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It is the time required to charge the capacitor, through the resistor, from an initial charge voltage of zero to approximately 63.2% of the value of an applied DC voltage, or to discharge the capacitor through the same resistor to approximately 36.8% of its initial charge voltage.
where C is the capacitance of the capacitor. Solving this equation for V yields the formula for exponential decay: =, where V 0 is the capacitor voltage at time t = 0. The time required for the voltage to fall to V 0 / e is called the RC time constant and is given by, [1]
This time constant determines the charge/discharge time. A 100 F capacitor with an internal resistance of 30 mΩ for example, has a time constant of 0.03 • 100 = 3 s. After 3 seconds charging with a current limited only by internal resistance, the capacitor has 63.2% of full charge (or is discharged to 36.8% of full charge).
The relaxation time is a measure of the time it takes for one object in the system (the "test star") to be significantly perturbed by other objects in the system (the "field stars"). It is most commonly defined as the time for the test star's velocity to change by of order itself. [6] Suppose that the test star has velocity v.
In the short-time limit, if the capacitor starts with a certain voltage V, since the voltage drop on the capacitor is known at this instant, we can replace it with an ideal voltage source of voltage V. Specifically, if V=0 (capacitor is uncharged), the short-time equivalence of a capacitor is a short circuit.
The equation is a good approximation if d is small compared to the other dimensions of the plates so that the electric field in the capacitor area is uniform, and the so-called fringing field around the periphery provides only a small contribution to the capacitance.
In the same vein, a resistor in parallel with the capacitor in a series LC circuit can be used to represent a capacitor with a lossy dielectric. This configuration is shown in Figure 5. The resonant frequency (frequency at which the impedance has zero imaginary part) in this case is given by [22]
A capacitor connected to a voltage supply initially causes a current as it accumulates charge; this current will however decay in time as the capacitor fills, eventually falling to zero. A capacitor will therefore not permit a steady state current, but instead blocks it. [57]: 216–20