Ferrite Chip Inductors

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1.ferrite chip inductors

2.leaching resistant rerminations due to metal teb electrodes

3.Coils encapsulated in heat-proof

Features

1.Ferrite chip inductors

2.leaching resistant rerminations due to metal teb electrodes.

3.Coils encapsulated in heat-proof resin make high accurate

4.dimesions and resostant dimensionsand resostant to mechanical shock or pressure.

5.High resistance to heat and humidity.

Applications

Personal computers. Disk drives and comuter peripherals. Telecommunications devices. VCD, DVD and TV circuits,

Test equipment. Electronic control control boards for automobiles.

Q: If the design of an inductor calls for 80 turns of .5mm wire around a 40mm OD iron torroidal core, how do I calculate the inductance in mH or uH?This isn't a homework problem, I'm building a circuit that calls for this inductor and I want to know what the resulting inductance is so that I don't have to build the inductor myself. I want to purchase one pre-wound.: Inductance is given by Luo ur N^2 Ae/le where uo4 pi E-7, urrelative permeability, Nnumber of turns, Ae is cross sectional area of core [m^2], lelength of core [m].

Q: A 0.56-m H inductor stores 3.5×10-5 J when carrying a DC current. What is the magnitude of that current?: 3.5×10-5 J (1/2) 0.56 x10^-3 i^2 i 0.356 A

Q: Cant seem to get my head around this question so any advice would be greatA circuit with an inductor(with both resistance and inductance) Vrl (200+j300)V connected in series to a capacitor c150 micraF across a power supply with a 50Hz supply voltage. The supply current is 25lt;0 degreesCalculatea) impedance of the inductor expressed in complex formb) the values of R and L of the inductorc) the value of the supply voltage, expressed in polar formd) the phase angle of the circuite) the apparent and active powers of the circuitf) draw a scaled argand diagram showing the supply current and all voltages: you better to see the transmission lines and waveguide book

Q: If ig(t) 0.5 cos 2000t A, find the average power absorbed by each element ((a) 130-Ω resistor, (b) 40-Ω resistor, (c) source, (d) inductor, (e) capacitor) in the circuit in the figure below.: Let's just go through it the straight way. You know, from your source, that: ? ? ? ? ω 2000 ? ? ? ? Ip ?A, ∴ Irms Ip ? (√2) You can compute the reactance of your capacitor (C12.5μF) and inductor (L60mH) as: ? ? ? ? XC 1 ? (ω?C) 40?, ∴ Z?C -40j ? ? ? ? XL ω?L 120?, ∴ Z?L 120j Let's say that: ? ? ? ? R? 130? ? ? ? ? R? 40? You have two parallel legs, each with a series pair of devices. The total impedance is in the capacitive leg is: ? ? ? ? Z? R? + Z?C 130 - 40j, ∴ |Z?| 10√185 In the inductive leg it is: ? ? ? ? Z? R? + Z?L 40 + 120j, ∴ |Z?| 40√10 So the total parallel impedance of these two legs is: ? ? ? ? Ztot Z??Z? ? (Z? + Z?) 200?(141 + 79j) ? 353 ? ? ? ? ∴ |Ztot| ≈ 91.57112 You know that: ? ? ? ? Vrms Irms ? |Ztot| ? ? ? ? Irms? Irms ? |Ztot| / |Z?| ? ? ? ? Irms? Irms ? |Ztot| / |Z?| ? ? ? ? P? Irms???R? (Irms ? |Ztot| / |Z?|)? ? R? 7??????? W ≈ 7.365 W ? ? ? ? P? Irms???R? (Irms ? |Ztot| / |Z?|)? ? R? 2??????? W ≈ 2.620 W ? ? ? ? Ptot P? + P? 9??????? W ≈ 9.986 W Real power isn't dissipated in ideal capacitors or inductors, so (a) ≈ 7.365 W, (b) ≈ 2.620 W, (c) ≈ 9.986 W, (d) 0 W, and (e) 0 W. Since the problem you show doesn't illustrate VAR units on it, I don't think (d) and (e) should use reactive power. ? ? ? ? P? Irms???R? (Irms ? |Ztot| / |Z?|)? ? R? 7??????? W ≈ 7.365 W ? ? ? ? P? Irms???R? (Irms ? |Ztot| / |Z?|)? ? R? 2??????? W ≈ 2.620 W The above was validated with Spice.

Q: Any idea please. Thanks: Read this in detail.

Q: as we know ,the resistor of wire is called resistor but for inductor it is called induction why?: The RESISTANCE of wire makes a resistor, the current running through a wire makes the wire an inductor. The inductor is usually a coiled wire when used for practical purposes.

Q: at which the maximum current flows. Explain briefly.: Have u lost ur mind???,,,,I do know your feeble attempt .but grow up and contact me if you need answers.

Q: and what are the possible components.: An inductor is a component that is made from a coil of wire. If a DC current is applied to it, it will pass straight through, since it's just a bunch of wire. However, when the current passes through the wire, it will create a magnetic field around the wire. This will not do anything to a DC current, but if it's an AC current, where the curernt direction is changing and the amount of current rises and falls with the voltage, that magnetic field will also expand and contract. If the wire was straight, this would have no effect, but since it's coiled, the magnetic field caused by the current flow will pass through other coils of wire in that same inductor. Remember that a current flowing through a wire creates a magnetic field? The reverse is also true - a magnetic field crossing a wire will create a current flow, just like in a generator. As the AC current rises and falls in the inductor, the magnetic field expands and contracts, crossing the wires of the inductor, and creating its own currents, which interfere with the input current, which creates resistance to that AC current. The end result is this - a DC current passes through an inductor with no resistance, but an AC current passing through causes resistance. Because the resistance is only for AC, it isn't measured in Ohms, it's measured in Henries.

Q: Hello, i need some help solving. please show steps. thanks in advance. A 3.5 mH inductor and a 4.5 mH inductor are connected in parallel. The equivalent inductance is:A) 2.0 mHB) .51 mHC) .13 mHD) 1.0 mHE) 8.0 mH: B. Like with parallel resistors, the parallel inductors will result in an harmonic average inductance: 1/L 1/(3.5mH) + 1/(4.5mH)

Q: A parallel RLC circuit consists of a 3.7 mH inductor, a 4.7 ?F capacitor, and a 1.4 ? resistor driven at 150 Hz:1.) Calculate the reactance of each component at the driving frequency (f 150 Hz, but youprobably want ω).2.) Calculate the impedance Z for this circuit. Remember that you need to use the AC versionof Ohm’s law appropriately, since the current is common.: L 3.7 mH C 4.7 μF R 1.4 Ω f 150 Hz Reactance of i) Capacitor X_c 1/(2πfC) 1/(2*3.14*150*4.7*10^-6) 225.87 Ω ii) Inductor X_? 2πfL 2*3.14*3.7*150*10^-3 3.4854 Ω iii) Resitor X_r R 1.4 Ω ___________________________ Impedance of circuit Z sqrt((X_? - X_c)^2 - R^2) Z sqrt((3.4854 - 225.87)^2 - 1.4^2) 222.38 Ω

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1. Manufacturer Overview

Location	Guangdong,China (Mainland)
Year Established	2010
Annual Output Value	US$10 Million - US$50 Million
Main Markets	North America; South America; Eastern Europe; Southeast Asia; Africa; Oceania; Mid East; Eastern Asia; Western Europe
Company Certifications	ISO 9001:2000

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Export Percentage	41% - 50%
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