DATE: 17-MAY-12
THE CAPACITIVE REACTANCE VARIES INVERSELY WITH THE AMOUNT OF THE PRODUCT OF MAGNITUDE OF THE SUPPLY FREQUENCY, TWICE THE VALUE OF CONSTANT PI AND THE MAGNITUDE OF CAPACITANCE. IN OTHER WORDS, EVERY
TIME YOU INCREASE ANY OF THESE FACTORS OR IF SAY YOU FIX THE VALUE OF CAPACITANCE TO SAY 1 MICROFARAD AND OTHER FACTOR (i.e. 2 x PI) AS CONSTANT ALSO, BUT THEN WHEN DOUBLING THE FREQUENCY, SO THEN THE EFFECT TO THE CAPACITIVE REACTANCE WILL BE INVERSELY OR HALVED. THE MORE YOU INCREASE THE FREQUENCY, THE SMALLER THIS VALUE WILL BE. WHICH IN EFFECT ALLOWING MORE HIGH FREQUENCY CURRENT OR VOLTAGE TO PASS THRU THE CIRCUIT. AND THE REVERSE IS TRUE, MEANING ONCE YOU ALLOW LOW FREQUENCY CURRENT, THEN THE CAPAICTIVE REACTANCE WILL INCREASE AS A RESULT.
BY FMSJr. / ABU DHABI, UAE
There is no such term as 'inductance reactance'; the correct term is 'inductive reactance'. This is the opposition to the flow of a.c. current, due to the inductance of the load, and the frequency of the supply, and is measured in ohms.Inductive reactance is directly proportional to both the supply frequency and the load's inductance.
Your question is rather vague, but what you may be asking is, "What happens in a circuit if the supply frequency is increased?"Well, circuits have some degree of natural resistance, inductance, and capacitance, which may be modified with resistors, inductors, and capacitors. Frequency affects each of these, as follows:Resistance -Resistance is inversely-proportional to a conductor's cross-sectional area. In a DC circuit, charge flow distributes itself across the full cross section of the conductor. However, with AC currents, an effect called 'skin effect' comes into play -this describes the tendency of charge carriers to move closer to the surface of the conductor, essentially reducing the effective cross-sectional area of the conductor, and increasing its resistance. We call this the 'AC resistance' of the conductor; at normal supply frequencies (50/60 Hz) this is insignificant, however it increases significantly with frequency.Inductance -Inductive reactance opposes the flow of AC current, and is directly proportional to the circuit's inductance and to the frequency of the supply. So, as frequency increases, the circuit's inductive reactance increases.Capacitance -Capacitive reactance opposes the flow of AC current, and is inversely proportional to the circuit's capacitance and to the frequency of the supply. So, as the frequency increases, the circuit's capacitive reactance falls.
-- it has larger capacitance -- it can store more charge at the same voltage -- it can store more energy at the same voltage -- it has lower capacitive reactance at any given frequency -- in combination with a given resistor, the cut-off frequency of a high- or low-pass filter is lower -- in combination with a given inductor, their resonant frequency is lower
AnswerThe 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.
The properties of a series alternating-current L-R-C circuit at resonance are:the only opposition to current flow is resistance of the circuitthe current flowing through the circuit is maximumthe voltage across the resistive component of the circuit is equal to the supply voltagethe individual voltages across the inductive and capacitive components of the circuit are equal, but act in the opposite sense to each otherthe voltage appearing across both the inductive and capacitive components of the circuit is zeroif the resistance is low, then the individual voltages appearing across the inductive and capacitive components of the circuit may be significantly higher than the supply voltage
This isn't necessarily the case. It depends upon the value of resistance (which, at resonance, determines the current), and the values of the inductive- and capacitive-reactance.At resonance, the impedance of the circuit is equal to its resistance. This is because the vector sum of resistance, inductive reactance, and capacitive reactance, is equal the the resistance. This happens because, at resonance, the inductive- and capacitive-reactance are equal but opposite. Although they still actually exist, individually.If the resistance is low in comparison to the inductive and capacitive reactance, then the large current will cause a large voltage drop across the inductive reactance and a large voltage drop across the capacitive reactance. Because these two voltage drops are equal, but act the opposite sense to each other, the net reactive voltage drop is zero.So, at (series) resonance:a. the circuit's impedance is its resistance (Z = R)b. the current is maximumc. the voltage drop across the resistive component is equal to the supply voltaged. the voltage drop across the inductive-reactance component is the product of the supply current and the inductive reactancee. the voltage drop across the capacitive-reactance component is the product of the supply current and the capacitive reactancef. the voltage drop across both inductive- and capacitive-reactance is zero.
The speed halves.
If the frequency of a sound is doubled, the wavelength would be halved. This is because wavelength and frequency have an inverse relationship: as one increases, the other decreases.
The wavelength is halved.
There is no such term as 'inductance reactance'; the correct term is 'inductive reactance'. This is the opposition to the flow of a.c. current, due to the inductance of the load, and the frequency of the supply, and is measured in ohms.Inductive reactance is directly proportional to both the supply frequency and the load's inductance.
No, as 100% efficiency is not possible.AnswerYes, it occurs at resonance. That is, when a circuit's inductive reactance is exactly equal to its capacitive reactance. This can be achieved by adjusting the frequency of the supply until resonance is achieved. Incidentally, power factor has nothing to do with 'efficiency'.
When the frequency is doubled, the resistance of a circuit remains unchanged. Resistance in a circuit is independent of frequency and is determined by the material and physical dimensions of the resistor.
Nothing happens
The speed halves.
Wavelength = 1/frequency. If you double the frequency, the wavelength drops to half.
The frequency also doubles of the wave length stays the same. Remember that Velocity = (the wavelength) x (the frequency)
When the frequency of a wave is doubled, the wavelength is halved. This is because the speed of a wave is constant in a given medium, so an increase in frequency results in a decrease in wavelength to maintain a constant speed.