What is a displacement current in a capacitor?
In a capacitor, you always have a displacement current and never a conduction current under normal conditions (i.e. you apply a potential difference across it which is below it's specified max. voltage). Conduction currents is when electrons actually move. But in displacement current, no charge carriers are involved. It's just the variations in the electric field, which are imagined to be equivalent to a current.Capacitor contains an insulating material (called dielectric) sandwiched between two conductors. Since insulators can carry only an electric field but not moving carriers, the above paragraph is justified.However, if you apply a huge voltage across a capacitor, it behaves differently. Under sufficiently large potential differences, many insulators stop insulating. i.e. they conduct electricity. So if you apply a large voltage beyond the specified limit, the dielectric behaves as a conductor. So you get a conduction current in the capacitor. This happens just like a lightning strike, when the potential difference between the clouds and the earth becomes so large, that the atmosphere is forced to conduct and the electric flash strikes the ground. This is called breaking down of a capacitorWhen a capacitor breaks down, it no longer carries displacement current. Because it's now a conductor! So, just like any other conductor, the electric field inside it is zero, and so is the displacement current. These days, you get very compact capacitors. Due to their compactness, their dielectric breaks down rather easily. For example, many of today's tiny capacitors are manufactured to function properly only up to 60V, beyond which their breakdown occurs.
Transformer Leakage Current & Efficiency?
I'm trying to develop my understanding of transformers and have run into a question that puzzles me. In tinkering with a microwave oven transformer, I found that the 120V primary winding measures just 56mH (at small signal level). That seems to imply a leakage current of 6 amps! Judging from the size of the spark when I energized the primary, that's about what it's drawing at no-load. I realize microwave oven designers have no incentive to make them efficient, but this really seems a bit much. In any case, I started thinking about resonating the primary with a motor-run capacitor to minimize the leakage current. Then a question occurred to me. All this leakage current is creating flux in the core. That flux is available to the secondary winding for power transfer. Is this leakage current just an issue under no-load conditions? Does it go to good use, does efficiency improve under load conditions? And, to the point, would resonating with a motor-run capacitor actually decrease at-load efficiency while it improves no-load leakage? I know what you're thinking. Tinkering with high voltage, high current transformers can be dangerous. It sure can. I've been working with high power transmitters professionally for thirty years. This is really just a question about the magnetics of it, I suppose I could apply various loads and measure the efficiency to answer this question, but I don't have many load resistors in the hundreds of watt range. Someone who understands can probably answer from intuition? My application involves using the transformer to build a 600V 350ma supply for the plate supply of a 200 watt ham transmitter. I anticipate drifting out the magnetic shunts and peeling off about 2/3 of the secondary turns to get the required output voltage. Using this 1200 watt transformer at the 200 watt level should keep it nice and cool.
How do you calculate the leakage current of a parallel plate capacitor?
A capacitor’s leakage current acts to discharge it. Sampling the terminal voltage at intervals is a measure of the value of the leakage resistance. In one CR time constant, the terminal voltage decays to 37% of its initial value. The leakage current decays in the same way.
Does direct current pass through a capacitor?
It depends on what you mean by "blocks" DC. Certainly when a capacitor charges, current goes in one terminal and out the other terminal until the capacitor has charged to the applied voltage. It behaves as though the current went "through" it but no electrons which entered one terminal came out the other terminal, so in that sense it "blocked" the current. A better way to think about DC blocking is to imagine a voltage source which has both an AC and a DC component, connected in series with a capacitor and resistor. The capacitor will charge up until its average voltage is equal to the DC component and this charging current will pass through the resistor until charging is complete. From then on, the current in the resistor will correspond to just the AC component of the voltage source and the DC component will be across the capacitor. In that sense, the DC component has been blocked from the load resistor. If the DC component of the input voltage changes, the capacitor will charge or discharge additionally until its average voltage is equal to the average of the applied voltage and this transient current will flow in the resistor and appear across it as a transient voltage. In that sense, the DC was not blocked. It is worth noting that, since a capacitor in series with a resistor cannot charge instantly, the voltage across the capacitor cannot change instantly. That means that in this example, if a step change is made in the DC component of the input voltage, that voltage step will appear instantly across the load resistor, gradually dying off as the capacitor charges over time to the new average DC voltage. In that sense, the capacitor did not block the change in DC voltage at first but only over time. You get into the question of what do you call DC voltage since a step change in the DC component of the input voltage is not really DC, it is really a transient.
