PBP Series Shielded SMD Power Inductor

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1. pbp series shielded smd power Inductor

2. Rated current:0.5-10A

3. Inductance0.5~6000uH

4. Quality assured

5. competitive price

smallest possible size and high performance they are with high energy storage

Feartures:

The Surface Mount Inductors are designed for the smallest possible size and high performance,

They are with high energy storage and very low resistance making them the ideal inductors for

DC-DC conversion in the following application.

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

Q: A 440 resistor, 45 ?F capacitor, and 810 mH inductor are each connected across 6.3 V rms, 60 Hz AC power sources. Find the rms current in each path.a) resistorb) capacitorc) inductorI keep getting the wrong answer I am using all the right formulas can someone let me know how to do this problemplease!!!: They can all be considered separately since they don't interact. A. I(R) V/R 0.0143182 A B. w 2pi*60 376.99 rad/s X(C) 1/(wC) 58.9463 ohms; I(C) V/X(C) 0.106877 A C. X(L) wL 305.363 ohms; I(L) V/X(L) 0.026312 A

Q: Hi,I'm planning to build an audio amplifier and for the crossover, I've been looking in to using resistors instead of inductors?I know what each one does. Simply put resistors do not hold an electrical charge like inductors in an electromagnetic field.What is the role of the inductor in this circuit? I'm not asking for the definition, I'm asking what it's doing to the actual audio signal.Also I've put a capacitor in series for the top frequency and after trying it out with different caps, the audio gets distorted in the speaker component (the cone moves to its maximum rating). Am I right in saying that I'd need a bigger speaker, well a bigger speaker cone in fact, to increase the rating of volume and power of the signal to reduce distortion/noise?Thanks,Rish: Crossover Inductors

Q: I plan on making a 500mH inductor, but I'm not sure how big it needs to be to allow the space for the number of turns it needs with a reasonable size wire. (Reasonable size meaning it is readily available and not so small soldering burns through the wire.): 500mh Inductor

Q: An inductor in an LC circuit has a maximum current of 2.2 A and a maximum energy of 60 mJWhen the current in the inductor is 1.1 A , what is the energy stored in the capacitor? U _____ mJ: We know that he energy in an LC circuit oscillates back and forth between the inductor and capacitor. When Vc(t) 0, VL(t) is at it's maximum and when VL(t) 0, Vc(t) is at it's maximum. At iLmax 2.2A all the energy is in L and there is no energy in C (Vc 0) Solve for L E ?Li? 0.06J -----L 0.12/2.2? 0.0248H 24.8mH We know that ?Li? + ?CV? 0.06J for all t Subtract the energy in L from the max energy to find the energy in C: ?CV? 0.06 - ?Li? 0.06 - ?(0.12/2.2?)*1.1? 0.045J 45mJ This makes sense because the energy in the inductor is proportional to i? 0.06*(1.1/2.2)? 0.06/4 0.015 J so when the current is half in L, the energy is 1/4 meaning the other 3/4 of the energy must be in the capacitor.

Q: An RLC circuit has a 400 ohm resistance a 2.5 mH inductor and a capacitor. If the circuit is in resonance at 30kHz and is attached to a 20mVrms power supply, what is the power dissipated by the circuit?: I'm assuming that the power supply is at the resonant frequency. Therefore, by definition, at resonance the series impedance of the inductor and capacitor equals zero. The power supply therefore sees just a 400 ohm resistor. Therefore the power dissipation is given by V^2/R, which equals 1 microwatt.

Q: A 20.0 uF capacitor is charged by a 170.0 V power supply, then disconnected from the power and connected in series with a 0.270 mH inductor.: The natural (radian) frequency of oscillation is: ω 1/sqrt(L*C) 1/sqrt(270*10^-6 * 20*10^-6) 13.61*10^3 radian/sec At the start of the transient, the inductor current is zero, and all of the energy is stored as electric field energy in the capacitor. The current in the inductor will be sinusoidal with zero crossings at every N*π where N 0, 1, 2, 3 The angle of the sinusoidal inductor current at 1.40ms is: θ ω*t 13.61*10^3 * 1.4*10^-3 19.05 radians Divide this by 2*π to find out how many complete periods are included in the 19.05 radians: N 19.05/(2*π) 3.03 This is probably close enough to 3 complete periods of the inductor current to conclude that for practical purposes, all of the energy has been returned to the capacitor, and only a very small amount of energy (within the significant figures of the component values given) would be in the inductor. I'd answer: approximately zero

Q: (a) How long will it take for the current to reach 67% of its maximum value?(b) What is the maximum energy stored in the inductor?(c) How long will it for the energy stored in the inductor to reach 67% of its maximum value?: oubaas: are you able to describe why for the inductor you do not use the a million-e^even if and for the resistor you do? i'm purely attempting to appreciate the thoughts previously for my midterm. thanks!

Q: I'm a little confused about AC power.If a circuit has only resisters in it, then it's easy to calculate the total resistance of the circuit and from that the current and power. But does putting capacitors or inductors in a circuit change the resistance of the circuit, and thus change the average power being dissipated? I think the answer is yes, but it's a little confusing because I read that capacitors and inductors are reactive elements and draw no power?: Reactive power does not register on your wathour meter. You don't pay for it. But it adds to the total KVA of the power company;s system. As such, power is wasted in the transmission lines. It doesn't ride free. The power company is penalized by low power factors because they have to furnish the equipment and fuel to supply the KVAR component of their power lines. You, as a customer of the power company, pay for this wasted energy as the power company passes the cost on to you. The resistance is not changed when capacitors are added to the circuit. Resistance applies to the KW portion of the power triangle. The reactance of inductors and capacitors are 180 degrees out of phase. When capacitors are installed, the total impedance of the circuit is reduced, as well as the KVA of the circuit. The capacitor stores energy in one half of the cycle and releases it during the next half cycle. This energy circulates around the circuit supplying the inductance for motor and other inductive equipment.

Q: 5.00 H. Find the current through the resistor, inductor, and the total current in the circuit.: Actually you should specify the resistance of the 5 H inductor, but in this case X(L)R so the impedance can be estimated as the inductive reactance, where 5 H @ 60hz by X(L)2piFL6.28*60*5 1884 ohms. Resistor current equals 117/1000 .117 A. Inductor current 117/1884 .062A

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Location	Guangdong,China (Mainland)
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