Low Frequency Cheap Filter/Iductor

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*low frequency Inductor/filter

*High power storage

*Easy insertion,low loose

*Used in various electronic and industry products






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Q:A 45-mH inductor is connected to an ac source (Physics Help)?
The reactance (XL) of an inductor is given by: XL 2πfL where f is the frequency in Hz and L is the inductance in henrys. XL 2π * 400 * 45/1000 XL 113 ohms Now it is just Ohm's Law i v / XL i 20 / 113 i 0.177 A i ≈ 0.18 A { I suspect a typo for answer (B) it should be 0.18 A }
Q:inductor capacitor series circuit?
Vrl 200 + j 300 Vrl √[200² + j300]² 360.56 Vrms 56.31 degrees R Zr 200 V / 25A 8 Ohm Is 25A 0 degrees Zrl 360.56 Vrms 56.31 degrees / 25 A 0 degrees Zrl 14.42 Ohm 56.31 degrees Zrl 8 Ohm + j12 Ohm f 50Hz C 150 μF Zc -1/2πfC Zc -1/(100π(.00015)) Zc -j21.22 Ohm Zr + Zc Zrc Zr 8 + j00.00 Ohm Zc 0 - j21.22 Ohm Zrc 8 - j21.22 Ohm Zrl - Zrc Zl Zrl 8 + j12.00 Ohm Zrc 8 - j21.22 Ohm Zl 0 + j33.22 Ohm Zl 33.22 Ohm Zl 2 π f L L Zl/ 2 π f 33.22 Ohm /(100 x π) .105743 L 105.743 mH Zt Zr + Zl + Zc Zr 8 + j00.00 Ohm Zc 0 - j21.22 Ohm Zl 0 + j33.22 Ohm Zt 8 + j12.00 Ohm Zt 14.42 56.31 degrees Is 25A 0 degrees Vt (14.42 56.31) x (25A 0) 360.5 Vrms 56.31 degrees Pt 360.5 Vrms 56.31 degrees x 25A 0 degrees 9012.5 56.31 W Is² 25A² 0 degrees 625 A² 0 Pr 8 Ohm x 625A² 5000 W Pc - j21.22 Ohm x 625A² -13262.5 VAR Pl j33.22 Ohm x 625A² 20762.5 VAR Pt 5000 W + 7500 VAR Pt 9013.9 VA 56.31 degrees a) impedance of the inductor expressed in complex form Zl 0 Ohm + 33.22 j Ohm b) the values of R and L of the inductor R Zr 200 V / 25A 8 Ohm L 105.743 mH c) the value of the supply voltage, expressed in polar form Vt (14.42 56.31) x (25A 0) 360.5 Vrms 56.31 degrees d) the phase angle of the circuit φ 56.31 degrees e) the apparent and active powers of the circuit Pr 8 x 625A² 5000 W Pt 5000 W + 7500 VAR Pt 9013.9 VA 56.31 degrees f) draw a scaled argand diagram showing the supply current and all voltages Let me goI won't make itYou go on ahead without me :)
Q:Physics inductor in LC circuit question?
We know that he energy in an LC circuit oscillates back and forth between the inductor and capacitor. When Vc(t) 0, VL(t) is at it's maximum and when VL(t) 0, Vc(t) is at it's maximum. At iLmax 2.2A all the energy is in L and there is no energy in C (Vc 0) Solve for L E ½Li² 0.06J -----L 0.12/2.2² 0.0248H 24.8mH We know that ½Li² + ½CV² 0.06J for all t Subtract the energy in L from the max energy to find the energy in C: ½CV² 0.06 - ½Li² 0.06 - ½(0.12/2.2²)*1.1² 0.045J 45mJ This makes sense because the energy in the inductor is proportional to i² 0.06*(1.1/2.2)² 0.06/4 0.015 J so when the current is half in L, the energy is 1/4 meaning the other 3/4 of the energy must be in the capacitor.
Q:inductor capacitor resistor max current voltage physics?
The reactance of the inductor is XL2 * Pi * f * L XL 2 * 3.1416 * 42 * 0.4 XL 105.55776 ohms The reactance of the capacitor is XC 1 / (2 * Pi * f * C) XC 1 / (2 * 3.1416 * 42 * 0.00000443) XC 855.3938 ohms The total reactance of the circuit is X XL - XC X 105.55776 - 855.3938 X - 749.83604 ohms The reactance is actually 749.83604 ohms. The minus sign indicates the power factor is a leading power factor. The impedance of the circuit is Z - Sqrt(R^2 + X^2) Z Sqrt(500^2 + 749.83604^2) Z Sqrt(812254.08688) Z 901.2514 ohms The voltage to produce 200 mA in the circuit is E I * Z E 0.2 * 901.2514 E 180.25 V (This is RMS voltage) The peak voltage will be E / 0.707 180.25 / 0.707 254.95 V The power factor of the circuit is PF R / Z PF 500 / 901.2514 PF 0.554784 leading The phase angle is the angle having a cosine value of 0.554784. This is 56.30415 degrees. I suggest you check my calculations. TexMav
