Low Frequency Copper Wire Only Chock Coil
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low frequency chock coil
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3.Radial Fixed Leaded Inductors with Inductance ranging from 0.30uH to 100mH.
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- Q: A real inductor can be modeled as an ideal inductor in series with an internal resistance as illustrated in the figure above.A time-varying current I is passed through a real inductor. At a time when I +4 A and dI/dt +14 A/s, the voltage drop across the inductor is observed to be Va - Vb +185 V. At another time when I +4 A and dI/dt -14 A/s, the observed voltage drop is V'a- V'b +43 V.
- 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: 1. how to calculate the resistance of an inductor in a circuit?for example a resistor connected series with an inductor, how to calculate the overall resistance of the inductor and resistor together?2.if inductors make induced e.m.f, why do they behave as short circuits in DC circuits3.if inductors are strong enough what don't they seem to short cicuit like in strong electromagnets?
- 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 √(R? + X?) 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 m? A πr?, r is radius of wire in m resistivity Cu 17.2e-9 Ω-m or 17.2e-6 ohm-mm .
- Q: A 1.15 k-ohms resistor and a 560mH inductor are connected in series to a 1450Hz generator with an rms voltage of 12.6 V. A)What is the rms current in the circuit? B)What capacitance must be inserted in series with the resistor and inductor to reduce the rms current to half the value found in part A?i would realli appreciate a walk through of this problem and what answer i should expect. thanks
- A) the reactance of the inductor is XL 2π*f*L 2π*1450*0.560 5102Ω So the impedance Z sqrt(1150^2 + 5102^2) 5230Ω Therefore the rms current V/Z 12.6V/5230Ω 2.41x10^-3A 2.41 mA B) If the current is reduced to 1/2 then then impedance must double So Z sqrt(R^2 + (XlL- XC)^2) 10460 So 1150^2 + (5102 - XC)^2 10460^2 or 5102 - XC +-sqrt(10460^2 - 1150^2) +-10397 or XC 5102 + 10397 15499 Now XC 1/2πf*C or C 1/(2π*1450*15499) 7.08x10^-9F 7.08nF
- Q: An ac generator has a frequency of 5.0 kHz and a voltage of 22 V. When an inductor is conencted between the terminals of this generator, the current in the inductor is 10 mA. What is the inductance of the inductor?
- The reactance is 22/10ma 2200 ohms X? 2πfL 2200 2π(5000)L L 0.070 H or 70 mH .
- Q: An inductor is plugged into a 120V /60Hz wall outlet in the U.S. Would the peak current be larger, smaller, or unchanged if this inductor were plugged into a wall outlet in a country where the voltage is 120V at 50Hz ? Explain.
- Current is larger because inductance XL became less with 50c/s. I 120 / XL
- Q: Find the equivalent inductance for each of the series and parallel combinations
- It's the same as parallel and series resistors. 1) has 2H in series with 2H, making 4H, in parallel with 12, making 3, which is in series with 2, making 5 total. 2) Re the wire, it shorts out the 6H and the 3H inductors, so that leg just has 18H. Other leg has 10 and 15 in parallel, equivalent to 6H. In series with 3, that makes 9H. so you have 9 in parallel with 18, and that combo in series with 1 .
- Q: A 150 Omega resistor is connected in series with a 0.250-H inductor. The voltage across the resistor is v_R(3.80 V) *[cos(720*rad/ s)*t]Derive an expression for the circuit current. and Derive an expression for the voltage v_L across the inductor.I got by the equation I v/R.0253A*cos((720*rad/s)*t)and for the other i got by the equation v_LL* (di/dt) -(4.55*V)*sin((720 rad/s)*t)
- The voltage across a resistor is in phase with the current and proportional to its resistor, or V_R I*R The voltage drop across an inductor will be value of the inductance, multiplied by the time derivative of the current. In math notation, this becomes V_L L*dI/dt. In a series RL circuit, the current through the inductor will be the same as that through the resistor. We are given the voltage across the resistor. The current through the resistor, and through the inductor as well, will be given by the expression I V_R / R [3.80 * cos (720t)] / 150 0.0253 cos 720t amperes The voltage across the inductor will then be V_L L*dI/dt 0.250 * d[3.80 * cos (720t) / 150] / dt 0.250/150 * 3.80 * d[cos (720t)]/dt 0.250/150*3.80*(-720)sin (720t) -4.56 sin (720t) volts, which is what you got. A note on units, breaking them down into their fundamental MKSA components: volts watts / amps (kg?m?/s?)/s / A kg?m? / (A?s?) Ohms V/I volts/amps kg?m? / (A??s?) Farads Q/V A?s / volts A??s^4 / (kg?m?) Henries V / (dI/dt) volts / (amps/s) [kg?m? / (A?s?)]?s / A kg?m? / (A??s?)
- Q: Act as a hpf?And why does everyone use inductors in series with speakers for low pass filters and not a cap in parallel? Caps are a fraction of the cost
- A cap in parallel will put a greater load on the amplifier
- Q: A 33.0-mH inductor has a reactance of 2.20 kΩ. (a) What is the frequency of the ac current that passes through the inductor?(b) What is the capacitance of a capacitor that has the same reactance at this frequency?(c) The frequency is tripled, so that the reactances of the inductor and capacitor are no longer equal. What is the new reactance of the inductor?(d) What is the new reactance of the capacitor?I've been stuck on this physics problem. Please Help!! Thanks so much!!
- 2200 Ohms 2 Pi (f Hz) (0.033 Henry), thus f 10610 Hz 2200 Ohms 1 / (2 Pi (10610 Hz) (C Farad)), thus C 6.818 nF Continue with these two relationships to answer (c) and (d)
- Q: Hello. in basic electronics i learnt that : In DC (analysis) , Capacitor lt; open switch. and inductor lt; short circuitIn AC analysis, Capacitor lt; short circuitBut i also learnt that in AC current, Capacitor 's current is i CdV/dt and Zc jXcI also learnt that when at t0, the uncharged Capacitor acts as short-circuit (why?) and after charge(t--infinite), acts as an open circuit.And at t0 , the inductor acts as an open circuit and at t--infinite , inductor acts as short circuitCan you make an explanation to these?
- Read your textbook. In particular study the part about exponential charge/discharge. Trying to re-write your textbook here would be nearly impossible (and unnecessary). Doug
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Low Frequency Copper Wire Only Chock Coil
- Ref Price:
-
- Loading Port:
- China Main Port
- Payment Terms:
- TT or LC
- Min Order Qty:
- 3000 Pieces pc
- Supply Capability:
- 300000 Pieces per Month pc/month
OKorder Service Pledge
Quality Product, Order Online Tracking, Timely Delivery
OKorder Financial Service
Credit Rating, Credit Services, Credit Purchasing
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