Low Frequency High Quality Coil/Inductor

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1.Radial Fixed Leaded Inductors with Inductance ranging from 0.30uH to 100mH.

2.ROHS compliance.

3.Good quality and low price

4.low loose


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Q:What would be the voltage across an inductor just after the battery is disconnected?
After a long time an inductor acts like a short circuit, meaning all the current flows through the element, and there is no effective resistance. Therefore, the voltage across it would be V 0, as there is no voltage across a short circuit.
Q:Lamp & inductor ?
Since you have a battery the inductor will have only a transient influence on the circuits current. An equivalent circuit for an inductor is its resistance in series with its inductance. V(t)R i(t) + L di(t)/dt + R(lamp) i(t) R- resistance of the inductor R(lamp)- resistace of the lamp L - inductance Once the steady state is established we will have V(t)R i(t) + R(lamp) i(t) Opening or shortening the winding will collapse or induce the field. Fast collapse of the field can produce a high voltage spike However I would not recommend an inductor a dimming control in a DC circuit.
Q:When you have an inductor in parallel in a circuit, does it act as a short circuit?
If you're working with DC, and inductor is a short circuit. If you're working with AC, the inductor has an impedance and is not a short. The higher the frequency the higher the equivalent impedance. Z jwL (where w 2*pi*f)
Q:What is starting AC current on an inductor?
pbolton seems to think you mean a motor starting current. If you've expressed yourself clearly (which I believe to be case) then you mean the starting current on a pure inductance. As you say you might apply the voltage as sin. (staring from zero) then the steady state current would be -cos. However at the start we are faced with the problem that a current in an inductance cannot change suddenly. What happens is that a dc current component of +1 appears instantaneously in the coil to cancel the -cos, giving a net starting current of zero. The steady state -cos wave then proceeds superimposed on the +1 dc. The latter (starting transient) then slowly decays at the time constant of the coil (L/R) since it has no voltage to sustain it, leaving the -cos as steady state value. If the inductor and supply voltage had no resistance the d.c component would persist indefinitely. If you start the voltage at different points in the cycle the size of the transient dc component will vary but is always somewhere in the range -1 to +1.
Q:how to design inductor and capacitor for boost converter circuit?
The way to do this is to chose the cap to get the desired ripple current and voltage. Chose the inductor to provide the same peak o peak ripple current as the cap.
Q:If frequency selective circuit can be built by using capacitors and resistors than why inductors are used?
first, op amps don't work well at RF frequencies. Those that do are expensive. second, a frequency filter using only R's and C's has a pretty broad bandwidth, and would not select adjacent channels very well.
Q:an inductor of 0.5H inductance and 90 ohms resistance is connected in parallel with a 20 microF capacitor.A voltage of 230v at 50 hz is main?
Xl 2pi fl 157 Ohms Z of L sqrt [(90^2) + (157^2)] 180.97 Ohms I of L I of R 230V/180.97 Ohm 1.27 Amps Total power taken from source (I^2)*(R) (!.27A^2)*(90 Ohm) 145 Watts
Q:what are the specs needed to make a 5 mH inductor?
You okorder
Q:Frequency fixed by capacitor-inductor tuning circuit?
First R1 is adjusted so that the LED puts out pulses at 5 kilohertz. f 1.44/((2*R1 + 10,000)(0.0047 uF)). where f is 5,000 hertz (2*R1 + 10,000)(0.0047 uF) 2.88 x 10^-4 seconds which is 1.44/5,000 hertz R1*0.0094u sec + 4.7 x 10^-5 2.88 x 10^-4 seconds R1* 9.4 x 10^-9 2.41 x 10^-4 seconds R1 is about 25.6 kohms, this sets the transmit frequency The 100 k ohm resistor across the LC tank circuit is used to flatten the response. It DOESN'T modulate the 5 kilohertz frequency. It appears to be optional and is most likely not needed. Try the circuit without R3, R3 will reduce the peak response of the tank circuit. If the circuit doesn't ever seem to trip when the beam is broken, it may need to be incorporated. Use a linear not log pot and set it to the halfway point i.e. 50 kohm The NPN photo transistor conducts when the LED is emitting photons. The current pulses from the photo transistor will be 5 kilohertz pulses i.e. will occur every 200 microseconds. I would leave R3 out and see how the circuit responds. As long as LC satisfies this equation the circuit should work fine f 1/(2*π*√(L*C)) where f is 5,000 hertz √(LC) 1/(6.28*5,000) 31.8 micro seconds LC 1 x 10^-9 s^2 C 1 x 10^-7 farads L 1 x 10^-2 Henry L should be 10 milli Henry not 10 micro Henry So change L to 10 mH, the circuit will not work with a 10 uH inductor.
Q:Simple AC Circuit Analysis of Inductor?
The magnitude of the circuit impedance is equal to the inductive reactance provided by the inductance: L X ωL, but ω 200 rad/s so that X 200(0.25) 50 ohm while V 100 volts the phasor current for the circuit: I V/Z where V is the voltage phasor V 100 angle 0º and the complex impedance: Z jX j50 Therefore: I 100/(jX) -j2 2 angle -90º The phasor current lags the phasor voltage by 90º.
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Location Shenzhen, Guangdong, China (Mainland)
Year Established 2006
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