency domain is based on Ohm's Law, which is capacitance times voltage is equal to current. The higher frequency, the lower the capacitance and vice versa.

capacitive reactance is inversely proportional to the capacitance of the capacitor and frequency of the AC line reactance (in ohms) = 1/(capacitance * frequency)

It can take a lot of capacitance to present a low impedance to a low frequency. Electrolytics offer lots of capacitance for a low price.

reactance due to the capacitance of a capacitor or circuit,equal to the inverse of the product of the capacitance and the angular frequency.

The relationship between resistance and capacitance in a clc circuit is the capacitive reactance given by XC.

You seem to be mixing up your terminology. There is no such thing as 'self-capacitance of an inductor'! If you know the frequency and equivalent capacitance for two capacitors, then you can find the equivalent capacitive reactance of the capacitors, but that's not what you seem to be asking! I suggest you rephrase the question.

First, capacitance is the resistance of something to a change in voltage. And capacitance exists anywhere there is a conductor that is insulated from another conductor. With that definition, anything has capacitance. And that's correct. It is also the key to understanding the capacitance in high frequency (radio frequency or RF) circuits. The fact that a circuit had conductive pathways and component leads and such means that there is a lot of little bits of capacitance distributed around the circuit. The capacitance is already there; it isn't "added" later as might be inferred. Normally, this bit of capacitance isn't a problem. But at higher and higher frequencies, it is. Remember that the higher the frequency of an AC signal, the better it goes through a given cap. So at higher and higher frequencies, the distributed capacitance in the circuit "shorts the signal to ground" and takes it out of the circuit. The RF is said to be coupled out of the circuit through the distributed capacitance in that circuit. The higher the frequency a given circuit is asked to deal with, the more signal will be lost to this effect. It's just that simple. Design considerations and proper component selection minimize the distributed capacitance in a circuit.

the circuit will pass waves of a lower frequency

An oscillator has a tuned circuit (inductance+capacitance) to determine the frequency. When the inductor is tapped to give the required phase-shift for oscillation it is a Hartley oscillator. When the capacitance is tapped it is a Colpitts.