XC = -1 / (2 pi f C)
XC = about -2653 ohms
The minus sign indicates the current is leading in the this case. Treat it as if the sign were not there.
In the basic configuration, a capacitor is constructed with two parallel conductor plates with a layer of insulating material in between. When the cap is hooked up to the AC power supply, the voltage (v) across the plates and the charge (q) induced on the plates follow this capacitance expression: C = dq/dv or i = C dv/dt, where C is determined by the properties of the insulating material and the geometry of the cap (in the case of the parallel plates, the separation between the two electrodes (t). For the parallel plates, C can be written as (dielectric constant * plate area / t). Electrically, the change in the charge induced on the plates (dq), is directly related to the change in voltage difference (dv) between the two plates, since C is a constant. Theoretically, no energy is lost by charging and discharging the cap with an AC current. When the cap absorbs electrical energy from the power supply, it stores the energy in the electric field in the insulator. When discharging, the cap gives the stored energy back to the circuit -- hence, no energy loss. In a circuit, we use the cap to prolong/smoothen/resist any voltage change in time or to absorb a sudden energy surge (electrostatic discharge and power-line glitches, for example).
It is 100+j(500-300) ohm = (100+j200) ohm = 223.6<630 ohm
The reactance (X_L) of an inductor is calculated using the formula (X_L = 2\pi f L), where (f) is the frequency in hertz and (L) is the inductance in henries. For a 3-H inductor at a frequency of 100 Hz, the reactance is (X_L = 2\pi (100)(3) \approx 1884.96 , \Omega). Thus, the reactance of the 3-H inductor at 100 Hz is approximately 1885 ohms.
1/(2 Pi sqrt(LC))
The total impedance ( Z_t ) of a circuit with a resistor (R) and inductor (L) in series can be calculated using the formula ( Z_t = R + jX_L ), where ( X_L = \omega L ) is the inductive reactance. For a 100-ohm resistor and a 100-ohm inductor, the impedance will depend on the frequency ( f ) of the AC current. If the inductance and frequency are such that ( X_L = 100 ) ohms (e.g., at a specific frequency), then ( Z_t = 100 + j100 ) ohms, resulting in a magnitude of ( Z_t = \sqrt{100^2 + 100^2} = 100\sqrt{2} ) ohms.
The reactance of a capacitor depends on its capacitanceand the frequency of the voltage across it.In general, the magnitude of capacitive reactance is . . .1 / (2pi x frequency x capacitance)At 100 Hz, that would be0.00159 / (capacitance) in Farads .
If current and voltage of an AC are in phase, then the "power factor" is 100%, and the load is a pure resistance, with no inductive or capacitive reactance (at least at the operating frequency of the AC).
In the basic configuration, a capacitor is constructed with two parallel conductor plates with a layer of insulating material in between. When the cap is hooked up to the AC power supply, the voltage (v) across the plates and the charge (q) induced on the plates follow this capacitance expression: C = dq/dv or i = C dv/dt, where C is determined by the properties of the insulating material and the geometry of the cap (in the case of the parallel plates, the separation between the two electrodes (t). For the parallel plates, C can be written as (dielectric constant * plate area / t). Electrically, the change in the charge induced on the plates (dq), is directly related to the change in voltage difference (dv) between the two plates, since C is a constant. Theoretically, no energy is lost by charging and discharging the cap with an AC current. When the cap absorbs electrical energy from the power supply, it stores the energy in the electric field in the insulator. When discharging, the cap gives the stored energy back to the circuit -- hence, no energy loss. In a circuit, we use the cap to prolong/smoothen/resist any voltage change in time or to absorb a sudden energy surge (electrostatic discharge and power-line glitches, for example).
It is 100+j(500-300) ohm = (100+j200) ohm = 223.6<630 ohm
Impedance (Z) = Resistance (R) + Reactance (X)Resistance is real. Reactance is imaginary.Z = R + jXAn ideal capacitor has only reactance.Z = 0 + jXReactance of a capacitor is given by X = 1/(jwC).Where:Angular frequency (w) = 2*pi*frequencyCapacitance (C) = 12 microfarads (uF) in our caseZ = 1/( j*2*pi*100 Hz*10^-6 uF ) = 13.26/j uF = j13.26/( j*j ) uF = -j13.26 uF
The reactance (X_L) of an inductor is calculated using the formula (X_L = 2\pi f L), where (f) is the frequency in hertz and (L) is the inductance in henries. For a 3-H inductor at a frequency of 100 Hz, the reactance is (X_L = 2\pi (100)(3) \approx 1884.96 , \Omega). Thus, the reactance of the 3-H inductor at 100 Hz is approximately 1885 ohms.
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'.
The reactance of an inductor is calculated as Xl = 2πfL, where Xl is the inductive reactance, f is the frequency, and L is the inductance. Substituting the given values of 100 microhenries for inductance and 400 Hz for frequency into the formula gives Xl = 2 * π * 400 * 100 * 10^-6 which equals approximately 251.3 ohms.
1/(2 Pi sqrt(LC))
You want a power factor of 1 or 100%, which is a purely resistive circuit. If you have a motor or some other inductive load in a circuit the total voltage and total current in the circuit will not be in phase (phase shift), your power factor will be less than 1. By adding a capacitor (180 degrees out of phase with inductive load) to the circuit that has a capacitive reactance equal to the inductive reactance of the motor, you can cancel the phase shift and have an ideal power factor (no wasted power). Anything above .9 would be good.
100 WVDC means 100 working volts DC. That is the maximum operating voltage that the capacitor is certified to have across it. Exceeding that rating could puncture a hole in the dielectric, leading to catastrophic failure of the capacitor.
Percentage reactance of a transformer (or in general, a circuit) is the percentage of phase voltage drop when full load current flows through it, i.e %X=(IX/V)*100. Now Short Circuit Current is V/X So short Circuit current is I*(100/%X).