Low Frequency Toroidal Needle Insert PCB Mounted Transformer

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1.Professional engineer

2.Qualified material competative price

3.small loss

4.usage:power phase:single coil

5.structure:toroidal

6.core dimension:adjustable by clients

7.size:follow customer's request

8.Remark:we can produce it according client's requirement

Our products have gained the international certifications, such as CQC, CE, RoHS, UL and so on, from internationally powerful authorities. We have got ISO9001 certificate.We promise to offer the best products to our clients.We look forward to cooperating with all friends for more mutual benefits.

Q: current that passes through the capacitor is directly proportional to the frequency of the currentas dc has zero frequency so it cannot passplz explain it like ths if you can :): Some of the energy of the current goes to build the magnetic field around the inductorthe inductor resists being charged?once the magnetic field is build that factor goes away. inductors pass DC For AC when the current changes direction the magnetic field collapses causing a voltage out of the inductor that is opposite to the current trying to charge the inductorthe inductor voltage pushes back against the current voltage resisting the current the inductor opposes an AC current at a frequency that results in the forward and backward voltage being in sync

Q: A 10.2*10^-6H inductor with negligible resistance and a resistor are the only elements in a circuit in which a current is flowing.What must be the value of the resistor so that the current will decrease to 50.0% of its maximum value in 1.10ms ?: I Io e^(-t/τ) and τ L/R Given I/Io 0.5 e^(-1.1e-3/τ) Taking log to the base e and Solving we get τ L/R 0.00158690 R 10.2 /0.00158690 6428 ?

Q: An inductor coil of 1H carrying a current of 2A in the circuit . To prevent sparking , when the current is switched off a capacitor which can withstand a potential difference of 400 V is used . What must be the minimum capacitance ?Please show the working steps .: NOTE: that is a LARGE inductor and a LARGE current. The energy of the magnetic field in the inductor is the following: MPE 1/2*L*I^2 This energy will be conserved when it is dis-current-ed in to charging the capacitor. MPE EPE EPE 1/2*C*V^2 Substitute: 1/2*L*I^2 1/2*C*V^2 Solve for C: C L*I^2/V^2 Calculate: C25 microfarads

Q: I need help answering this question i have no idea how to do the steps if you could lead me through the steps or tell me what to do that'd be fantastic.An inductor is fabricated with 500 turn of wire, a diameter of 1 centimeter and a length of 2 centimeters, and an air core with a permeability of 1.25X10^-6 H/m. A DC current of 500 mA is made to flow through the inductor.Find:A) The inductanceB) The energy stored in the field:C) The energy stored in the field if a steel core, with apermittivity of 4000 is inserted into the coil.: A)From physics, you know that: B μNI / h where μ is the permeability h is the height of the loop B is the magnetic flux N number of loops B 0.03125 wb An the inductance L is equal to L B/I L 0.0625 H B) Energy stored 1/2 LI^2 0.007813

Q: Find the induced emf when the current in a 47.5 mH inductor increases from 0 to 521 mA in 17.1 ms. Any help would be appreciated. If you are able to answer this one, please show your work. Thanks!: Find the induced emf when the current in a 47.5 mH inductor increases from 0 to 521 mA in 17.1 ms. EMF is voltage. As the current increases, the magnetic field increases. This increasing magnetic field induces an emf. The emf is directly proportional to the rate of change of the current. EMF -L * (? I ÷ ?t) L 47 mH 0.047 Henry ? I 521 mA 0.521 amp ?t 17.1 ms 0.0171 seconds EMF -0.047 * (0.521 ÷0.0171) The unit is volts

Q: A 220-mH inductor carries 350mA. How much energy must be supplied to the inductor in raising the current to 800mA?: tricky problem. do a search with search engines like google. this may help!

Q: A certain circuit consists of an inductor of 55 mH in series with a resistor of 90 Ω. At a moment when the current in the circuit is 21 A, a switch in the circuit is opened. How long will it take for the current to fall to 7.14 A? Answer in units of s.I thought that this one was easy. first time around I used i(Io)(e^-Rt/L) --- 7.1421e^(-90t/(55x10^-3)) 2.53926105e-4this is wrong. Then based on someone's advise I got on here, I used the equation i(Io)e^(t/-RL) and I got an answer of 5.34.also wrong! Please help! What am I doing wrong? What's the answer??: The power in an inductor is proportional to the rectangular of the present. At part the present it holds a million/four the volume of power. It held forty J on the starting, it now holds 10 J so the difference is 30 J

Q: A circuit consists of an inductor L and a resistor R connected in series to a battery via an open switch. After the switch is closed, how long must we wait for the current in the LR circuit to build up to 99.99% of its equilibrium value. Express the answer in terms of the number of “time constants,” L/R.Additional DetailsAnswer is 9.2 but i have no idea how to arrive at the answer :(: the present would be in section with the utilized voltage at resonance. that's while the capacitive reactance, Xc equals the Inductive reactance, XL. Xc XL a million/(2 x pi x f x C) 2 x pi x f x L f a million/(2 x pi x sq. Root of (L x C)) F 419 Hz. The resistor performs no area different than to limit the present at resonance to: utilized voltage /R and likewise to regulate 'Q'.

Q: 2.An inductor passes 20 mA rms at 12 V rms and 1000 Hz. Calculate the inductance. (95 mH)3.Calculate the inductance of a coil 25 mm diameter, 100 mm long with 30 turns. The core has a relative permeability of 2000. (0.0111 H)Calculate the energy stored when 10 A d.c. flow. (0.555 J) Calculate the reactance for ac with a frequency of 100 Hz. (6.97 ?)Calculate the rms voltage needed to make 10 A rms flow. (69.7V rms): 3 question are U mad

Q: If there is an inductor of 600 mH in a 140 V, 45 Hz AC power line, what is the current in the circuit?: Inductive reactance (2pi FL) 169.67Ω. Remember L must be in H., and F in Hz. Current (E/R) 140/169.67, 0.825A., or 825mA. approx.

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1. Manufacturer Overview

Location	Shenzhen, Guangdong, China (Mainland)
Year Established	2006
Annual Output Value	US$2.5 Million - US$5 Million
Main Markets	North America; South America; Eastern Europe; Southeast Asia; Africa; Oceania; Mid East; Eastern Asia; Western Europe; Central America; Northern Europe; Southern Europe; South Asia; Domestic Market
Company Certifications	CE Certificates

2. Manufacturer Certificates

a) Certification Name
Range
Reference
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3. Manufacturer Capability

a) Trade Capacity
Nearest Port	Shekou,Yantian
Export Percentage	51% - 60%
No.of Employees in Trade Department	3-5 People
Language Spoken:	English, Chinese
b) Factory Information
Factory Size:	3,000-5,000 square meters
No. of Production Lines	9
Contract Manufacturing	OEM Service Offered Design Service Offered Buyer Label Offered
Product Price Range	Average