XC = -1 / (2 pi f C)XC = about -2653 ohmsThe minus sign indicates the current is leading in the this case. Treat it as if the sign were not there.
The relationship between resistance and capacitance in a clc circuit is the capacitive reactance given by XC.
The frequency at which the impedance of the circuit becomes zero is known as resonance frequency. Actually at resonance resistance only presence in the circuit. That means the impedance of the inductor and capacitor will automatically vanish.
in a dc frequency=0 impedance is xc=1/2*pi*f*c so impedance infinite
In case of DC: because the frequency of dc current is zero capacitive reactance ( Xc) = 1/2(22/7)fc f= frequency c=capacitence and when Xc=0 this implies that I=0 In case of AC: Xc >0 => I>0
the impedance of the capacitor is given by Xc=1/jwC where w=2*pi*f and for DC source f=0 hence Xc=infinity ie, the capacitor will provide infinite impedance for DC, or its Open circuit
The total impedance of a circuit with a capacitor in parallel with a resistor is calculated using the formula Z 1 / (1/R 1/Xc), where Z is the total impedance, R is the resistance of the resistor, and Xc is the reactance of the capacitor. This formula takes into account the combined effects of resistance and reactance in the circuit.
The equivalent impedance of a resistor and capacitor in parallel is calculated using the formula Z 1 / (1/R 1/Xc), where Z is the total impedance, R is the resistance of the resistor, and Xc is the reactance of the capacitor. This formula takes into account the combined effects of resistance and capacitance in the circuit.
In dc equivalent circuit of an amplifier all capacitors are replaced by open circuit because capacitor block dc. As , Xc=1/2πfC We know that that frequency of dc is zero so Xc will infinite so we replace all capacitors with open circuit.
XL=XC
Because reactance of capacitor is inversly proportional to the frequency i.e- Xc=1/(2*pie*f*c) where f is frequency and c is capacitance of capacitor.
At high frequency, capacitor can be considered as 1. Short Circuit in AC analysis. 2. Open Circuit in DC analysis. {because Xc= 1/(2*f*pi) where f= supply frequency,pi=3.14} As at high frequencies, in DC analysis, capacitor will be open circuited & can block the DC signal while AC signal is allowed to pass through.. Hence, this capacitor will act as a blocking capacitor for DC supply.
what parameter stay the same in LCR circuit ?
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).
Nice data. What's your question ?Could you possibly be asking for the frequency ???Xc = 1 / (2 pi f C)f = 1 / (2 pi Xc C)= 106 / (2 pi Xc 0.15)= 106 / (600.264)= 1665.9 Hz (rounded)
A capacitor will oppose the flow of a.c. due to its capacitive reactance (Xc), expressed in ohms.The capacitive reactance for a given capacitor is inversely-proportional to the frequency of the supply; in other words, the higher the frequency, to lower the capacitive reactance.
Actually, neither d.c. nor a.c. current passes through a capacitor. A.C. current 'appears' to flow through a capacitor but, in reality, it is only flowing through the connecting circuit while what is known as a 'displacement current' (actually, a distortion of the atoms' shells rather than a conduction current) occurs within the dielectric separating the capacitor's plates. D.C. current is 'blocked' by a capacitor because, when the capacitor is fully charged, the potential difference appearing across its plates is equal and opposite that of the supply. With no net potential difference in the circuit, no current can flow.