How are Inductors and Capacitors Constructed?
930475
An inductor is a coil of wire sometimes with an iron core.  \n\nA capacitor is two parallel sheets of conductor separated by a thin dielectric (could be air or vacuum).\n\nThe physical shape or size depends on the application.
1050520
Think about the simpliest capacitor. It is clear that two metal plates that don't touch won't let a DC current pass, right? But the spacing is very thin so that the plate's charges can effect one another. When you first put current (even DC) to a discharged capacitor, it charges up and for a brief time it does conduct current. How long? Well it will charge to 2/3 of the total voltage in RC seconds (where R is the resistance of the circuit). So when AC current is "fast" enough, the cap is constantly in a charge/discharge state and current flows. The effective resistance (reactance) of the cap is 1/(2*pi*F*C). So the resistance goes down as the capacitance and the frequency rises.\n\nThe inductor, on the other hand is just a coil of wire. Again, it is easy to imagine that it will conduct DC. But when you first apply current (even DC) to a coil, the magnetic field will interact with the coil and produce an opposing voltage. This only occurs when the magnetic field is forming or collapsing (remember, magnetic fields that do not move do not produce electricity). So right when you change the current, the inductor will "resist" the change, but if the current doesn't change, it is just a wire. The reactance of a coil is 2*pi*F*L, so the resistance increases with frequency and inductance.\n\nNow, combining inductors and capacitors is another interesting subject, but you didn't ask about that....
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I have already answered part 1. in another question you've asked. I think you probably didn't use the right units (like cm needs to be convered to meters) and got the wrong answer.\n\nNow for part 2:\n\nInduced EMF, ℰ = -L(di/dt)\n\nwhere i is the current and L is the inductance\n\nWe are given that the magnitude of the induced current,\n\ndi = 47 mA\n\nand the oscillation frequency is 100 Hz. This means that dt = 1/f = 0.01 sec\n\nWe have L = 8.44 × 10-^3 henries from before\n\nSo |ℰ| = L(di/dt) = -(8.44 × 10-^3)(0.047)/(0.01) = 0.0397 volts\n\n------------------------------------------------------\n\nNow part 3 has to do with mutual inductance:\n\nVs = VpNs / Np\n\nwhere \n\nVs is the voltage across the secondary inductor, \nVp is the voltage across the primary inductor (the one connected to a power source), \nNs is the number of turns in the secondary inductor, and \nNp is the number of turns in the primary inductor. \n\nTherefore Vp = VsNp/Ns and Vs = 0.0397 volts as calculated before, so\n\nVp = (0.0397)(197)/(2920) = 0.00268 volts
230874
Hey, ive built some circuits....check out:\nwww.techlib.com\nwww.home.maine.rr.com\nwww.hanssummers.com\nwww.angelfire.com\nif they don't work try http:\nthey are all good website to build transmitters/recievers.......or google it
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V=It/C\n\nwhere\nI is current in amps\nV is voltage in volts\nt is time constant in seconds\n\nIn an ideal capacitor, this holds true.  Theoretically, an ideal capacitor has no leakage current and no inductance.
52883
A moving charged particle will create a magnetic field.  Conversely, a moving magentic field can make an electric current.  The orientation of the magnetic field goes around the current.  So if you coil a copper wire and run a current through it, all of the magnetic field will add together in the middle of the coil and make a magnetic field.  You can wrap a wire around a steel rod, and make an electromagnet in this way.  But in electronics, what they can do is put two coils in close proximity.  As you run current through one coil, it creates a magnetic field.  The creation of the new magnetic field in the vicinity of the other coil will induce a current in that second coil.  This is called and inductor, and that's how induction works.
1097336
solenoid connected to a timing capacitor, cutting of the connection until another quarter is entered. \n\nsupport hallitubes: http://www.hallisystem.bravehost.com
639035
Not easily. IC fabrication techniques are essentially planar and can't be used to create helical structures in an IC. \n\nWhat can be done is to simulate inductors using capacitors (small value capacitors can be fabricted on ICs), and active circuitry. Sometimes this is possible, but sometimes it's not enough. Particularly at low frequencies (eg, audio signals from near DC to perhaps 20KHz), external inductors may be the only choice.
