Color Loop Inductor

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4.coil structure:axial

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Q:Capacitor & Inductor ?
A charged perfect capacitor connected to a perfect inductor will have an oscillating voltage: discharge to 0, recharge to initial charge value with opposite polarity, back to 0, etc. forever. This is a sinusoidal oscillation which takes place at a frequency 1/sqrt(LC) rad/s. The reason for the oscillation is explained in the ref., which discusses such a circuit with added resistance. The resistance absorbs energy and causes the oscillation to damp out over time, but the behavior of the capacitor and inductor are the same without resistance.
Q:17: A certain circuit consists of an inductor of 5 mH in series with a resistor of 200. At a moment when?
First find the time constant: T L/R.005/200 take this answer and plug it in this equation; Second find time t: t T LN(Io/If) T LN(19/6.46). Don't answer in scientific notation.
Q:rewiring an inductor with 0.50mm wire?
If you use the same number of turns and length of winding the inductance will be the same. The resistance will be higher.
Q:Inductor connected to AC source?
Use the equation X 2pi*f*L. You are given X indirectly, through V and I. Since I V/X, X V/I. Once you solve for X you can put this in the first equation and solve for f. X V/I 149/.83 179.52 X/(2pi*L) f 179.52/(2pi*.004) 7142.797 Hz
Q:How much energy is stored in the inductor?
PI*R^2 (electric power wiki)
Q:Is this inductor example correct?
The answer is incorrect. The voltage waveform is incorrect. The voltage must remain constant during intervals in which the slope of the current is constant. The voltage should be -2.5V between 2 and 6ms. The stored energy is also incorrect. At t3ms, i 6.25ma; so the energy stored at that time should be 6.25^2 X 2H / 2 39.06 micro-Joules.
Q:A 90 mH inductor, a 60 uF Capacitor, a 6 Ohm resistor and a 60 uF Capacitor are connected in series,find: Zt?
X of 60uF 1 /2pi fC 37.894 Ohm (0 - j37.894) Ohm X of 90mH 2pi fL 39.584 (0 + j39.584) Ohm R 6 Ohm (6 + j0) Ohm Zt R + XL - 2XC (6 - j36.2) Ohm 36.7 |-_80.7` I am a little shaky on keeping up with the angles from here on so I desinated the angles as I don't know. Maybe it will give you an idea on how to proceed. i V/Z 141.4213V/36.7 Ohm 3.8534 Amps |_?` V of L (i)*(XL) (3.8534A)*(39.584 Ohm) 152.535 V |_?` V of L in rectangular form [152.535 cos |_?` + j152.535 sin |_?`] Volts
Q:HOW WOULD YOU BUILD A 0.1H INDUCTOR?!?
Let me split your question into two parts. How do you build an inductor? How to make it 0.1H? Building an inductor is very simple. Wind a few turns of wire and you have an air cored inductor. That means, you basically need two things to build an inductor. A conductor having a certain number of turns and a core. The core can be air, silicon steel, nickel, soft ferrite, hard ferrite, powdered iron How to make it 0.1 Henry (100mH)? So an inductor has a certain inductance. The inductance depends on the number of turns and the material of the core. Apart from the two, you need to know the following factors: 1. Current through the conductor, 2. Type, size, area and permeability of the core 3. The frequency of operation. The current is important, because it should not saturate the core. Read info regarding the B/H curve. To keep it simple, for a single layer AIR CORED INDUCTOR the formula is: L ((r^2)(N^2)) / (9r+10l) where L is inductance in microhenrys, r is radius of coil in inches, l is the length of coil in inches, and N is the number of turns in the coil.
Q:Inductors and Inductive Reactance?
ok for the first one: reactance is 2pifL Xl reactance Solve for f which is frequency b. then u use 1/2pifC Xc for part b, and use this equation to get C c. Just do what it says for c and d. and use the above equations.
Q:Does voltage have any affect on the E.M. field that it produces when it puts a current through an inductor?
The variable that you haven't mentioned is the resistance of the inductor. At steady-state (DC), the voltage needed will be the current multiplied by the resistance. The equations for magnetic field rely only on current for simple structures, and voltage enters the picture through the resistance for DC circuits. In an AC circuit (which is what your water pump is), the current is usually established by the inductance and the amplitude and frequency of the voltage source. The resistance usually represents an undesired loss, but it will affect the current amplitude. In your water pump, you are trying to deliver mechanical power to a load (the water pressure), and this is represented as a resistance in the motor equivalent circuit. But instead of being a loss, the resistance models the delivery of power to a mechanical load. An unloaded motor draws relatively little current compared to full load, and the current that it does draw is out-of-phase with the applied voltage, which results in little real power actually consumed. Under no-load conditions, most of the power consumed is caused by the line current flowing in the winding resistance. As the mechanical load on the motor is increased, the line current increases, and becomes more in-phase, resulting in significant power consumption, although most of the power is not consumed in the motor, but actually delivered to the load. Your water pump has two sets of windings, and these are either wired in series, or parallel to determine the voltage and current needed. The magnetic field is established by the number of turns of a coil, multiplied by the current in the coil. The power rating of the motor is the constant, so at 230V, the current drawn would be half of that drawn at 115V. So at 230V, the number of turns in the coil is doubled, and the current is halved, resulting in exactly the same peak magnetic field amplitude as the 115V case.
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Location Shenzhen, Guangdong, China (Mainland)
Year Established 2006
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