Capacitance: Xc(Ω) = 1/2πfC
Inductance: XL(Ω) = 2πfL
π: 3.14
f: power supply frequency
C: capacitance (F)
L: inductance (H)
Expanded Information. Strong> Extended information:
The capacitive reactance is denoted by XC, the capacitance is denoted by C (F), and the frequency is denoted by f (Hz), so Xc = 1/2πfc The unit of capacitive reactance is the ohm. Knowing the frequency f and the capacitance C of the alternating current, the capacitive reactance can be worked out using the above equation.?
The inductive reactance is expressed as XL, the inductance is expressed as L(H), and the frequency is expressed as f(Hz), then XL=2πfL The unit of inductive reactance is ohm. Knowing the frequency of the alternating current, f, and the inductance of the coil, L, the inductive reactance can be calculated using the above equation.
Knowing the capacitive reactance and inductive reactance, the corresponding voltage and current can be calculated using Ohm's law, if the capacitance is used together with the resistance and inductance, it is necessary to take into account the phase relationship.
The greater the capacitance of a capacitor, the greater the capacitor's ability to store charge, the greater the amount of charge moved per unit of time in the circuit charging and discharging under a certain voltage, the greater the current, so the capacitance of the alternating current of the obstruction of the role of the lesser, i.e., the lower the capacitance of the capacitance of the reactance.
In the alternating current voltage is certain, the higher the frequency of the alternating current, the more frequent charging and discharging in the circuit, the greater the rate of charge movement per unit of time, the greater the current, the capacitance of the alternating current of the obstruction of the smaller, that is, the smaller the capacitive impedance. Expression: Xc=1/(2πfC)
①In a pure capacitive circuit, when the power supply is turned on, the voltage of the power supply makes the free charge in the wire in a certain direction for directional movement, due to the accumulation of charge on the capacitor poles in the process of generating a potential difference, and thus resistance to the charge's continued movement, thus forming the capacitive resistance.
② For the same amount of capacitors, capacitance, the greater the potential difference between the two plates is smaller, so the capacitive resistance and capacitance is inversely proportional. The higher the frequency of the alternating current, the faster the charge and discharge, the smaller the capacitive resistance. Therefore, the capacitive reactance and frequency are also inversely proportional. That is, Xc = 1/ωC.
3 in ideal conditions, when ω = 0, because Xc = 1/ωC, Xc tends to infinity, which means that DC will not be able to pass through the capacitance, so the capacitor's role is to "pass through the alternating current, isolation". In AC circuits, often apply the frequency characteristics of capacitive resistance to "pass high-frequency AC, blocking low-frequency AC".
4 In a purely capacitive circuit, the capacitor plate on the charge and voltage relationship is q = CU. at the same time in △ t time on the capacitor plate charge change for △ q so the current in the circuit for I = △ q / △ t, in capacitive circuits, capacitance is the basic law of I = C - △ u / △ t. Because sinusoidal alternating current in a cycle of the voltage changes periodically, the rate of change of voltage ( △ u / △ t), the voltage of the change of the rate of change of the voltage (△ u / △ t), the voltage of the voltage (△ u / △ t). △u/△t) is changing.
It follows that when the voltage is zero, the rate of change of voltage (△u/△t) is maximum and the current in the circuit is maximum. Conversely, when the voltage is maximum, the rate of change of voltage (?u/?t) is zero and the current is also zero. So the phase of the current in the circuit overshoots the voltage across the capacitor by π/2. as shown in the figure.
5 The capacitor in a purely capacitive circuit does not consume electrical energy. Because in the charging process, the electric field established between the plates of the capacitor converts the electrical energy of the power supply into electric field energy, and in the discharging process, the electric field gradually disappears, and the stored electric field energy is converted into electrical energy and returned to the power supply. So the active power of the pure capacitor circuit is zero, no external work, and the maximum value of reactive power QL = (I^2)Xc.
Winding a small voltage transformer, the inductive impedance of the formula is deduced as follows:
2πfL = R Primary Load (1)
Which the R Primary Load includes the impedance of the transformer's primary coils and inductive impedance. Since I only have to wind about 10 turns, the impedance can be viewed as approximately 0; so the R primary load is mainly caused by the inductive reactance. Knowing the magnitude of the R primary load and f (frequency is known to be 500KHz), then:
L= R primary load/(2πf) (2)
So how do I get the value of the R primary load? This value is derived from the static current and the primary voltage:
R Primary Load= V Primary/ I Static (3)
The primary voltage is known, and the empirical value of the static current (the current present in the primary coil when the secondary is open) is:
I Static=5%*I Primary Full Load (4)
I Primary Full Load* V Primary= I secondary full load * V secondary (5)
Because the ratio of primary to secondary voltage is a known quantity, then the value of I primary full load can be known by knowing the value of I secondary full load. I want to make a transformer where the primary to secondary voltage ratio is 1:1.2 and the I secondary full load is 200 mA. Then I primary full load = 240 milliamps, and by bringing this value into equation (4), I can find that I static is about 10 milliamps.
The V primary is a known quantity, and here my transformer primary voltage is V primary = 5 V. Substituting V primary = 5 V and I static = 10 mA into equation (3) gives R primary load = 500 ohms. Substituting R primary load = 500 ohms into equation (2) gives:
L=500/(2πf)=500/(2π*500000)=159 (microhenry)
References: