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: A 15-mH inductor is connected to a standard electrical outlet (rms voltage 112 V, frequency f87 Hz). Determine the energy stored in the inductor at t 7.6 ms, assuming that this energy is zero at t 0. Answer suppose to be in Ji tried v(t) 112 sin (wt)but help please!: V L(di/dt), i (1/L)∫V(t)dt ; this is the current in an inductor If V 112 sin(2π87t) ∫112sin(2π87t)dt -0.2 cos(547t); this is the integral of V(t)dt i -13.3 cos(547t). this is the current in the inductor i -13.3 cos(547*0.0076) -13.26 amps at 7.6 milliseconds Energy in an inductor (1/2)(L)(i^2) E (1/2)(0.015)(-13.26)^2 1.32 joules

Q: An inductor that has an inductance of 24 H and a resistance of 41 ? is connected across a 120 V battery. What is the rate of increase of the current at 0 s. Answer in units of A/s.What is the rate of increase of the current at 0.7 s.Answer in units of A/s.: Back emf,e L.di/dt where L inductor, di/dt is the rate of change of current. And Z^2 R^2 + XL^2 where Z is impedance, R is the resistance and XL is the inductive reactance. Z^2 41^2 + (2pi*60*24)^2 [Assuming line frequency, f 60 Hz) Z^2 1681 + 8985.6 10666.5 Z 103.27 ohms. Now the total current ,I 120/ Z 120/ 103.27 1.162 amps. Again, T L/R where T Time constant 24/41 .58 sec which is one time constant' In this period, the current flows 63% of the full current 1.162 amp .73 amps. For full current of 1.62 to flow is the time, 5T .58*52.9 sec.And discharge current is 37% 1.162*.37 .59 amps, During this period, back emf occurs Hence the back emf,e .59*sqrt8985.6 or,e .59*94.7 55.8 volts at this voltage, then at .7 sec, di/dt .59/.7 .84 amp/sec at 0 sec, di/dt .84/0 infinity that is like a short circuit current. I think this is the answer. Thank you.

Q: A battery is connected in series with a 3.0 ohm resistor and a 12 mH inductor, the max current in the circuit is 150 mA. What is the time constant of this circuit, and what is the EMF of the battery?If the time constant (T) is equal to the time required to reach 63.2 % of the max current, I calculated it to be the time required to reach 0.0948 Amps. T (0.632)(0.00150 A) 0.0948 A. But this is as far as I got. The equation I found that I think I should use is I E/R (1-e ^(-t/T) Please help - thank you: The time constant is simply L/R. That's a standard result (Like RC is the time constant for a resistor/capacitor). Time constant 12x10^-3H / 3.0ohm 4 x10^-3s 4ms When the current is steady, the inductor - if it's perfect - behaves like a short-circuit - i.e. zero resistance. This is because the magnetic field inside it isn't changing so there is no induced emf. The toal circuit resistance is 3.0 ohms and the current is 150mA. So cell's emf IR 150x10^-3A x 3.0ohm 450x10^-3V 0.45V

Q: I am given the following circuit:: This is a simple RL circuit. The Thevenin equivalent of the current source is 1000 volts in series with 1k. (The 2.2k resistor in series with the current source has no effect, because the output resistance of the current source is infinite.) The initial current in the inductor is 5/100.05 amps. The final current is 1 amp. The inductor current changes exponentially between the initial value and the final value with a time constant of L/R. You should be able to determine the equation with the information I have given you.

