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What else can I help you with?
Inductor and capacitor?
An inductor is a magnetic device that resists a change in current. It is constructed with windings that can be backed by ferro-magnetic cores. The equation of an inductor is ... di/dt = V/L ... meaning that the rate of change of current per time is proportional to voltage and inversely proportional to inductance. Inductors, since they work on magnetic fields, can be coupled, as transformers, motors, and generators. A capacitor is a charge device that resists a change in voltage. It is constructed with parallel plates. The equation of a capacitor is ... dv/dt = I/C ... meaning that the rate of change of voltage per time is proportional to current and inversely proportional to capacitance. Inductors and capacitors, since they work in opposite phasor angles, can be coupled to make resonant filters, giving bandpass or bandcut to particular frequencies.
What is meaning of RLC?
RLC circuit(or LCR circuitorCRL circuitorRCL circuit) is anelectrical circuitconsisting of aresistor, aninductor, and acapacitor, connected in series or in parallel. The RLC part of the name is due to those letters being the usual electrical symbols forresistance,inductanceandcapacitancerespectively. The circuit forms aharmonic oscillatorfor current and willresonatein a similar way as anLC circuitwill. The main difference that the presence of the resistor makes is that any oscillation induced in the circuit will die away over time if it is not kept going by a source. This effect of the resistor is calleddamping. The presence of the resistance also reduces the peak resonant frequency somewhat. Some resistance is unavoidable in real circuits, even if a resistor is not specifically included as a component. A pure LC circuit is an ideal which really only exists in theory
Why does a switch spark when disconnecting a coil carrying high dc current?
An inductor resists a change in current. If you have a steady state current going through an inductor and you attempt to suddenly increase the current, the inductor will nearly instantaneously present a higher resistance so that the current does not immediately change. Its resistance, then, will start to decrease as the current ramps up to the new value. Similarly, if you have a steady state current going through an inductor and you attempt to suddenly decrease the current, the inductor will nearly instantaneously present a lower resistance so that the current does not immediately change. Its resistance, then, will start to increase as the current ramps down to the new value. That's all just background information so you can understand what an inductor is. If you have a steady state current going through an inductor and you attempt to suddenly decrease the current to zero by opening the circuit, the inductor will respond by attempting to maintain the current, but that current has nowhere to go. This creates a large negative voltage spike across the inductor. Think about it. Ohm's law says that voltage is current times resistance. You have some current; you have infinite resistance; therefore you must have infinite voltage. In fact, a theoretical pure inductor will do exactly that - generate an infinitely large negative voltage spike. That does not happen in practice, but it is very common to see transients of several hundred or thousand volts. This is why you need to have some kind of suppression circuit in place - otherwise that transient will go back and blow out whatever circuit is driving it.
What is going to wrong if you use adhesive around capacitor?
I don't see anything wrong with that. In industry capacitors are often glued onto circuit boards to hold them better.
What happens when a dc voltage is applied to an inductor?
The equation of an inductor is ...di/dt = V/L... meaning that the rate of change of current in amperes per second is proportional to voltage and inversely proportional to inductance in henries.If, for example, you connect a 200 millihenry inductor across a 12 volt battery, the current will increase at a rate of 60 amperes per second.Now, the question is, can the inductor, conductors, and/or battery handle that? The answer is no. Something is going to fail. The inductor will rather quickly look like a short circuit across the battery.This example does not take resistance into account. Practical inductors, conductors, and batteries have resistance, and that will place an upper limit on current but, still, this is not an appropriate way to connect an inductor to a battery.DO NOT TRY IT IN THE LAB - THERE IS RISK OF EXPLOSION.
Does the current in a pure capacitive circuit lead or lag the applied voltage?
Answer #1Just remember our friend ELI the ICE man! E is voltage, I is current, L is inductance, and C is capacitance. ELI In an inductively reactive circuit (L), the voltage (E) comes first, then the current (I) lags behind. ICE In a capacitively reactive circuit (C), the current (I) leads, then the voltage (E) comes later. Note that while your assumption (in the stated question) is correct, an engineer or electrician would not say it that way. The voltage waveform is the constant, and the current waveform is said to lead or lag. This is because reactive or non-linear loads distort the current. If you look at a power-factor meter, and it says leading or lagging, it is referring to the current. It would be more accurate to re-phrase your question: In a capacitive reactive circuit does the current lead the voltage? Yes! Answer #2: Another method I learned from one of my EE professor is that in an inductor the current lags the voltage because the electrons get dizzy going through all of those loops (coils) in the inductor and "lag" behind the voltage.
What size capacitor can i replace a 1000uf 50v with if i don't have the exact one i am working on a project and it calls for 1000uf 50v cap I know i need 50v plus but can i use 1200uf 50v or 2000uf?
When replacing a capacitor in a circuit, it is generally safe to increase the capacitance value while keeping the voltage rating the same or higher. In this case, you can replace a 1000uF 50V capacitor with a 1200uF 50V capacitor without any issues. However, using a 2000uF capacitor may not be ideal as it significantly deviates from the original capacitance value, potentially affecting the circuit's performance. It is recommended to stick as close as possible to the original specifications for optimal results.
Why does the current lead the voltage across a capacitor by 90 degree rather than lag it?
