An ideal capacitor is characterized by a single constant value for its capacitance.
The effective resistance of the capacitor reduces the ripple current through the capacitor making it less effective in its function of smoothing the voltage. But if the capacitor filter is fed by a transformer and diodes, the resistance of the transformer exceeds that of the capacitor.
At resonance...a parallel tank circuit matches the applied sine voltage so close that there is almost 0 current flow from the source...i.e., max impedance at resonance...the capacitor and inductor are swapping energy with each other in tune with the source... visualize it...in order to have 0 current flow for an incoming varying voltage...that would mean that the tank voltage would be varying exactly at the same frequency and voltage! Thus...you have effectively "tuned" into a voltage which would be critical in 'tuned' voltage amplifier... A series resonant circuit does not tune into a voltage...in fact at resonance the voltage across the inductor capacitor will be 0!...a short or minimum impedance condition Of course my discussion assumed ideal components...in the real world there will be 'stray' resistances which will alter the results in magnitude to the size of the resistance... Hope this helps
When you apply DC directly to a capacitor, it charges to the value of the DC potential, and then there is (nearly1) zero current flow through the capacitor. If the capacitance is large enough, though, and the DC source has a low enough impedance, the current flow can be quite substantial, damaging things.The reason the equilibrium current is zero is that a capacitor resists a change in voltage, proportional to current and inversely proportional to capacitance...dv/dt = i/c... which makes the capacitor essentially a high pass filter, and a DC blocker.1 The equilibrium current is "nearly" zero because, in our non-ideal world, every capacitor has some leakage current. Practically, the current is zero - from a purist perspective, it is not.
1. Compressor 2.A fan 3.A gas, like freon, which vapourises @ a temp lower than room temp 4.The tubing/Piping An AC is a heat transfer operation.The gas(in the form of a liquid) travels through the piping and it evaporates very quickly cooling the tube.The fan blows air across the tubes into the room, now to turn back the evaporated gas into a liquid so that it can be evaporated again, it is led into the compressor, the compressor is basically a mechanical component, which squeezes the gas back into liquid so that it can be evaporated again, this happens continously, and during compression, a lot of heat is generated, which is thrown into the outside atmosphere.This is the basic working of any refrigating device.
The Sun or heat sources can be a greatideal energy source.
A: A voltage source Will charge a capacitor to 63% of its input value, The value to get there is stated a Resistance time capacitor as time. Mathematically it will never get there but engineering consider 5 times RC time constant as close enough,
Yes. A capacitor stores charge from any source, including AC.The difference between DC and AC, however, is that the capacitor will be constantly changing in charge, in step with the AC. Due to the nature of the capacitor, the current through the capacitor will lead the voltage by some amount, depending on capacitance and resistance. {In the ideal case of a perfect capacitor, conductors, and AC power source, the current will lead the voltage by 90 degrees phase angle.}This is called capacitive reactance.Another way for a capacitor to store charge from an AC source, of course, is to place a rectifier diode in front of the capacitor. This, then, becomes an AC to DC converter.
basically a capacitor will charge to the input DC level however it will mathematically never happen since capacitors charge at a certain rate the voltage drop across a capacitor will follow the R C time constant or 63% of the applied voltage for a unit time.AnswerIn the case of an a.c. supply, yes, there will be a voltage drop across a capacitor. In the case of an 'ideal' capacitor, this will be the product of the load current and the capacitive reactance of the capacitor.
Because a capacitor is ideal for storing energy over short periods, as in a reservoir capacitor in a power supply.
The effective resistance of the capacitor reduces the ripple current through the capacitor making it less effective in its function of smoothing the voltage. But if the capacitor filter is fed by a transformer and diodes, the resistance of the transformer exceeds that of the capacitor.
Voltage source inverters use the dc voltage (e.g a capacitor in parallel) as a source while the current source inverer (inductor in series) use the dc current as a source. Please note that voltage can not be changed abruptly in capacitor as current can not be changed abruptly in inductor.
An ideal voltage source has no internal resistance, and a constant voltage output. In reality, all voltage sources (battery, generator, etc.) have some internal resistance, and their voltage may degrade or change over time.Ans 2: An ideal voltage source will have zero input impedance and the voltage can rise to infinity to supply the current.Read more: What_does_an_ideal_voltage_controled_voltage_sources_do
Ripple Voltage is voltage variation across the load and it is the AC component. To answer this question, consider a Half Wave rectifier with a smoothing capacitor: This rectifier will consist of a sinusoidal voltage source, an ideal diode, a capacitor in parallel with the load. At t=0, the voltage across capacitor = load voltage When the circuit is switched on, the capacitor is fully charged as the sinusoidal source reaches its peak. However, the sinusoidal nature causes the source voltage to decline after reaching the peak. This means that no current will flow through the diode. But the capacitor is still charged. So this will supply current to the load while it discharges. But during the discharging period (till the sinusoidal picks up again), the load voltage is an exponential function = peak voltage *exp-[(t - t')*resistance of load*capacitance] Now a key point is that the pulsating current is flowing through the diode to recharge the capacitor. Because of this constant charge and discharge of the capacitor in the cycle, the load voltage has AC ripples. At the same time load current is never zero and is directly prop to load voltage. The dc component >> ac component and the ripple voltage is greatly reduced by the capacitance esp a large one. You can minimize these by choosing a large capacitance. This is how a capacitor accounts for AC ripples. You can never actually rid these ripples even if you use a full-wave rectifier! Google search half - wave rectifier graphs on the ripples to understand this!! --- Sona
A current source varies the output voltage to maintain the desired current. A voltage source has a constant output regardless of the current draw (up to the capacity of the supply, of course).
yes
when the magnitude of voltage of a source is controlled by another small voltage source in the circuit the former is called voltage controlled voltage source and the later is called controller voltage source.
zener diode