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: Power Transformer 100kV
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: jbk high voltage transformer power transformer
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: JBK5 power transformer high voltage transformer manufacture
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: Oil immersed Power Transformer 66kV
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: 5000AAC Photovoltaic Solar Power Current Transformer CE
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: 68MVA/500kV standby transformer power plant
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: 32MVA/220kV main transformer power plant
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: 400MVA/220kV main transformer power plant
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: 270MVA/525kV single phase transformer power plant
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: 500KW Solar Inverter For Solar Power Plant or Solar Power System without Transformer
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: SZ9-1600-20000/33KV Three Phase Oil Immersed On-load Tap Power Transformer
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: 360MVA/63kV main transformer for Hydro power station
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: 250MVA/242kV three phase combined shell type transformer for hydro power station
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: Oil immersed transformer
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: 63MVA/220KV startup/standby transformer for factory
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: High Quality Silicon steel sheet of transformer
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: JDCF-110,132,220,230 Voltage Transformer
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: S9-M-30~2500/10KV Three Phase Oil Immersed All-sealed Power Transformer
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: MRD-NP Discharging Gap for Transformer Neutral
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: KS 9 series power transformer for mining
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: 360MVA/242kV three phase water cooling main transformer for hydro power station
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: High voltage transformer
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: S9-M-50~2500/33KV Three Phase Oil-immersed No-excitation Voltage-regulating Power Transformer
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: JBK3 transformer high voltage transformer
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: 720MVA/550kV main transformer Power plant
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: 500MVA/765kV substation auto-transformer
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: 370MVA/236kV main transformer power plant
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: 250MVA/220kV standby transformer for factory
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Hello everyone, Anyone knows how to findout the rating of 3R3 inductor which is broken and needed to be replaced for a tablet xtouch X906? Please help.Thank youBernard: 3R3 is typically a 3.3 mH but could be a 3.3 uH depending on the case size. Most are surface mount and if you can match it pictorially on digikey, you might have a chance in getting the same part. It is always best to have the manufacturer replace the part as they can test other areas that may be broken and need repair. Also you can do more damage if the proper part is not installed properly or if you overheat sensitive parts.s

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?: your barking up the wrong tree trying to use magnetic flux. V Ldi/dt instantaneous V across the inductor 12*6/14 5.14 V/L di/dt 5.14/0.45 11.4 A/s Power 58.74W 58.74 J/s in 20ms, E 58.74*0.02 1.17J This answer makes a number of assumptions, including a constant voltage and assumes di/dt is linear which may not be true.s

An LC circuit consists of a capacitor, C 1.72 μF, and an inductor, L 5.32 mH. The capacitor is fully charged using a battery and then connected to the inductor. An oscilloscope is used to measure the frequency of the oscillations in the circuit. Next, the circuit is opened, and a resistor, R, is inserted in series with the inductor and the capacitor. The capacitor is again fully charged using the same battery and then connected to the circuit. The angular frequency of the damped oscillations in the RLC circuit is found to be 19.5% less than the angular frequency of the oscillations in the LC circuit. a) Determine the resistance of the resistor. b) How long after the capacitor is reconnected in the circuit will the amplitude of the damped current through the circuit be 19.0% of the initial amplitude? c) How many complete damped oscillations will have occurred in that time?: We know Wo 1/sqrt(LC) 10,454 rad/s Wd is given to be 0.805*Wo Looking at the reference alpha^2 Wo^2 - Wd^2 and Wd 0.805*Wo alpha^2 Wo^2(1-0.805^2)38,465,981 alpha 6202.1 R/2L R 66 Ohms b) we can write 0.19 e^(-6202.1*t) Ln(0.19)/(-6202.1) t 0.268 milliseconds c) wd 8415.5 2pi/T T 0.747ms So it appears the current is at 19% before even one oscillation has finished unless I have an error. In an RLC ckt when the charged cap is connected, i(0) 0 All of the voltage is dropped across the inductor meaning VR0 at t0 which means i(0) 0 So not sure what initial amplitude they are talking abouts

