SMD Multilayer Ferrite Chip Bead 0805

Ref Price:

Loading Port:: China Main Port

Payment Terms:: TT or LC

Min Order Qty:: 4000 Pieces pc

Supply Capability:: 4000 Pieces per Day pc/month

Inquire Now

Add to My Favorites

OKorder Service Pledge

Quality Product, Order Online Tracking, Timely Delivery

OKorder Financial Service

Credit Rating, Credit Services, Credit Purchasing

You Might Also Like

Latest Auto 90W Slim Universal Adapter with 8 DC Tips

Home Office Appliance Single Modified Sine Wave 200 Watt 300 Watt UPS Plug

AC Motor Drive

MV Voltage 3KV 6KV 10KV Frequency Inverter

High Voltage 11KV Variable Frequency Drive

AC/DC Adapter Charger 5V 2A for PC/Mobile Phone CE GS Certificat

Variable Frequency Drive 3.3KV 630KW VFD

Single Crystal Furnace HF Switching Power Supply

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: 2.In the figure, after the switch is closed at time t0, the emf of the source is automatically adjusted to maintain a constant current I through S. a) Find the current through the inductor as function of time. b) At what time the current through the resistor equal to the current through the inductorthe circuit looks like this:a constant current source with a switch in series and then splits into a resistor and inductor in parallel: the means in an inductor is proportional to the oblong of the present. At area the present it holds a million/4 the quantity of means. It held 40 J on the commencing, it now holds 10 J so the distinction is 30 J

Q: A 100 mH inductor is in series with a resistance R 1.1 Ω.If a current I Acos(ωt) is applied, to this circuit, find the frequency at which:The voltage amplitude across the resistance is 1% of the voltage amplitude across the inductor.Please show me how to do this question with the working out and explanation because i have tried for hours and am getting very annoyed i really need help.thanks: It is where I * R I * ω * L / 100 or R 2 pi f * L / 100 or f 100 * R / (2 * pi * L) 110 / (6.2832 * 0.10) 175.1 Hz

Q: 1)50v 1 khz 2)50v 100hz 3)50v dc: 3. Because 1 and 2 are AC which introduce inductive reactance into circuit, ( impedance, or AC resistance) which will reduce the voltage across the lamp and current in the circuit. Inductive reactance XL 2pi x L x F. Where L is inductance and F frequency. So the greater the frequency, the greater the reactance XL. AC resistance (Impedance Z) sq.rt (R sq + xL sq + xC sq). xC Capacitive reactance.

Q: A circuit containing a 39-H inductor has an alternating current supplied to it. If is and is find the frequency of the signal.0.033 Hz0.41 Hz0.0052 Hz0.21 Hz: Look at the question again- you will see it des not make sense. To find the frequency we need to know the series resistance of the inductor, the voltage across the component and the current through it.

Q: A series circuit consists of a 0.45 H inductor with internalresistance 6.0 ohms connected in series with a 8.0 ohm resistor, aswitch, and a 12 volt battery. Twenty milliseconds after the switch isclosed, how much energy is stored by the inductor?I understand that energy 1/2LI^2 Lmagnetic flux/I energy1/2 (magnetic flux)*Iand that’s as far as I can get. Any help?: The charging time constant T? L ? R (0.45 H) ? (14 ?) 32.1 msec ?мax (12 V) ? (14 ?) 0.857 Amps ???i (?мax) ? [1 ? e^ˉ(t ? T?)] ???i (0,857) ? [1 ? e^ˉ(0.020 ? 0.0321)] ???i 0.4 Amps ???????????????Energy (?) ? L ? i? ???????????????Energy (?) ? (0.45) ? (0.4)? ???????????????Energy 0.036 Joules

Q: A 1.24-?H inductor is connected in series with a variable capacitor in the tuning section of a short wave radio set. What capacitance tunes the circuit to the signal from a transmitter broadcasting at 6.40 MHz?: frequency 1/(2 pi sqrt(L C)) (where L is the inductance) check it out here

Q: A capacitor C charged to voltage V is discharged into an inductor L. What is the voltage on C at the instant when its stored energy and inductor's energy are equal?: energy stored in charged capacitor is 1/2(c*v^2) and energy stored in inductor is given by 1/2(l*i^2).so when the energies are equal then vsquareroot(l*i/c)

Q: if i have coupled inductors with mutual inductance M ,how would i go about replacing them with 1 equivilant inductor and what would be its value ?: An equivalent single X value for the two initial inductances X1and X2, having mutual inductance M, very much depends on how X1 and X2 are connected into a circuit and upon how X is to be connected. There is not a unique answer independent of circuit connection.

Q: how to calculate SRF of a inductor?: Inductor Srf

Q: A 16.0 mH inductor, with internal resistance of 20.0 ohms, is connected to a 110 V rms source. If the average power dissipated in the circuit is 40.0 W, what is the frequency? (Model the inductor as an ideal inductor in series with a resistor.) ____ HzI keep getting 467 Hz but this is incorrect and have no idea what I am doing wrong. I've tried doing it a bunch of different ways but keep getting lost somewhere along the way.Any suggestions on how to accomplish this problem?: 1 mH 10^(-3) H current power / voltage 40 W / 110 V 0.3636 Amps voltage Henries * current / time 110 V 1.6 * 10 ^ (-2) H * 0.3636 Amps / time Isolate time: time 1.6 * 10^(-2) H * 0.3636 Amps / 110 V 5.2887 * 10^(-5) seconds frequency inverse of time 1/(5.2887 * 10^(-5)) 18908 Hz I probably made a mistake in there somewhere. I don't remember how to factor in internal resistance or if that's even necessary for this problem.

Our products have highest quality and competitive prices.Our well-equipped facilities and excellent quality control throughout all stages of production enable us to guarantee total customer satisfaction. As a result of our high quality products and outstanding customer service, we have gained a global sales network.

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

2. Manufacturer Certificates

a) Certification Name
Range
Reference
Validity Period

3. Manufacturer Capability

a) Trade Capacity
Nearest Port
Export Percentage	41% - 50%
No.of Employees in Trade Department
Language Spoken:
b) Factory Information
Factory Size:
No. of Production Lines
Contract Manufacturing	OEM Service Offered Design Service Offered Buyer Label Offered
Product Price Range