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:Two Inductors in Parallel?
In L1: Vs r i(t) + L1 di/dt di/dt Vs/L1 5/8 0.625 A/s In L2: di/dt Vs/L2 5 A/s Assuming R_inductors 0 steady current in L1: i1 (V/R)/2 0.5 A steady current in L2: i2 (V/R)/2 0.5 A
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:Frequency fixed by capacitor-inductor tuning circuit?
First R1 is adjusted so that the LED puts out pulses at 5 kilohertz. f 1.44/((2*R1 + 10,000)(0.0047 uF)). where f is 5,000 hertz (2*R1 + 10,000)(0.0047 uF) 2.88 x 10^-4 seconds which is 1.44/5,000 hertz R1*0.0094u sec + 4.7 x 10^-5 2.88 x 10^-4 seconds R1* 9.4 x 10^-9 2.41 x 10^-4 seconds R1 is about 25.6 kohms, this sets the transmit frequency The 100 k ohm resistor across the LC tank circuit is used to flatten the response. It DOESN'T modulate the 5 kilohertz frequency. It appears to be optional and is most likely not needed. Try the circuit without R3, R3 will reduce the peak response of the tank circuit. If the circuit doesn't ever seem to trip when the beam is broken, it may need to be incorporated. Use a linear not log pot and set it to the halfway point i.e. 50 kohm The NPN photo transistor conducts when the LED is emitting photons. The current pulses from the photo transistor will be 5 kilohertz pulses i.e. will occur every 200 microseconds. I would leave R3 out and see how the circuit responds. As long as LC satisfies this equation the circuit should work fine f 1/(2*π*√(L*C)) where f is 5,000 hertz √(LC) 1/(6.28*5,000) 31.8 micro seconds LC 1 x 10^-9 s^2 C 1 x 10^-7 farads L 1 x 10^-2 Henry L should be 10 milli Henry not 10 micro Henry So change L to 10 mH, the circuit will not work with a 10 uH inductor.
Q:A circuit consists of 6ohm resistor; an inductor of 18 mH is connected in parallel. The total?
Given R13.Zero, R210.0, L7.0mH, Vs5.8V due to the fact that the present via an inductor cannot alternate instantly, i.Lzero when change closes. V.R1 VsR1/(R1+R2) 1.3V V.R2 VsR2/(R1+R2) 4.5V lengthy after switch is closed, inductor is a brief circuit. V.R1 Vs 5.8V V.R1 0V
Q:what is capacitor,inductor,voltage?
Voltage is a measurement of the amount of Electro Motive Force (E.M.F.) pushing the electrons along; just as pounds per square inch (Or in Australia, Kilopascals) pushing the water. Amps are a measurement of how many electrons are being pushed past a given point. Just as with water we say flow rate or gallons per minute (or on Oz Litres per second.) An inductor is simply a coil of wire. If a current Is passed through the inductor ( coil of wire) s a magnetic field is created. This magnet field can be used to A create movement i.e. electric motor. If two inductors (coils of wire) are wrapped around the same core (say a piece of iron passing an ac current through one inductor will create an electric current in the other inductor. This is the principle of a transformer. An inductor is the solenoid that opens and shut the remote locks on cars. Any electromagnet is actually an inductor. A capacitor is simply two metal plates that are separated by a dielectric insulator. These plates can hold a charge and can discharge back into a circuit when required. Think of an electricity bucket holding electrons. A capacitors uses are many from correcting power factor , used in a timing circuits. Used in a defibrillator, used in an electric fence. Filters for power supplies and many more uses. And I have no idea re a dependant source.
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