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.
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.
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
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.
The reactance of a capacitor is influenced by its capacitance value and the frequency of the alternating current passing through it. Higher capacitance results in lower reactance, while higher frequency leads to higher reactance. Temperature and the material used in the capacitor can also affect its reactance.
The relationship between amperage and capacitance is indirect. Capacitance stores and releases electrical energy, affecting the flow of current in a circuit. Higher capacitance can lead to slower changes in current (i.e., lower frequency), while lower capacitance can result in faster changes in current.
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.
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.
The two factors that determine the capacitive reactance of a capacitor are the frequency of the alternating current passing through the capacitor and the capacitance value of the capacitor. Capacitive reactance (Xc) is inversely proportional to the frequency (f) and directly proportional to the capacitance (C), as calculated using the formula Xc = 1 / (2πfC).
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