If a charge is placed between 2 metal plates applied with +V and -V voltage, how much time will the charge take to reach the plate?
By ‘charge’, I assume you mean charged particle. I assume you mean in a vacuum with no other forces (eg. gravity, magnetic) and starting from rest.The electric field is given by E= ∆V/r (r is the distance between the two plates and ∆V= 2V, given by the asker)The force would be F=qE (absent other forces)The acceleration is given by a=F/m (where m is the mass of the particle)Rearranging the standard distance formulae, d = [math]tv_{average} [/math]and [math]v_{max}=2v_{average}[/math][math]=at[/math],we gett= [math]\sqrt{2d/a}[/math]where d is the distance from the -V or "bottom" plate)Therefore the total time, under all the assumptions above, is surprisingly succinct:t=[math]\sqrt{dmr/qV}[/math]Interestingly, this result is important for Milikan and Fletcher’s Oil Drop Experiment
What is the Breakdown voltage of any capacitor and how is it determined?
Capacitors used on Low voltage printed circuitboards generally have a voltage rating written on them.This voltage rating on a capacitor is the maximum amount of voltage that a capacitor can safely be exposed to and can store. The capacitor won’t store or handle this voltage unless the circuit in which this capacitor is connected handles it. A capacitor's voltage rating implies the maximum voltage it can handle or store safely Its not the voltage that the capacitor will charge up to. The capacitor could breakdown if the capacitor voltage is less than the voltage in the circuit. A DC capacitor (electrolytic) like the one shown above could also fail if polarities are connected in the wrong way!Power capacitors are normally rated for kVAR for a said voltage. Generally capacitors are rated 10–20% greater than the system voltages they are handling.When handling voltages closer to capacitor rated voltage, an increase in leakage current occurs. This causes bubbles in the electrolyte and may boil due to high power dissipation (along with high current usage). Eventually the capacitors could explode.
How much does 1/4 steel plate weigh?
Here is an approximate value. It will depend on the type of steel. 1/4" = 1/48' 4' x 8' x 1/48' = .667 ft³. The density of steel is around 490 lbs per ft³ .667 ft³ x 490 lbs / ft³ = 327 lbs.
Why does concrete kill a lead-acid battery?
What it does is cause the battery to discharge because of moisture trapped under the battery. The moisture provides a high resistance path for current to flow without an electrical connection. Concrete is actually porous to a certain degree, and absorbs moisture much like a sponge, unless the surface is sealed with something like paint or plastic. Putting a battery on a bare concrete floor is like putting a battery in a puddle of water. It WILL discharge because of the moisture in a week or two. The problem is when it is discharged THEN the plates "shed" lead oxide, and THAT is what destroys the battery, NOT the fact it sat on concrete. Sitting on concrete simply discharged the battery, and it was sitting in a discharged state without a charge that actually destroyed the battery. Merely sitting a battery on concrete does NOT destroy he battery, sitting in a discharged condition for a long time does. It does not matter where the discharged battery sits. It could sit on your kitchen table, and if in a discharged state for a period of time, it will die, die, die...
How long will it take for a 6 amp battery charger to charge a 90 amp hour car battery?
Depending on the charge of the battery, it could be an hour to all night. The charger starts out with 6 amp and will slowly drop as the battery becomes charged, since there is some backward resistance from the battery. As it nears full charge, it will be tapered off to 1 amp or so, and just enough to keep it charged, So stating a time for complete charge is pretty difficult. If the battery is weak, leave it on over night, at least 8 to 10 hours, and it will taper off on its own, or should. Most chargers will. Make sure yours does, or it will over charge the battery, and cause hydrogen gas to be emitted. Hydrogen gas will come out anyway, but not as bad if using a taper type charger. Don't smoke or make a spark close to the battery after charging it. Also, let it set for a while, before installing it. Check the voltage right after charging, and then an hour later, check it again. See if it is holding charge. If it drops, your battery is on its way to Battery Heaven.
What is the difference between a battery and a capacitor?
a battery produces electricity and a capacitor stores it.