• PBH Series SMD Power Inductor System 1
PBH Series SMD Power Inductor

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:a fully charged capacitor is connected to an inductor and a resistor. what is the value of tfe charge on the capacitor when time becomes large?
Depends on what the total circuit looks like. If both ends of the capacitor are connected to each other thought the resistor, and maybe the inductor, then the large-time charge will be zero.
Q:Find the equivalent inductance for each of the series and parallel combinations
It's the same as parallel and series resistors. 1) has 2H in series with 2H, making 4H, in parallel with 12, making 3, which is in series with 2, making 5 total. 2) Re the wire, it shorts out the 6H and the 3H inductors, so that leg just has 18H. Other leg has 10 and 15 in parallel, equivalent to 6H. In series with 3, that makes 9H. so you have 9 in parallel with 18, and that combo in series with 1 .
Q:An inductor has a 59.6-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 46.0-Hz source that produces a 115-V rms voltage. What is the maximum current in the inductor?____ A
XL(1) 2π fL 59.6 Ω L 59.6/2π 60 0.158 H XL(2) 2π*46.0*158 45.67 Ω XL(2) V/A A V/XL(2) 115/46.67 ? 2.25 A
Q:help me to know the basics of tapping an inductor
For small air-core inductors there are a couple of methods that I have used. 1. Wind 10 turns and carry the end lead out about 1 inch away from the coil, then carry the wire back next to the 1 inch lead, back to the coil and continue winding for another 30 turns. Scrape off the insulation of the double-wire lead and use that as the tap. 2. Wind 10 turns. Scrape off the insulation at the 10th turn, leaving an opening to solder a wire on the coil. Continue winding 40 turns. Solder a wire onto the scraped off opening for use as the tap. I like the first method best, but both will work. .
Q:Can anyone tell the different types of inductors.
I'd like to add, that no-one uses inductors if at all possible to design them out lossy, costly, imprecise (too much resistance for a given size) Filters use R and C's (and opamps) or DSP in digital In power supplies, frequencies are moving into the 1MHz area so that they can use flat coils made into the pcb You can still buy inductors, basically two types tiny minature inductors with relatively high resistance for tuning and old fashioned filters or power inductors for switching regulators Powetr factor control or control of harmonics in power circuits or ballast (as in lighting)
Q:How much DC current flows through a resistor-inductor series circuit after the inductor has fluxed (i.e., after the transient)?
At an 'infinite' time after applying a DC source, the current will settle to a value determined by the resistance of the inductor a la ohms law. But if its an ideal inductor, the current will simply continue to ramp up linearly for ever
Q:and is increasing at the rate of 100 mA/s. a.) What is the initial energy stored in the inductor if the inductance is known to be 61.0 mH? Jb.)How long does it take for the energy to increase by a factor of 10 from the initial value? s
If you look up inductor on Wikipedia, you'll find that. E?LI? Where Eenergy LInductance Icurrent So finding the energy isn't hard. To find the time for part b, first find the current needed for 10 times the enrgy (also using E?LI?, but this time with 10 times the energy). Then, if we assume that the rate of increase is constant (seems to be true from the problem statement) finding the time is a matter of dividing the current difference by the rate of change (the 100 mA/s).
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 6.0 and the driving frequency is increased by a factor of 8.0?
X wL where X is the inductor's Reactance and w is the frequency in rads/sec the reactance will increase by 6 x 8 48 so the current will decrease by the same factor
Q:A 100-turn, air-core inductor has an area of 200 square micro-meters and is 200 milli-meters long. What is its inductance?A. 12.6 micro-henrysB. 1.26 milli-henrysC. 12.6 milli-henrysD. 126 micro-henrys
Lmuo N^2 area/length where muo4 pi*10^(-7) and area is in meters squared and length is in meters. You can do the math.
Q:find the inductive reactance of the inductor if there are 240 mA of current flowing through it and 1.44 x 10 negative 4 joules of potential energy stored inside its magnetic field.
Potential energy in an inductor is equal to (1/2)Li^2. Solving for L with algebra will give you the inductance. 1.44 x 10^-4 (1/2)L(.240)^2 L .005 henrys Inductive reactance is simply given by X wL. Since w 2(pi)f, plug in 120 Hz for f and get w 240pi. So X 240pi(.005) 1.2 pi 3.77 ohms reactive
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Location Guangdong,China (Mainland)
Year Established 2010
Annual Output Value US$10 Million - US$50 Million
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