Low Frequency Toroidal Needle Insert PCB Mounted Transformer
- Ref Price:
-
- Loading Port:
- China Main Port
- Payment Terms:
- TT or LC
- Min Order Qty:
- 1000 Pieces pc
- Supply Capability:
- 30000 Pieces per Month pc/month
OKorder Service Pledge
Quality Product, Order Online Tracking, Timely Delivery
OKorder Financial Service
Credit Rating, Credit Services, Credit Purchasing
You Might Also Like
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:Having a bit of trouble with this question; I'm asked to calculate the time in ms it takes to dissipate 40% of the initial stored energy of an Inductor.I know that the initial stored energy is 62.5 JThe time constant is 50msV 200e^(-20t)I 12.5e^(-20t)I've worted out the above values, but I cannot see how I can work out the time taken for it to discharge 40%, as i keep getting a negative time value.I would be very grateful of any help :)(P.S. I also need the same for a capacitor when I have similar values)
- 0.4 e^(-20t) ln(0.4) -20t t ln(0.4)/(-20) t (-0.916)/(-20) t 0.0458 46ms
- 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:Determine the constant value of the inductorI cant find constant value in my text books nor do I believe my lecturer has covered it. :(This is for electrical theory.
- No such term! I think you are leaving out important details. The inductance of an inductor is constant. .
- Q:Ex. Negatively charged vinyl strip touches sphere of an electroscope. What happens to the leaves and positions of charges?
- Then, positive charges interact with the negatively charged inductor Good luck!
- 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
- Q: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?
- a million - particular. many times extra desirable than one moving interior a similar path. 2 - a moving value (a modern) continually generates a magnetic field in accordance to Lorentz regulation. 3 - a variable magnetic field induces a modern right into a conductor. 4 - possibly you meant capacitors, not conductors. a capacitor is a gadget in a position to generate a persevering with electric field. an inductor is a gadget in a position to generate a persevering with magnetic field.
- Q:current under one volt?
- Variable inductors are not used very often, but a powdered iron toroid could be used. I would be built something like an auto transformer. Most diodes have less than a volt when forward biased.
- Q:how does the energy gets stored in the inductor if we are connecting it to an rlc circuit either in parallel or in seriesi also wanted to ask about the phenomena of resonance ocuring in this circiut in details,i know that when the inductor has potential energy the capacitor doesnt have and vice versa,thats why its called electrical resonance,but i want someone to tell me the reason of this phenomena,an athountic answer please.thank u buddy
- Energy is stored in the inductance as a magnetic field and at any time E (1/2)Li^2. In a series circuit the impedance is Z R + j(wL - 1/(wC)) ?Z? SQRT[R^2 + (wL - 1/(wC))^2] Resonance occurs when wL 1/wC w^2 LC w SQRT(LC) where w is angular frequency and j^2 -1
- Q:Do i have to find the equivalent L and use this equation?i(t)1/L* integrate(v(t) dt+i(t_0),i(t),i(t_0))
- Your equation looks correct, but I don't understand the stuff after dt+. (The i(t_0) looks like it might be initial conditions, but the other terms don't make sense.) In your circuit and equation i(t) and L can be either i1(t) or i2(t) and L1 or L2. The current add to get i(t).
- Q:I understand that power is that rate at which work is done and that because of this the power is equal to d/dt (1/2Li^2). I also understand that the power is also equal to Li di/dt since Ldi/dt is v and v*i is power. I understand that since the power is equal to both of these equations that they are equal to each other. The part that I don't get is mathematically how to get from one to the other.
- you have written the steps yourself, from what i can tell d/dt [ (1/2)Li^2 ]. L is constant, i d/dt [ q(t) ], not constant using the power rule for differentiation d/dt [ (1/2)Li^2 ] (2)(1/2)Li*(di/dt). implicit differentiation if that step did not make sense, review the calculus techniques from first semester calc,, i dq/dt v -dΦ/dt -d/dt ( BAcosθ ) || B || (μ0 / 4π) * i visualize the charge (q) running through wire approaching inductor, as it flows into the coil, a B-field is induced, which tries to resist the CHANGE in current, this is where the work occurs
Our goods sell very well and gain a good reputation from the clients with our production scale, environmentally friendly products, excellent product quality, first-class enterprise management, the most competitive price and perfect service.Our products have gained the international certifications, such as CQC, CE, RoHS, UL and so on.
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 | |
| Validity Period | |
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 |
Send your message to us
Low Frequency Toroidal Needle Insert PCB Mounted Transformer
- Ref Price:
-
- Loading Port:
- China Main Port
- Payment Terms:
- TT or LC
- Min Order Qty:
- 1000 Pieces pc
- Supply Capability:
- 30000 Pieces per Month pc/month
OKorder Service Pledge
Quality Product, Order Online Tracking, Timely Delivery
OKorder Financial Service
Credit Rating, Credit Services, Credit Purchasing
Similar products
New products
Hot products
Hot Searches
Related keywords
