Low Frequency Copper Wire Only Chock Coil

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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.

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Q:A 4 mH inductor is connected to an AC voltage source of 141 V rms. If the rms current in the circuit is 0.85 A?
reactance X V/I 141/.85 165.88 ohms X 2(pi)fL f X/2(pi)L 165.88/2x3.14x4x10^ --3 6603 (check the calculation)
Q:A 12.6 V battery is in series with a 30.0 mH inductor and 0.150 ohm resistor connected through a switch. When?
Let V_s the voltage of the battery 12.6 V Let i the current through the series circuit Let R the resistance of the resistor 0.150 Ω Let L the inductance of the inductor 0.03 H Let V_r the voltage across the resistor (i)R Let V_l the voltage across the inductor L(di/dt) The source voltage must equal the sum of the voltages across the components: V_s V_r + V_l 12.6 V (i)R + L(di/dt) di/dt + (i)(R/L) (12.6 V)/L The integrating factor for this is e^{∫ (R/L)dt} e^{(R/L)t} e^{(R/L)t}di/dt + e^{(R/L)t}(i)(R/L) e^{(R/L)t}(12.6 V)/L The left side integrates as the reverse of the product rule and the right side integrates with the reciprocal of the coefficient with a constant, C: e^{(R/L)t}(i) e^{(R/L)t}(12.6 V)/R + C Multiply both sides by e^{-(R/L)t}: (i) (12.6 V)/R + Ce^{-(R/L)t} i (12.6 V)/0.150 Ω + Ce^{-(R/L)t} i 84 A + Ce^{-(R/L)t} We find the value of C by knowing that i 0 at t 0 0 84 A + Ce^0 C - 84 A i (84 A)(1 - e^{-(R/L)t}) To find the time constant set (R/L)t 1: t L/R 0.03/0.150 0.2 s One time constant means that -(R/L)t -1 i (84 A)(1 - e^-1) ≈ 53.1 A The current is 0 at t 0 so V_r R(0) 0 The current is 53.1 A at t 0.2 s so V_r (0.150 Ω)(53.1 A) ≈ 7.97 V The charge rate is di/dt and we have an equation involving that: di/dt + (i)(R/L) (12.6 V)/L Solve for di/dt: di/dt (12.6 V)/L - (i)(R/L) di/dt 12.6 V/0.03 H - (53.1 A)(0.150 Ω/0.03 H) di/dt 154.5 A/s
Q:Physics Question regarding inductors. Calculating current and inductance Confused with a couple parts my attempt is included?
You ve calculated the RMS voltage of the inductor. Now you need to find the impedance of the inductor. XL ωL 2πfL Now the voltage equation for the inductor XL * i(RMS) V(RMS) 2πfL * i V L V / (2πfi) L 79.37 / (2 * π * 60 * 1.8) L ≈ 0.117 H 117 mH
Q:Inductor connected to dc source?
Yes, the current will rise until it is limited by the source or until the wire of the inductor fuses open.
Q:if an inductor has a parasitic resistance does it effect the resonance frequency, if so how?
no. .
Q:Help with inductor and sinusoidal waves.?
You do not need the actual frequencies to solve this problem. You can use the frequency ratios. Since the current is given as 3.0 Amps peak I will assume the 120 Volts is peak also. (b1) XL at original frequency .(707 x 120V) / (.707 x 3A) 40 Ohms XL at (2) x (original frequency) 2pi x 2f x L (b2) But 2pi and L are both constant therefore XL at 2f 80 Ohms (a) Maximum current when frequency is doubled V / XL 120V / 80 Ohm 1.5 Amps
Q:the current through an inductor?
The inductor integrates the voltage across it and that results in a current. Since the current is 0 when the voltage is applied, I0 at t0.
Q:Ideal inductor in AC?
inductors and capacitors will offset the current vs voltage by 90 degrees. With an inductor, (if my memory is correct) you will find the current will be 90 degrees ahead of the voltage. It seems strange, but if you sit and think how the inductor (or capacitor) reacts to a changing voltage, it might make some sense.
Q:questions about inductors?
You may be getting resistance confused with impedance. 1. But taking your question as it stands, the resistance of an inductor can only be calculated if you know the size of wire used and the length of wire. Formula is below. For total resistance of that in series with a discrete resistor, just add them together like any two resistors in series. If you mean impedance or reactance, the inductive reactance of an inductor is XL 2πfL where L is the inductance in henrys and f is the frequency. From that you can calculate the impedance: Impedance Z √(R² + X²) where X XL – Xc 2. Because DC does not induce any voltage. And they are not a short circuit, see #1 above. 3. This question makes no sense, sorry. What do you mean by strong. Resistance of a wire in Ω R ρL/A ρ is resistivity of the material in Ω-m L is length in meters A is cross-sectional area in m² A πr², r is radius of wire in m resistivity Cu 17.2e-9 Ω-m or 17.2e-6 ohm-mm .
Q:how does an inductor affects a circuit?
An inductor stores energy in a magnetic field induced, around it, by the current flowing thought it. This magnetic field takes time to build up and time to decay. Such an induced magnetic field, resists a change in the current flowing. Thus, if the current flowing though the inductor changes the magnetic field induces a back emf, which resists the change in the current flow. The back emf is given by the equation: - V -LdI __ dt Where 'L' is the self inductance of the inductor with a unit of the henry (H). Hence, an inductor 'resists' a change in current flow so that the voltage across it is always ahead of the current change. This may be remembered using the mnemonic CIVIL, or in a capacitor the current is ahead of the applied voltage and in an inductor the applied voltage is ahead of the current change. Hence, an inductor allows a steady state current to flow but delays the change in any increasing or decaying current! Thus, after a current peaks at maximum, the inductor will cause it to lag behind in its decay to minimum. This means that in an AC circuit an inductor may be coupled with a capacitor to create an oscillatory or resonant circuit.
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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
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