A 9.5 V battery, a 4.92 resistor, and a 9.7 H inductor are connected in series:(a) After the current in the circuit has reached its maximum value, calculate the power being supplied by the battery.(b) Calculate the power being delivered to the resistor.(c) What is the power being delivered to the inductor?(d) What is the energy stored in the magnetic field of the inductor?: Once the current has reached its max. then all the power will go to the resistor, so: a) p v^2/r 9.5^2/4.92 b) see a) above c) 0 d) I don't remember these equations, sorry.s

An ac circuit has a 20 ?F capacitor, an 80 mH inductor, and a 60 Ω resistor in series. If the voltage source has an angular frequency of 1000 rad/s, what is the phase constant for the circuit?: First calculate the reactance of the inductor and capacitor XL ωL 1000*80x10^-3 80Ω XC 1/ωC 1/(1000*20x10^-6) 50Ω So the phase constant φ arctan((XL - XC)/R) arctan((80-50)/60) 26.6os

An inductor has a 54.0Ohms reactance at 60.0 Hz. What willbe the maximum current if this inductor is connected to a50.0-Hz source that produces a 100-V rms voltage?: the impedance it given to you from the formula sq. root R^2+(XC-XL)^2 which could provide you 360.40 4 ohms for this reason Z multiplies via the amp of 233mA is 80 3.98V wish this helpss

For example, if you put 10 amps through an inductor using 10 volts, is the field any stronger than a field produced by 10 amps going through the same inductor at 1 volt? I've been told that the only factors that effect the strength of an electromagnet are the physical specs, i.e. turns, core, guage, etc., and one electrical factor, current. I know that can't be true, because my house's water pump is able to substitute current for voltage in order to create the same output.More specifically, it runs on either 5A x 230V, or 10A x 115v. Which means that voltage must play a factor in the strength of the magnetic fields created by it's armatures. I theorized that the current going through an inductor is what determines the size, or far-reachingness of the field, while the voltage is what backs up the magnetic field, or gives it it's rigidity. So what's the answer??: The variable that you haven't mentioned is the resistance of the inductor. At steady-state (DC), the voltage needed will be the current multiplied by the resistance. The equations for magnetic field rely only on current for simple structures, and voltage enters the picture through the resistance for DC circuits. In an AC circuit (which is what your water pump is), the current is usually established by the inductance and the amplitude and frequency of the voltage source. The resistance usually represents an undesired loss, but it will affect the current amplitude. In your water pump, you are trying to deliver mechanical power to a load (the water pressure), and this is represented as a resistance in the motor equivalent circuit. But instead of being a loss, the resistance models the delivery of power to a mechanical load. An unloaded motor draws relatively little current compared to full load, and the current that it does draw is out-of-phase with the applied voltage, which results in little real power actually consumed. Under no-load conditions, most of the power consumed is caused by the line current flowing in the winding resistance. As the mechanical load on the motor is increased, the line current increases, and becomes more in-phase, resulting in significant power consumption, although most of the power is not consumed in the motor, but actually delivered to the load. Your water pump has two sets of windings, and these are either wired in series, or parallel to determine the voltage and current needed. The magnetic field is established by the number of turns of a coil, multiplied by the current in the coil. The power rating of the motor is the constant, so at 230V, the current drawn would be half of that drawn at 115V. So at 230V, the number of turns in the coil is doubled, and the current is halved, resulting in exactly the same peak magnetic field amplitude as the 115V case.s

Do you think that drill instructors in the marines and Drill Sergeants in the army be allow to hit recruits like during the Vietnam War and before. War is much more cruel then boot camp and if you can't survive a punch to the gut because you made a mistake there how will you survive in war. Also I've heard from my cousins in the marines that Drill instructors will still hit you if you mess up in boot camp when the media isn't around is that true?: While war is always going to have its ugly moments, training for it has become harder to do since the violence and hazing is strictly forbidden these days. Recruits today will NEVER go through what I went through in 1982 to pass boot camp. Discipline and good order in 2013 also means with a gentler, kinder hand - and yes, it might be making us a bit softer. But out-and-out physical violence is not ever going to be a part of future basic training.s

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