SMD Multilayer Ferrite Chip Bead

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low frequency Inductor/filter

1.SMD multilayer ferrite chip bead

2.Low Crosstalk/DCR Features

3.competitive price

4.by customer's requests

Features:

1.Low crosstalk/DCR, high reliability

2.Low crosstalk between adjacent circuits

3.Single MZA series chip provides noise attenuation for four lines, ideal for various highly miniaturized I/D lines

4.Internal electrodes feature low DC resistance, minimizing wasteful power consumption

5.Electroplated terminal electrodes accommodate flow and reflow soldering

6.Monolithic structure ensures high reliability

7.Operating temperature range: from -25 to 85 degree Celsius

Applications:

1.High-frequency noise counter measured in computer

2.Printers

3.Portable telephones and other equipments

4.VCRs

5.Televisions

Q: An inductor has an impedance of 30.0 Ω and a resistance of 20.0 Ω at a frequency of 80.0 Hz. What is the inductance? (Model the inductor as an ideal inductor in series with a resistor.)?: Impedance for an ideal inductor is: Z j*omega*L For a real inductor with a parasitic resistance modeled in series: Z R + j*omega*L j refers to the imaginary number unit. omega is the angular frequency of the signal. We were given the cycle frequency and we can convert: omgea 2*Pi*f I assume your 30 ohms refers to the magnitude of impedance. Combine both components of the complex number using the Pythagorean theorem. Zmag sqrt(R^2 + (omega*L)^2) And with a substitution: Zmag sqrt(R^2 + (2*Pi*f*L)^2) Square both sides: Zmag^2 R^2 + (2*Pi*f*L)^2 Solve for L: (2*Pi*f*L)^2 Zmag^2 - R^2 2*Pi*f*L sqrt(Zmag^2 - R^2) Resulting expression: L sqrt(Zmag^2 - R^2)/(2*Pi*f) Data: Zmag:30.0 Ohms; R:20.0 Ohms; f:80.0 Hz; Result: L 0.04449 Henries Or, with a convenient prefix: L 44.49 milliHenries

Q: A 24 mH inductor is connected to a outlet where the rms voltage is 123 V and thefrequency is 60 Hz.Determine the energy stored in the inductorat t 5.2 ms, assuming that this energy iszero at t 0.Answer in units of J.Please help!!!: Energy stored in the inductor 1/2 L * I^2

Q: Hello, i really need some help solving. please show all steps. i really appreciate it! thanks in advance.An 8.0-mH inductor and a 2.0-Ω resistor are wired in series to a 20-V ideal battery. A switch in the circuit is closed at time 0, at which time the current is zero. After a long time the current in the resistor and current in the inductor are: A) 0,0B) 10A, 10AC) 2.5A, 2.5AD) 10A, 2.5AE) 10A, 0: for an ideal inductor the impedance for D.C.(supplied by battery) is zero as (angular frequency omega is zero). now when such a combination is connected to D.C transient current flows into the circuit. and the value of current at any instant is given by. IV/R(1-e^(-Rt/L)) as t is long the current is given by, lim t--infinity V/R(1-e^(-Rt/L)) V/R(1-(1/e)^infinity) as 1/e 0, (1/e)^infinity tends to zero. so IV/R(1-0) IV/R20/210 Amperes note that transient current either growing or decaying lies for a very short while. so if long time is given. you can directly put impedance of ideal inductor as zero and of ideal capacitor as infinite. hope this helps. so answer is B) 10A, 10A

Q: A coil has an inductance of 0.80 H and a resistance of 41 . The coil is connected to a 5.0 V ideal battery. (a) When the current reaches half its maximum value, at what rate is magnetic energy being stored in the inductor? (b) When the current reaches half its maximum value, at what rate is energy being dissipated? (c) When the current reaches half its maximum value, what is the total power that the battery supplies?: An inductor might no longer likely keep magnectic capability. It converts electric powered capability flowing with the aid of it to magnectic capability. If the electricity flow is stopped, the magnectic field colapses (like Jis4Jenius Jerald pronounced above). it extremely is termed an inductor using fact a changing magnetic field around the cord (or coil) induces electricity into it.