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if u know how to build a cricuit i can explain you .!!!!\n\nbut if u dont know how to make a circuit... u r in wrong way ... u probably need to learn circuit making... or better yu buy a remote from a shop ;-)
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See Resources below\n\n***WARNING: When working with power supplies, be sure to read up on power supply safety.  It is quite dangerous (possibly FATAL) to work on power supplies if you do NOT have knowledge about how to SAFELY work with power supplies***
772728
A resistor will resist current. A capacitor will push current in a certain derection, depending on how charge is built up on it. An inductor will resist change in current. No matter how much current is flowing through a circuit, the inductore will not resist that current until the current begins to change. This is the basis of RLC, alternating current.\n\nThe capacitor pushes current in one direction. The inductor resists this, but the capacitor wins. Then the capacitor eventually charges down and holds no potential difference, but has created a current in the circuit. The capacitor is no longer pushing the current forward, so the current would just stop if it were not for the inductor. The inductor keeps the current from changing too fast and causes the capacitor to build up another potential difference. This causes a current to move back and forth in an alternating fashion.
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It is assumed you have completed all of the tutorials on basic electronics because without that basic grounding you will go nowhere here. It is assumed you have a good understanding of ohms law to be able to follow the necessary calculations here.\n\nMany people are interested in Audio Amplifier design. If this is your area of interest then the first thing I sincerely recommend you do is buy this excellent reference book through my amazon affliate program.\n\nHigh-Power Audio Amplifier Construction Manual - G. Randy Slone, Randy Slone - highly recommended. Paperback and 476 pages.\n\n"Design and build awesome audio amps. Amateur and professional audiophiles alike can now design and construct superior quality amplifiers at a fraction of comparable retail prices with step-by-step instruction from the High-Power Audio Amplifier Construction Manual. Randy Slone, professional audio writer and electronics supply marketer, delivers the nuts-and-bolts know-how you need to optimize performance for any audio system--from home entertainment to musical instrument to sound stage".\n\n"Build a few simple projects or delve into the physics of audio amplifier operation and design. This easy to understand guide walks you through: Building the optimum audio power supply; Audio amplifier power supplies and construction: Amplifier and loudspeaker protection methods; Stability, distortion, and performance; Audio amplifier cookbook designs; Construction techniques; Diagnostic equipment and testing procedures; Output stage configurations, classes, and device types; Crossover distortion physics; Mirror-image input stage topologies".
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Differential equations is very useful to electronics inductries. This is commonly used in designing electrical circuit and their responses to the signals. Input current response, voltage responses and even time decay is critical to the computation of integrated circuit responses. Without is certain electrical functions such as response time, triggering mode, active and inactive function regions will not be possible.
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The basic components are:\n\nAn antenna (to "capture" the RF signal)\nA filter (LC circuit to tune to the frequency of interest)\nAn amplifier (to amplify the weak signal)\nAn FM demodulator (to downconvert and "decode" the FM signal and get back the baseband signal which is what our ears interepret as sound)\nAnother amplifer (make it variable so you can change the volume)\nA speaker\nOh, and of course the power supply (battery) to bias all the active components.\n\nI know you asked for a wiring diagram...but maybe try purchasing each of the above components seperately as modules and put the thing together... just do a google search... but congrads for trying to build an FM radio... It shouldn't be too bad... Make sure you have a bread board....
835611
Inductors are used , usually in combination with capacitors, to filter or shape electrical signals (voltage or current).  The inductor functions by storing electrical energy in its magnetic field.\n\nHere are some examples of the applications of inductors:\n1. The station tuning circuit in most radios\n2. In the "crossover" network in speaker systems (to filter out and send the high frequency content to the "tweeter" speaker)\n\n3. In most modern switching power supplies as an electrical energy storage device.
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1. Solving AC circuits that have resistors, inductors, and capacitors in them.\n2. Signal processing\n3. Solving differential equations.\n\nA smart guy by the name of Euler showed that e^(i*x) = cos(x) + i*sin(x), thus allowing the conversion of a complex number in rectangular form (a + b*i) into polar form (r*e^(i*x)). It also allows us to write sinusoids in terms of complex exponentials, as in:\ncos(x) = (1/2)(e^(i*x)-e^(-i*x))\nIt turns out, perhaps unexpectedly, that complex exponentials are oftentimes easier to deal with then sinusoids. It also turns out, according to a guy by the name of Fourier, that most "well-behaved" functions can be written in terms of sinusoids (and thus can be written in terms of complex exponentials as well). This paved the way for a lot of neat applications. Google "complex exponential" (or any of the terms or names I dropped here) if you want more details.
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Electronic engineering - circuit analysis:\n\nSolving a circuit for the current and voltage according to Kirchhoff's Laws and these device equations:\nResistor   : v = R * i\nCapacitor: i = C * dv/dt\nInductor   : v = L * di/dt\n\nAnything complicated you generally solve by using the Laplace transform, in which case the above equations simplify to\nfor V,I = Laplace transforms of v,i:\nResistor   : V = RI\nCapacitor: I = sCV\nInductor   : V = sL\n\nThen you solve the system of equations for V,I and apply a reverse Laplace transform.