SMD Multilayer Ferrite Chip Bead 0805

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1.SMD multilayer ferrite chip bead

2.High Impedance Characteristics

3.size: 0805

4.RoHs

5.competitive price

Features:

*Low crosstalk/DCR, high reliability

*Low crosstalk between adjacent circuits

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

*Internal electrodes feature low DC resistance, minimizing wasteful power consumption

*Electroplated terminal electrodes accommodate flow and reflow soldering

*Mololithic structure ensures high reliability

*Operating temperature range from -25 to 85℃

Applications:

*High-frequency noise counter measured in computer

*Printers

*Portable telephones and other equipments

*VCRs

*Televisions

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 student wants to figure out if an unlabeled item is a capacitor or inductor. He applies a voltage with diferent frequencies and find that as the frequency goes up, the current through the item goes down. Is this a capcitor or an inductor? Why? Thanks for the help! Please explain in detail.: An easy way to remember the answer is to apply DC (frequency 0) and see what happens: In inductor is simply a coil of wire. If DC is applied, it will be nearly a short circuit, and very high current will flow. As the frequency increases, the current will get lower. A capacitor is simply 2 parallel plates that don't touch, separated by a dielectric. If DC is applied, you will have an open circuit and no current will flow. As the frequency increases, the current will also increase.

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?: We first need the time at half maximum current. I (V/R)[1 - e^(-Rt/L)] Max current V/R so V/(2R) V/R[1 - e^(-Rt/L)] e^(-Rt/L) 0.5 -Rt/L ln(0.5) t -L.ln(0.5)/R -0.8 ln(0.5)/41 0.014 s Magnetic energy stored: E (1/2)LI^2 dE/dt LI dI/dt . dI/dt (V/L)e^(Rt/L) 3.05, t 0.014 dE/dt {( LV)/2R} x 3.05 0.15 Js^-1 Electrical power dissipated P I^2 R (V^2/4R^2)R 0.15 Js^-1 Power drawn from battery 0.15 + 0.15 0.3 W

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.: This is a step-value function and will need to be solved piece-wise. For t 0 to 5us: Since I (1/L) Vdt therefore:: I (1 / 2E-3) * 10V * 5us 500 * 10V * 5us 25mA For t 5us to 15us I (1 / 2E-3) * -10V * 10us 500 * -10V * 10us -50mA For t 15us to 20us I (1 / 2E-3) * 10V * 5us 500 * -10V * 5us 25mA The total current from t 0 thru t 20us 25mA -50mA +25mA 0 mA

Q: I am making an inductor coil. But I don't have any enameled copper wire, I just have solid copper wire, will this work as a improvisation?: It will work provided you keep the turns apart. Depending on how big your coil has to be you could use a hair comb, or two glued back to back, as a coil former to hold the turns apart by winding the turns into the spaces between the teeth.

Q: Hi, I'm working on homework. The circuit has three inductors, L1, L2, amp; L3, in parallel with a resistor, R. I already know to calculate the total inductance when the inductors are in parallel. The problem is part B of the question has added another inductor in series with L1, L2, L3, amp; the load, R.I don't know how to figure L total in a series-parallel inductor combination circuit.: JUST ADD L4 VALUE TO THE LAST RESULT.

Q: If the voltage waveform is applied across the terminals of a 5-H inductor, calculate the current through the inductor. Assume i(0) -1A.: since t 0.use equation I(t) Im * e ^ (-tR/L) if it sdischarging or if its charging, replace (e^(-tR/L)) with (1-e^(-tR/L)). It depends on the circuit though

Q: Assume that we set up a circuit with a resistor, a capacitor and an inductor all in serial. In which condition is the magnitude of the current is maximum over the resistor?: The current is a maximum at resonance, when the reactances of the cap and the inductor cancel out. But the current will not exceed the value you would get with the resistor alone. resonance frequency: f 1/(2π√(LC)) .

Q: can anyone give out any ideas of what happens upon switch is closed rather than just charging ? any ideas would be greatly appreciated. here's the diagram: a) in position A, the C will charge up towards 8 volts with a time constant of 560 x 22u 12 ms. After 5 or so time constants, it will be very close to 8 volts. In position B it will discharge towards ground with a time constant of 820 X 22u or 18 ms. After a while it will be close to zero volts. .

Q: Can anyone tell the different types of inductors.: 1 il inductor 2.parallel inductor 3.toroidal inductor are the some types of inductor

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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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