In a capacitor, the current LEADS the voltage by 90 degrees, or to put it the other way, the voltage LAGS the current by 90 degrees.
This is because the current in a capacitor depends on the RATE OF CHANGE in voltage across it, and the greatest rate of change is when the voltage is passing through zero (the sine-wave is at its steepest). So current will peak when the voltage is zero, and will be zero when the rate of change of voltage is zero - at the peak of the voltage waveform, when the waveform has stopped rising, and is about to start falling towards zero.
90 degrees. In an AC circuit with a pure capacitance, the current leads the voltage by 90 degrees. This is because the current in a capacitor is proportional to the rate of change of voltage across it, leading to this phase relationship.
the voltage number on the capacitor indicates that the capacitor can with stand to that particular voltage across it.generally during design, the value of capacitor will be selected in such a way that this voltage rating should be double than what really we get in the circuit
Ripple voltage, in the presence of a filter capacitor, is inversely proportional to load resistance. If the load were zero (resistance infinite), then there would be no ripple voltage. As the load increases (resistance decreases), the ripple voltage increases. The ripple waveform will appear to be sawtooth, with the rising edge following the input AC from the diode's conductioin cycle, and with the falling edge either being linear or logarithmic, depending on load. If the load is resistive, without a regulator, the falling edge will be logarithmic. If the load is constant current, such as with a regulator, the falling edge will be linear.
The change in potential difference across a capacitor is determined by the amount of charge stored on the capacitor and the capacitance of the capacitor. The relationship is given by V = Q/C, where V is the potential difference, Q is the charge stored on the capacitor, and C is the capacitance.
An LED is a diode that emits light; diodes allow current to flow only one direction. The voltage applied to the diode attempts to force current to flow in a specific direction. If the voltage polarity is reversed, and current was flowing before (so there was a small voltage drop across the diode), current will cease to flow (assuming the voltage is not too high for the diode to handle), and (almost) all the voltage will be dropped across the diode (a small leakage current may flow, which means some of the voltage will not be dropped across the diode, but this is in the milli or micro range). I would never define a diode as a "voltage controller" or "current controller". It could be either or both, from the above description.
The relationship between capacitor current and voltage in an electrical circuit is that the current through a capacitor is directly proportional to the rate of change of voltage across it. This means that when the voltage across a capacitor changes, a current flows to either charge or discharge the capacitor. The relationship is described by the equation I C dV/dt, where I is the current, C is the capacitance of the capacitor, and dV/dt is the rate of change of voltage with respect to time.
When a current flows through a capacitor, the voltage across it increases or decreases depending on the rate of change of the current. If the current is constant, the voltage remains steady. If the current changes rapidly, the voltage across the capacitor changes quickly as well.
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
The voltage drop across a capacitor is directly proportional to the amount of charge stored in it. This means that as the charge stored in a capacitor increases, the voltage drop across it also increases.
In the ac waveform of a capacitor the current waveform leads the voltage waveformcurrent is large to start until capacitor fills with it's voltage charge if that helpsAnswerThe terms 'leading' and 'lagging', used when describing power factor, are defined in terms of whether the load current is leading or lagging the supply voltage.In a capacitive circuit, the load current leads the supply voltage, so the power factor is leading.
Capacitors can pass alternating current provided the current and the voltage are within the capacitor's rating. Very often there is a dc bias voltage across the capacitor as well as the ac voltage, so the peak voltage must not exceed the limit. Electrolytic capacitors must not have a reverse voltage across them in any circumstances, because this can cause failure.
It might mean that the voltage across a capacitor cannot change instantanteously because that would demand an infinite current. The current in a capacitor is C.dV/dt so with a finite current dV/dt must be finite and therefore the voltage cannot have a discontinuity.
No, the voltage across a capacitor cannot change instantaneously. It takes time for the voltage across a capacitor to change due to the storage and release of electrical energy in the capacitor.
Yes, voltage matters when charging a capacitor. Capacitor charge rate is proportional to current and inversely proportional to capacitance. dv/dt = i/c So, voltage matters in terms of charge rate, if you are simply using a resistor to limit the current flow, because a larger voltage will attempt to charge faster, and sometimes there is a limit on the current through a capacitor. There is also a limit on voltage across a capacitor, so a larger voltage could potentially damage the capacitor.
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.
90 degrees. In an AC circuit with a pure capacitance, the current leads the voltage by 90 degrees. This is because the current in a capacitor is proportional to the rate of change of voltage across it, leading to this phase relationship.
One interpretation is that if you take a capacitor that is not charged, it needs to take some current before any voltage appears across it. Therefore the current must precede the voltage.