A: Because a capacitor have to have time to charge to the voltage In a capacitor, the current depends on the voltage difference across it. On AC, this makes it charge, if the voltage is increasing above zero, and discharge if the voltage is reducing towards zero. Because a capacitor has almost no internal resistance, and most loads that it is connected to have only very small resistances in series with the capacitor, the charging and discharging currents depend pretty much on the rate at which the voltage is changing. At the zero crossing point of the sine-wave, when the voltage is actually zero, the rate of change of voltage is very high (the sine-wave is at its steepest), so the current is also very high. If the voltage is positive-going, the current is positive, and if the voltage is negative-going, the current is negative. At the peak of the voltage waveform, the rate of change of voltage is zero or very low (the sine-wave is flat, and not really changing its voltage) so the current is zero, too. Since the maximum positive current occurs when the voltage is passing through zero, going positive, and the maximum negative current happens when the voltage is passing through zero, going negative, the current peaks happen 90 degrees before the voltage peaks, so the current is said to lead the voltage. This is the same as saying the voltage lags the current by 90 degrees.
Is there any electrical current storage device?
Yes, once current starts going through an inductor, it tends to continue going through that inductor in the same direction -- even if that inductor is completely disconnected from everything else. Unfortunately, all available inductors ( even superconducting magnetic energy storage systems -- see link below ) store relatively little energy. You can, however, store the means to produce a current. A battery is such a storage device: when charged, it can produce a current in a conductor. Also, a capacitor stores an electric charge, which when released will produce a current.
Does a capacitor store energy in form of an magnetic field?
Current stops going into a capacitor when it's voltage is equal to the supply voltage. From then there is no flow of current, so there is no magnetic field. Yet the capacitor remains charged and has energy to release if required.
Would a 450V parallel plate capacitor made from copper or aluminum foil melt when discharged through a circuit of about 1 ohm?
There's no reason for the capacitor to heat up, because it's only storing energy,not dissipating it. When you discharge the capacitor, the energy flows out andthrough the external circuit, and that's where it dissipates. If anything is going tomelt or explode, it's going to be something outside of the capacitor, through whichyou try to jam the energy.Which brings us to the 1-ohm resistor . . .You have said that you have a "450V" capacitor. The rating marked on a capacitorisn't the voltage across it when it's charged ... that can be whatever you make it.The marking is the maximum that the capacitor can hold without arcing acrossbetween the plates ... the number is called the "maximum working voltage".You can try to charge a capacitor to whatever voltage you want, but it won'thold any more than the number marked on it.In order to discuss the fate of that 1-ohm resistor, we have to know what theinitial charge is on the capacitor. The only number given in the question is 450V,and even though that's more likely the "max working voltage" of the capacitor,let's assume for the moment that the capacitor is actually charged up to 450 volts DC.The charging is complete, the staff retreats behind their bullet-proof plexiglassbunker, puts on their dark glasses, and prepares to push the button that willremotely close the circuit and discharge the capacitor through the 1-ohmresistor. The button is pushed, and here's what the high-speed camera revealsafter the smoke clears and the fire units have departed:As the energy drains from the capacitor, the voltage on it steadily dwindles.How fast it dwindles depends on the resistance of the external circuit, and onthe "capacitance" of the capacitor (which tells us how much energy it takes tocharge it up to any given voltage).What we do know about this circuit is that the capacitor is charged initially to450 volts, and that it discharges through 1 ohm. This is enough for us tocalculate the initial current at the instant the switch is closed ...I = (450 volts)/(1 ohm) = 450 Amps.We can also calculate the power dissipated by the resistor at that instant:P = I2 R = (450)2 x 1 = 202.5 kilowatts ... roughly the power that would bedemanded of a car battery if it had to start 85 cars all at the same time!Regardless of the capacitance, and how quickly the charge on the capacitordwindles down from 450 volts, it's likely that this initial surge through the1-ohm resistor causes it to self-destruct like a pellet of plutonium on thetower at White Sands.To answer the question:Your capacitor is safe from harm. But if you're going to discharge it through 1 ohm,then please wear gloves, safety glasses, and a kevlar apron.
Why capacitor block direct current?
A capacitor opposes a change in voltage, but it will help to look at both the device and at a circuit up close to see what's going on. Any capacitor is two "plates" separated by a dielectric or insulator. Connect a wire to each plate and you've got the device. In a direct current circuit, the voltage source will cause current flow in only one direction. A common battery is a good example. Let's look further. When a capacitor is connected in a DC circuit and the circuit is energized, the voltage source will want to cause current to flow in only the one direction. In the initial moment when the power is switched on, electrons will flow in the circuit. Electrons will leave the negative terminal of the source and enter the positive terminal. The current flow will travel through the wire, and electrons will "pile up" on one of the plates of the capacitor. As electrons are "piling up" on one plate, their presence there will create an electric field across the dielectric to the other plate. This electric field will cause electrons on that other plate to leave. The capacitor is charging, and the voltage source will, for the first instant of time, think that things are "fine" and current will flow. But as the capacitor charges, current flow drops off, and it eventually stops when the voltage across the plates equals the source voltage. In review, as the DC power is switched on in a circuit with a capacitor in it, current will flow "normally" for the first instant. But as the first electrons arrive on one plate and force them off the other plate, current in the circuit will begin dropping off. The voltage developing across the plates of the capacitor opposes the battery voltage. Eventually the capacitor is charged and all current flow has stopped. There is some math that says something slightly different, but for all practical purposes, the capacitor is considered fully charged in a very short period of time. This will depend on circuit resistance and the ability of the source to deliver current, of course. But that capacitor will, when charged, not "pass" any more current. The voltage across the plates is equal to (an opposing) the source voltage, and no more electrons can get onto the negative plate to force more off the positive plate.
