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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.
What else can I help you with?
What is the element law of a capacitor in frequency domain?
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.
Why electrolytic capacitors are used for frequency response of BJT as amplifier?
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.
What is the relationship between resistance and capacitance in a clc circuit?
The relationship between resistance and capacitance in a clc circuit is the capacitive reactance given by XC.
What is the formula for calculating self capacitance of an inductor when the value of capacitance and frequency for two capacitors is given?
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.
Increasing the capacitance of the coupling capicitor in an RC coupled amplifier has what effect?
the circuit will pass waves of a lower frequency
'when frequency is increased why capacitance increases in a capacitor'?
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.
What is the relationship between capacitance and frequency?
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.
What factors affect the reactance of a capacitor?
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.
What is the relation between amperage and capacitance?
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.
What is the element law of a capacitor in frequency domain?
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.
Why electrolytic capacitors are used for frequency response of BJT as amplifier?
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.
What is the definition of capactive reactance?
reactance due to the capacitance of a capacitor or circuit,equal to the inverse of the product of the capacitance and the angular frequency.
What is the relationship between resistance and capacitance in a clc circuit?
The relationship between resistance and capacitance in a clc circuit is the capacitive reactance given by XC.
What is the formula for calculating self capacitance of an inductor when the value of capacitance and frequency for two capacitors is given?
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.
What is the relationship between potential difference and capacitance in a capacitor?
The relationship between potential difference and capacitance in a capacitor is that the potential difference across a capacitor is directly proportional to its capacitance. This means that as the capacitance of a capacitor increases, the potential difference across it also increases, and vice versa.
What two factors determine the capacitive reactance of a capacitor?
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).
What is the effect of capacitance in high frequency circuits and how it is getting added?
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.