Q: A battery, switch, resistor, and inductor are connected in series. When the switch is closed, the current rises to half its steady-state value in 2.0ms.How long does it take for the magnetic energy in the inductor to rise to half its steady-state value?My work:the time constant for the circuit is: L/R 2.9ms-I used the equation i(t) E/r ( 1 - e ^ -RT/L)It asks for time it takes for Magnetic Energy of inductor to rise to half its steady-state value.and the only equation i know that works is U .5LI^2 but i cant use it since i dont know L or I.Please help.: U (0.5) LI^2 (1) Question is to find time when it becomes U/2 Now from the above equation, you can find I when U becomes U/2. Let it be I' U/2 (0.5)LI'^2 (2) Taking ratio of (1) and (2), 2 (I/I')^2 I' I/√2 Now we have to find time when I becomes I/√2 given that I becomes I/2 in 2 minutes I/2 E/r [1 - e^(2/2.9)] and I/√2 E/r [1 - e^(t/2.9)] Taking ratio, √2 [1 - e^(t/2.9)] / [1 - e^(2/2.9)] 1 - e^(t/2.9) √2 [1 - e^(2/2.9)] - 1.40 e^(t/2.9) 2.40 t (2.9) ln (2.40) 2.54 minutes.

Q: A probe is attached to the Infrared LED and use the adjustable resistor R1 to calibrate the emitter to transmit a 5 kHz square wave.Data: The emitter circuit on the left produces light pulses at a frequency determined by the variable resistor R1. That resistor is adjusted to tune the frequency to the frequency peak of the narrow-band amplifier of the receiver on the right. That frequency is fixed by the capacitor-inductor tuning circuit (near ICA2).By other definitions: The variable resistor R3 near ICA2 across the LC circuit flattens the respons of the LC circuit and increases the IC2A filter respons bandwidth.So does that mean that I use the variable resistor R3 near ICA2 to modulate a frequency of 5kHz?If not, how is the 5kHz created in the receiver circuit and, what am I suposed to adjust the R3 according to? Is it suposed to be. R3 sets the overal sensitivity but what voltage or something else do I adjust it to?: If I keep in mind wisely, an LC clear out includes an Inductor (L) and Capacitor(C). On schematic diagrams, Inductors are frequently categorised L. The bleeder resistor removes any geared up up can charge left interior the capacitor after the enter skill has been got rid of. once you decrease the enter skill and the present stops, the inductor's magnetic container collaspes. This collaspe induces the capcitor to can charge up back. The bleeder resistor removes this would charge. i'm exceptionally constructive

Q: a) oscillatesb) makes x-raysc) is uselessd) shorts out one or bothe) stops all current: It will oscillate and form a series filter.

Q: I was in a surplus shop with all kinds of inductors, coils, chokes, transformers, etc The variety of these devices is amazing. How are these devices designed? How are they measured?How do the turns/radius/core determine the inductance?They are such basic devices, but I have never seen a standard multimeter capable of measuring it. Why?Just curious how I would know how to measure and design them from scratch?Thanks for any creative answers.: Inductor come in a large variety of forms, depending on their inductance value, current ratings, and applications. They all consist of coils of wire basically, and these coils can be wound around air cores, iron cores, ferrite cores. Some inductors have cores that can be moved in and out of the coil to vary the inductance. RLC meters measure inductance with an AC signal; some mulimeters measure the current rise with time with a constant applied voltage. There are also impedance bridges which match the inductance to reference components. The basic unit of inductance is the Henry. A current rise of 1 ampere per second with an applied constant voltage of 1 volt represents an inductance of 1 Henry. One Henry is a very large inductance, with most inductors valued in the microhenry or millihenry range. To design inductors there are formulas involving coil diameter, number of turns, and permeability of the core. For more information see

Q: Fo 1/2pi x sqr LC: lets take a look at a parallel circuit and assumed R is the parasitic resistance in series with the inductor, what we got is RL//C fo [ 1/2pi ]{ sqrt[ (1/(LC) - (R / L)^2 ] } there is a (R / L)^2 that will affect the resonance frequency fo the construction of the coil ideally R should be minimized, the trick is how? assuming R 0 fo [ 1/2pi ]{ sqrt[ 1 / (LC) ] }

Q: If R 7 ohms L 22 mH and V 95 volts, what is the voltage across the inductor at 1T?: TL/R. If you apply a 95 volt step to a series R L circuit the voltage, across the inductor will be 95 e^(-t/T). At tT the voltage will be 23.92 volts.

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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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