PBH Series SMD Power Inductor

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1. pbh series smd power inductor

2. Rated current: 1-10A

3. Inductance0.5~6000uH

4. ROHS

5. competitive price

Features:

1.SMD Power Inductor

2.Magnetic shieled surface mount inductor with high current rabing low D.C resistance

3.Excellent terminal strength

4.Packed in embossed carrier tape and can beused by automatic mounting machine.

5.Various hogh power inductors are superior to be high saturation for suiface mounting.

Applications:

Power supplu for VTR,OA equipment Digital camera, LCD television set notebook PC,

portable communication equip,ents, DC/DC converters, etc.

Q: Help! I am having trouble trying to compute the inductor current changes in a parallel R (1Kohms), and L (1mH). They are being driven by a current source of 1A. How do I verify that the current flows in the inductor changes by 63% of its initial value in one time constant.Also, How to,1. Analytical derivation of voltage and current across L2. How the specific values of R and L affect the transient behavior of voltage and current across the inductor3. Compare the analytical solution with the numeric results (I used LT SPICE to successfully obtain the voltage and current waveform across R and L)any help on any of the questions, particularly, the 63% question and Q1 and Q2 would be greatly appreciated. Thanks.: For this circuit, the current of the source is equal to the current through the resistor plus the inductor current. Therefore Ise/R+(1/L)integral(e dt), where e is the voltage across the current source. I'm not into solving equations like this any more, but the solution will be something like eI R e^-t/T where TL/R. e^-1.368 and 1-.36863.2%.

Q: A battery (V-9V) is connected with a resistor R1000 ohms and an inductor L 0.5mH. The circuit initially contains an open switch and at t 0, the switch is closed.a) what is the time circuit?b) Determine the Maximum current flowing in the circuit: The time constant for an RL circuit L/R (remember that L*Amps/second Amps*R so L/R has units of seconds). L/R 0.5e-3/1e3 0.5e-6 s R/L 2e6 2,000,000 After a long time an inductor is modeled as a short circuit when the current is a maximum V/R V/R 9/1000 9e-3 A 9 ma. iL(t) 9ma*[1-e^-t/(L/R)] 9ma*[1-e^-Rt/L] 9ma*[1-e^-2,000,000*t] After 5 time constants 2.5e-6 e^-2e6*2.5e-6 e^-5 iL(5*L/R) iL(2.5e-6) 9ma*(1-e^-5) 8.94ma which is 99.33% it's maximum value after only 2.5μs

Q: A 0.595 H inductor is connected in serieswith a ?uorescent lamp to limit the currentdrawn by the lamp.If the combination is connected to a57.3 Hz, 93.5 V line, and if the voltage acrossthe lamp is to be 26.9 V, what is the currentin the circuit? (The lamp is a pure resistiveload.)Answer in units of A.Please help!!!!: Calculate inductive reactance by formula X(L) 2 (pi) f L 2 *3.14 * 57.3 * 0.595 214.57 ohm Applied voltage 93.5 v. By Kirchoff;s law, this equals voltage drops in R and XL I*XL + I*R93.5 But these 2 voltages have phase difference of 90 deg so they must be added vectorially we know I*R 26.9 I*XL (93.5 ^2 -26.9^2)^0.5 i*XL89.54 i89.54/214.57 0.41733 Amp

Q: A variable inductor with negligible resistance is connected to an ac voltage source. By what factor does the current in the inductor decrease if the inductance is increased by a factor of 8.0 and the driving frequency is increased by a factor of 4.0?: For an inductor, Z j*2*pi*f*L If the inductance is increased by 8x, and the frequency is increased by 4x, the net increase in impedance is 32x. So the current drops to1/32 or 3.125% of the original magnitude.

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.: First, calculate L(eq1), equivalent inductance of the three parallel inductors, as you have already done. Then calculate L(eq2), the equivalent inductance of L(eq1) in series with L4 (L(eq1) + L4).

Q: I want to check and see if I understand all of this correctly, so I'll list what I am thinking happens and you can correct me if I'm wrong, Ok?I am talking about a very basic inductor in a very basic DC circuit, I am only wanting to understand the operation of an inductor in a circuit.1. When the cirucit is first energized the coil acts like a resistor, it drops the maximum amount of voltage ( the amount of the source) which declines rapidly as the field builds. The voltage it drops is opposite to the increase in polarity only, equal in value.So it is exhibiting decreasing resistance to current flow in this charging state.2 As current is decreased, the coil releases its energy by dropping a voltage of the same polarity to attempt and maintain the current's prior state. It is acting as a source in this state, but increasing its resistance to the change as its field deteriorates. 3. Steady state of current.NOT Sure?: Part 1 2 are correct. The steady state is what happens once the inductor is fully charged (or discharged depending on the action of the rest of the circuit). For part 1 when the inductor reaches the steady state there will be zero resistance, and the current flow will be determined from the rest of the circuit. For part 2 when the inductor reaches the steady state there is infinite resistance, and the current flow will be zero.

Q: I have simple rl circuit two resistance in parallel and a inductor - 5 H in parallel. I need to thevenize the circut but I do not know how to handle with the inductor, wich is in value of 5 Hplease help thank you: let's see what we can find :-) (by the way, what does thevenize mean? I've never heard of that before)

Q: A 220-mH inductor carries 350mA. How much energy must be supplied to the inductor in raising the current to 800mA?: for an inductor the energy is given by U (1/2) L i^2 , where 'i' is the current so that increase of energy dU (1/2) x 220 x10^-3 ( 0.8^2 - 0.35^2) 0.057 J (nearly)

Q: A constant voltage of 5.00 V has been observed over a certain time interval across a 1.10 H inductor. The current through the inductor, measured as 2.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 6.00 A at the end of the time interval. How long was this time interval?: Jag tycker det ?r lustigt hur du ska ?vers?tta detta f?r att se vad jag skrev. L?ser du fortfarande h?r? Du borde sluta nu.

Q: Calculate the inductance of an air-core coil with the following specifications: length 20 cm, #of turns 200 and diameter of core 2 cm.Could you show your work if you know how to do it? Thanks :): L 0.001 N?r? / (228r + 254l) where L is the inductance in henrys, r is the coil radius in metres, l is the coil length in metres (0.8r) and N is the number of turns. This formula applies at 'low' frequencies. At frequencies high enough for skin effect to occur a correction of up to about -2% is made. Old school version in Imperal measurement is La?n?/(9a+10l) a and l are in inches/

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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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