Low Frequency Electronic Transformer
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- Q: I want to know theory behined metal detector.Also that what makes a coil an inductor?what will happen if dc is applied on inductor?
- first one can be king kjors provided that he's ideal contributer of the wrestling area. 2d could be gary v simply by years he has watched wrestling. additionally others who provide stable consumers that do not junk mail or rfile questions for absalutley no motives.
- Q: An inductor has an impedance of 30.0 Ω and a resistance of 20.0 Ω at a frequency of 80.0 Hz. What is the inductance? (Model the inductor as an ideal inductor in series with a resistor.)?
- Impedance for an ideal inductor is: Z j*omega*L For a real inductor with a parasitic resistance modeled in series: Z R + j*omega*L j refers to the imaginary number unit. omega is the angular frequency of the signal. We were given the cycle frequency and we can convert: omgea 2*Pi*f I assume your 30 ohms refers to the magnitude of impedance. Combine both components of the complex number using the Pythagorean theorem. Zmag sqrt(R^2 + (omega*L)^2) And with a substitution: Zmag sqrt(R^2 + (2*Pi*f*L)^2) Square both sides: Zmag^2 R^2 + (2*Pi*f*L)^2 Solve for L: (2*Pi*f*L)^2 Zmag^2 - R^2 2*Pi*f*L sqrt(Zmag^2 - R^2) Resulting expression: L sqrt(Zmag^2 - R^2)/(2*Pi*f) Data: Zmag:30.0 Ohms; R:20.0 Ohms; f:80.0 Hz; Result: L 0.04449 Henries Or, with a convenient prefix: L 44.49 milliHenries
- Q: I'm building a circuit in a program called Cadence and the inductor has a little circle on one side of it but not on the other. So does the inductor have a direction? Should the part with the circle be connected to the voltage source or should the other side be connected to it?
- inductors do not have polarity. .
- Q: An 8.0 mH inductor and a 2.0 ohm resistor are wired in series an ideal battary. A switch in the circuit is closed at a time 0 sec. at which time current is zero the current reaches half its final value at a time of ofThe answer is 2.8 ms, How?????
- Let Vo be the voltage across the battery and VL be the voltage across the inductor then for your circuit: VL Vo e-(tR/L) If you divide both sides by R you get the current relationship instead (Ohms law) VL/R Vo/R e-(tR/L) this equals:- I Io e-(tR/L) or you can simply start with this line instead. for the current to reach half its value I/Io 1/2 I/Io e-(tR/L) 1/2 e-(tR/L) Now, to get rid of the exponential part we take NATURAL logs (ln on your calculator) of both sides giving:- ln(1/2) -(tR/L) this gives t as:- t-L/R ln(1/2) t -[(8x10^-3)/2 x (-0.6931) t 0.00277 t 2.8 ms
- 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: Building a BFO metal detectorReference coils calls for,AWG #30dimensions: 50 mm height x .5 mm width120 turnsI'll be using AWG #24. However, AWG #24 is .5 mm in diameter. If you do the math, that's120 turns x .5 mm 60 mmSo my coil will be 60 mm. THAT'S NOT WHAT THE ENGINEER CALLS FOR. I'm 10 mm too long, therefore I have an extra 20 winds (10 mm of 24 awg 20 winds)CAN I OVERLAP?Problem is the reference coil (this coil) is fitted into a water fitting. The water fitting is threaded on the outside fitted with a nut. When you move the nut up down the water fitting, it changes the frequencyWon't this be an issue?THanks
- The final inductance increases a little bit by overlap winding.
- Q: A series circuit contains a resistor with R 24 ohms, an inductor with L 2 H, a capacitor with C 0.005 F, and a generator producing a voltage of E(t) 12 sin(10t). The initial charge is Q 0.001 C and the initial current is 0. Find the charge at time t.Q(t) ??
- First thing to realize is that this circuit is in resonance. The characteristic frequency is equal to 1/sqrt(LC) 1/sqrt(2*.005) 10 This means the driving source won't provide net new charge to the capacitor and we can look at this in two pieces so for the output across the capacitor we have v-out 1/jωC/(R+jωL+1/(jωC))*v-in 1/(1-ω^2LC+jωRC) When we plug in values we get v-out1/(1-10^2*(2*0.005)+j10*24*.005) v-out1/(1-1+1.2j)*v-in -j/1.2*12sin(10t) 10sin(10t-π) so the change in Q across the capacitor from the driven source will be Q(t)-driven 0.005*10sin(10t-π) 0.0510sin(10t-π) C now we have to look at the initial conditions with a voltage of 0.2 V stored in the capacitor. We'll short out the driven source and solve for the decay in voltage through the resistor and inductor Because there is in essence a DC voltage stored on the capacitor and discharged through the resistor, the inductor has no role. The voltage discharge is 0.2exp(-t/RC)0.2exp(-8.33t). The charge is just this multiplied by C. or 0.2*0.005exp(-8.33t)0.001exp(-8.33t) the total will be Q(t)0.0510sin(10t-π)+0.001exp(-8.33t)
- Q: An inductor of 0.5 H and a resistance of 1.2 K ohms are connected in series with an ideal source with e.m.f. 6 V. The current expression in the circuit as a function of time is: 5.0*10^-3(1-e^-2398t ) . Calculate the potential difference across the inductor at t 0.417 ms. the correct answer is: 2.21v . can you tell me how to do it? :)
- Set t 0.417*10^-3 in the formula given. That gives you the current at t 0.417 ms ie 3.16*10^-3 ma In the 1.2k resistor, that drops v i*r 1.2*10^3*3.16*10^-3 volts ie 3.79 volts That leaves (6 - 3.79) volts ie 2.21 volts across the inductor.
- Q: How do you do this problem?Using the expression relating the rate at which current changes in an inductor to the voltage produced at it, find the voltage across the inductor when a switch supplying 6.8 A to a 6.5 mH inductor is suddenly opened. Assume that current drops linearly to zero in 3.4 milliseconds.
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- Q: If R 7 ohms L 22 mH and V 95 volts, what is the voltage across the inductor at 1T?
- TL/R. If you apply a 95 volt step to a series R L circuit the voltage, across the inductor will be 95 e^(-t/T). At tT the voltage will be 23.92 volts.
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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 |
| Company Certifications | CE Certificates |
2. Manufacturer Certificates
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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 |
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Low Frequency Electronic Transformer
- Ref Price:
-
- Loading Port:
- China Main Port
- Payment Terms:
- TT or LC
- Min Order Qty:
- 800 Pieces pc
- Supply Capability:
- 10000 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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