• Ferrite Chip Inductors System 1
Ferrite Chip Inductors

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:antenna ,is it act like an exitation source for loop antenna to generate oscillations.plz answer it
An inductive coupling link/loop can be used with a loop antenna. This works in the same way as a transformer, with the inductive link as the primary (for transmitting), but at the particular radio frequency. A loop antenna often represents some sort of resonant circuit, and this can be combined with the inductive link coupling to match the antenna impedance to the transmission line and transmitter impedances. With other types of antenna an inductor and/or capacitor arrangement (an LC network) may be used for matching the antenna impedance to the lines. The actual values and conficuration depends on the specific antenna. For example, teh model of a vertical quarter wave antenna is a small capacitor in series with the radiation resistance. By adding a series inductor the capacitance can be made to resonate, maximising the current in the radiation resistance.
Q:A 10 ohm resistor and 10 henry inductor are attached in series to a 10 volt DC supply. If the current i(0)2.5 amperes, what is the current after 1 second? Round in the tenths place.I got 1.55 but that isn't right.
What do you want to know?
Q:I need help answering this question i have no idea how to do the steps if you could lead me through the steps or tell me what to do that'd be fantastic.An inductor is fabricated with 500 turn of wire, a diameter of 1 centimeter and a length of 2 centimeters, and an air core with a permeability of 1.25X10^-6 H/m. A DC current of 500 mA is made to flow through the inductor.Find:A) The inductanceB) The energy stored in the field:C) The energy stored in the field if a steel core, with apermittivity of 4000 is inserted into the coil.
A)From physics, you know that: B μNI / h where μ is the permeability h is the height of the loop B is the magnetic flux N number of loops B 0.03125 wb An the inductance L is equal to L B/I L 0.0625 H B) Energy stored 1/2 LI^2 0.007813
Q:A 4 mH inductor is connected to an AC voltage source of 149 V rms. If the rms current in the circuit is 0.83 A, what is the frequency of the 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 does the energy gets stored in the inductor if we are connecting it to an rlc circuit either in parallel or in seriesi also wanted to ask about the phenomena of resonance ocuring in this circiut in details,i know that when the inductor has potential energy the capacitor doesnt have and vice versa,thats why its called electrical resonance,but i want someone to tell me the reason of this phenomena,an athountic answer please.thank u buddy
RLC Circuits are resonant circuits consisting of a resistor (R), a capacitor (C) and an inductor (L). The resonant frequency is given (in radians per second) : w1 / sqrt(LC) or in Hertz fw/2*pi 1/ (2*pi*sqrt(LC)). Resonance occurs when the complex impedance ZLC of the LC resonator becomes zero. The impedances: Zc1/jwC ; ZLjwL , where j^2-1 Attenuation is : a1/2RC (for the series and for the parallel configurations) For further study, you have to get a solid knowledge in Laplace and Fourier transforms.
Q:Hi, can somebody please help me with this inductor problem? I'm stuck on it and need some help i think that I have the general idea, but need some guidance.
I just told you how to do this problem. What don't you understand about it?
Q:guys is a axial inductor of value 56nH,1mH and 100nh available,can we use axial inductor to make LC circuits.
by axial, do you mean wound in a straight line, for example, using a resistor as a winding form? You can get this type of inductors in a wide range of values, from very small to medium sized. 1 mH may be difficult to implement this way, those are usually found on a donut shaped core. Yes, you can use them in resonant circuits. .
Q:A 9.5 V battery, a 4.92 resistor, and a 9.7 H inductor are connected in series:(a) After the current in the circuit has reached its maximum value, calculate the power being supplied by the battery.(b) Calculate the power being delivered to the resistor.(c) What is the power being delivered to the inductor?(d) What is the energy stored in the magnetic field of the inductor?
Once the current has reached its max. then all the power will go to the resistor, so: a) p v^2/r 9.5^2/4.92 b) see a) above c) 0 d) I don't remember these equations, sorry.
Q:An RLC circuit has a 400 ohm resistance a 2.5 mH inductor and a capacitor. If the circuit is in resonance at 30kHz and is attached to a 20mVrms power supply, what is the power dissipated by the circuit?
At resonance +jXL -jXC, therefore Z R +jXL -jXC R 400 ohms, since the reactances canel out. The power is only disspiated in the 400 ohm resistor, so P V^2 / R P (0.020V)^2 / 400 P 1E-6 Watt P 1 microWatt ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ EXTRA CREDIT The capacitor value for resonance with the 2.5 mH inductor at the frequency of 30 KHz will be f 1 / (2 * pi * SQRT (L * C) ) 30E3 Hz 1 / (2 * 3.1415 * SQRT (2.5E-3 H * C) ) 2 * 3.1415 * SQRT (2.5E-3 H * C) 1 / 30E3 Hz SQRT (2.5E-3 H * C) 5.3052E-6 (2.5E-3 H * C) (5.3052E-6)^2 2.5E-3 H * C 28.1448E-12 C 11.2579E-9 F C 0.0112579 microFarads (uF) 11.2579 nanoFarads (nF) 11,257.9 picoFarads (pF)
Q:A certain circuit consists of an inductor of 55 mH in series with a resistor of 90 Ω. At a moment when the current in the circuit is 21 A, a switch in the circuit is opened. How long will it take for the current to fall to 7.14 A? Answer in units of s.I thought that this one was easy. first time around I used i(Io)(e^-Rt/L) --- 7.1421e^(-90t/(55x10^-3)) 2.53926105e-4this is wrong. Then based on someone's advise I got on here, I used the equation i(Io)e^(t/-RL) and I got an answer of 5.34.also wrong! Please help! What am I doing wrong? What's the answer??
7.14 21 e^ - (R/L)t e^-(90/0.055) t 0.34 e^- (1636.36 t) 0.34 ln both sides : - 1636.36 t - 1.0788 hence t 6.59 x 10^-4 s
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Location Guangdong,China (Mainland)
Year Established 2010
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