Product Description:
low frequency chock coil
1.Good quality and low price
2.deliver the goods on schedule
3.Radial Fixed Leaded Inductors with Inductance ranging from 0.30uH to 100mH.
4.ROHS compliance.
5.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:To increase the Q factor of an inductor it would with?
 Q factor X_L / R_Eff Inductive Reactance / Effective Resistance. An inductor is made of wire, but wire has resistance, so to compare two inductors at a given frequency, the Q factor is used. A component of Effective Resistance is the resistance of the wire called R_DC. There is also R_AC, but this question has nothing to do with it. Wire resistance at a fixed temperature is R ρ ℓ / A. So resistance R is inversely proportional to cross sectional area A. By increasing A, the effective resistance R_EFF goes down, which will increase Q factor. Answer a. Thinner wire, decreases A making R go up. Longer wire, increases X_L, but R_Eff would go up proportionally, so Q factor would not change. Wire with heavy insulation, would not change Q factor.
 Q:First order circuit with two inductors help urgent?
 You cannot convert a T or Pi type filter into equivalent components and expect the same Q. Those designs aren't to use up excess components in a warehouse somewhere. So, you can build filters that are resonant at the same frequencies with fewer components, but the configuration changes and so do the circuit parameters.
 Q:Real Inductor. Find the internal resistor and inductance?
 Ldi/dt +4i Vab, L(14) + R(4) 185 L(14) + R(4) 43 System of linear eqns, add together; 8R 228, R 28.5 ohm
 Q:An inductor is plugged into a 120 /60 wall outlet in the U.S. Would the peak current be larger, smaller, or u?
 higher frequency means higher reactance, which means lower current.
 Q:Possible mistake in my book(Power dissipation in inductors) ?
 Yes, you're correct. Your picture shows the correct phase relationship between voltage and current in an inductor, and in that case your description is correct.
 Q:Physics  Energy in a typical Inductor?
 a) The energy stored in an inductor is given by U 1/2 L I^2 1/2*(10.2E3)*(1.15)^2 6.74E3J 6.74 mJ b) The current required to have 1 J stored in the inductor is then I sqrt (2*U/L) sqrt(2*1/10.2E3) 14. A That is not an unreasonable amount of current. The inductor needs to be made of wire with diameter larger than a 30 AWG otherwise it will melt. Not sure if a 10.2mH inductor can be made with such wire.
 Q:which device stores energy?
 B In the magnetic field Capacitors store energy in the electrostatic field
 Q:A 45mH ideal inductor is connected in series with a 60 ohm resistor through an ideal 15 VDC power supply an?
 The time constant T is T L/R. So for your problem T 45x10^3H/60 Ohm 7.5x10^4 s Now the current in the curcuit after a very long time is simply I0 V/R 15/60 0.25 A. The current as a function of time is given by I(t) I0*(1  e^(t/T)) set t 7x10^3 and solve for I I 0.25 A* (1  e^(7x10^3/(7.5x10^4))) 0.2499 A the current has been flowing for almost 10 time constants which means it is almost at full value.
 Q:A 4.40 mH inductor has an rms voltage of 24.9 V across it at a frequency of 230. Hz. What is the rms current i?
 The peak current is when the voltage is at it's peak divided by the reactance. The peak sinusodial voltage is 29.4V * SQRT (2) 29.4V * 1.414 41.5779 V (peak) The inductive reactance is 2 * pi * Freq * L 2 * 3.1415 * 230 * 4.4 * 10E3 6.3586 ohms Using ohms law V / R I ; 41.5779 V / 6.3586 ohms 6.5389 Amps (peak) To find the RMS voltage multiply by the square root of 2 divided by 2 or 0.707 So 6.5389 Amps * 0.707 4.6237 Amps This is also the RMS Voltage divided by the inductive reactance ! :b
 Q:Does voltage have any affect on the E.M. field that it produces when it puts a current through an inductor?
 The variable that you haven't mentioned is the resistance of the inductor. At steadystate (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 outofphase with the applied voltage, which results in little real power actually consumed. Under noload 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 inphase, 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.
Our goods sell very well and gain a good reputation from the clients with our production scale, environmentally friendly products, excellent product quality, firstclass 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 
35 People 
Language Spoken: 
English, Chinese 
b)Factory Information 

Factory Size: 
3,0005,000 square meters 
No. of Production Lines 
9 
Contract Manufacturing 
OEM Service Offered Design Service Offered Buyer Label Offered 
Product Price Range 
Average 