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.
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 voltage source is suddenly connected to an electrical circuit, causing a current to flow through a capacitor, the capacitor initially acts like a short circuit, allowing a large current to flow. As the capacitor charges up, the current decreases until it reaches a steady state where the capacitor is fully charged and no current flows through it.
Then I'll try this. Just as V=IR is the fundamental equation relating voltage, current and resistance for a resistor circuit, the following equation relates voltace, current and capacitance for a capacitor: Or, if you are not familiar with that calculus term with the derivative, you can think of it as: I(t) = C * (change of voltage per time) So when you have DC, there is no change of voltage with respect to time, so there is zero current. When you have an AC voltage signal that varies across the capacitor with time, that equation lets you calculate the current that results through the capacitor. A capacitor is two surfaces near each other, but not touching. A direct current "sees" a capacitor as an open switch. It cannot pass through. An alternating current "induces" a charge in a capacitor and can pass through.
The current through a material can be changed by varying the voltage applied across the material. By adjusting the voltage, you can increase or decrease the current flowing through the material. Additionally, changing the resistance of the material can also impact the current flowing through it.
It really depends on the experimental setup. If you have only a capacitor and a resistance in series, the current discharge from the capacitor will start high, then gradually go down. If you have a capacitor and an inductor in series, the current discharge will start being small, because the inductor will oppose any CHANGE in the current - that's how they work.
Capacitors resist a change in voltage, proportional to current and inversely proportional to capacitance. In a DC circuit, the voltage is not changing. Therefore, after equilibrium is reached, there is no current flowing through the capacitor.
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.
The voltage-current relationship for a capacitor is i = C dv/dt, where i is the current flowing through, C is the capacitance and dv/dt is the time rate change of the voltage across that capacitor. So, when a capacitor is fully charged, the voltage no longer changes with time (the derivative, dv/dt, is now 0). As can be seen from the equation, the current would therefore be 0. Anything with 0 current flowing through is an open circuit, and can be treated like a resistor with infinite resistance (in models, anyway). Practically speaking, capacitors aren't this perfect, but you will still have an extremely high resistance once fully charged (voltage changes negligibly after charging).
When a voltage source is suddenly connected to an electrical circuit, causing a current to flow through a capacitor, the capacitor initially acts like a short circuit, allowing a large current to flow. As the capacitor charges up, the current decreases until it reaches a steady state where the capacitor is fully charged and no current flows through it.
Then I'll try this. Just as V=IR is the fundamental equation relating voltage, current and resistance for a resistor circuit, the following equation relates voltace, current and capacitance for a capacitor: Or, if you are not familiar with that calculus term with the derivative, you can think of it as: I(t) = C * (change of voltage per time) So when you have DC, there is no change of voltage with respect to time, so there is zero current. When you have an AC voltage signal that varies across the capacitor with time, that equation lets you calculate the current that results through the capacitor. A capacitor is two surfaces near each other, but not touching. A direct current "sees" a capacitor as an open switch. It cannot pass through. An alternating current "induces" a charge in a capacitor and can pass through.
With a series RLC circuit the same current goes through all three components. The reactance of the capacitor and inductor are equal and opposite at the resonant frequency, so they cancel out and the supply voltage appears across the resistor. This means that the current is at its maximum, but that current, flowing through the inductor and the capacitor, produces a voltage across each that is equal to the current times the reactance. The voltage magnification is the 'Q factor', equal to the reactance divided by the resistance.
As long as you don't exceed the breakdown voltage of the capacitor ... which is marked right on it ... DC voltage on it produces NO current flow through it. Only AC 'appears' to flow through a capacitor, and even that appearance is bogus when you really get down to it.
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.
AC can pass through a capacitor. The higher the frequency of AC the lower the reactance (like resistance). The current and applied voltage are 90 degrees out of phase the current leading the voltage by this amount.
The current through a material can be changed by varying the voltage applied across the material. By adjusting the voltage, you can increase or decrease the current flowing through the material. Additionally, changing the resistance of the material can also impact the current flowing through it.
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.
It really depends on the experimental setup. If you have only a capacitor and a resistance in series, the current discharge from the capacitor will start high, then gradually go down. If you have a capacitor and an inductor in series, the current discharge will start being small, because the inductor will oppose any CHANGE in the current - that's how they work.