Q: here's the text which contains it,Referring to figure, it can be seen that no power is disspated in a pure inductor. In the first quarter of cycle, both V and I are positive so the power is positive, which means that energy is supplied to the inductor. In the second quarter, V is positive but I is negative. Now power is negative which implies that the energy is returned by the inductor(The figure looks something like this) ::: Yes, you're correct. Your picture shows the correct phase relationship between voltage and current in an inductor, and in that case your description is correct.

Q: 1. At what frequency will a 32.0-mH inductor have a reactance of 660 Ω?2. What is the reactance of a 9.2-μF capacitor at a frequency of (a) 60.0 Hz, (b) 1.00 MHz?3. A 3800-pF capacitor is connected in series to a 26.0-μH coil of resistance 2.00 Ω. Whatis the resonant frequency of this circuit?4. Determine the total impedance, phase angle, and current in an LRC circuit connected to a 10.0-kHz, 1025-V (peak value) source if L 32.0 mH, R 8.70 kΩ, and C 6250 pF.: I'll give you hints, not answers. 1. Z (impedance) of an inductor j2(pi)(f)L 2. Z of a capacitor 1/(j2(pi)(f)C). 3. Set Z capacitor equal to Z inductor and solve for f. But this time, add '2' (without a j) to the impedance of the inductor. 4. Use circuit analysis using the Z's given above. The phase angle is calculated as follows. Ztotal a+Bj phase angle arctan(B/a). Make sure the angle corresponds to the quadrant a and B are referring to. Good luck.

Q: I have 2 answers in terms of Joules, but not sure which one is correct.: At DC the current through the inductor will be 3*4/62 amps. The energy is .5*L*I^2.5*.5*2^21 Joule.

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.Okay, easy smeasy I thought. i made an expression 7.1421(1-e^(-90t/55*10^-3))Simply algebra, so I got t2.53926x10^-4. This is wrong! please show me the errors of my ways. Do the entire problem. Show work. I need to get this down to I can tackle the tougher problems.And I based my expression off of the equation i(Io)(1-e^(rt/L)). Thanks! will choose best answer!: This is easy. You used the wrong formula. The formula is for current INCREASING from zero and asymptoting to Io. But it isn't accurate even for that. The actual formula is i Io (e ^- t/(rL)) Note the features. The larger the value of r the less that is lost (or gained ) per second. The larger the value of L the less that is lost (or gained) per second so these must be in the denominator. The larger the time the more that is lost so it must be in the numerator. The negative sign in e ^ - x means that as x becomes larger ( more time taken) the answer becomes smaller. If we had 1 - e^ -x then it would asymptote to 1 not to zero. This would be the case for increasing the current to a maximum.

Q: An inductor of 0.5 H and a resistance of 1.2 K ohms are connected in series with an ideal source with e.m.f. 6 V. The current expression in the circuit as a function of time is: 5.0*10^-3(1-e^-2398t ) . Calculate the potential difference across the inductor at t 0.417 ms. the correct answer is: 2.21v . can you tell me how to do it? :): Set t 0.417*10^-3 in the formula given. That gives you the current at t 0.417 ms ie 3.16*10^-3 ma In the 1.2k resistor, that drops v i*r 1.2*10^3*3.16*10^-3 volts ie 3.79 volts That leaves (6 - 3.79) volts ie 2.21 volts across the inductor.

Q: Can anyone tell the different types of inductors.: Your question is vague, but I will try to cover things. Plain inductors, which is a coil of wire, can be air core or wrapped around some metallic core. Metallic cores vary when developing audio inductors for noise suppression. Sometimes ferrite beads are all that is necessary which is a bead over a wire - an inductor with one turn. Inductors can be used to vary the phase of the current. They can be used as current limiters by way of their inductive reactance. When inductors are coupled you get transformer action, which brings up a whole topic of various transformers.

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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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a) Trade Capacity
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Export Percentage	41% - 50%
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