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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.
So the capacitor will be charged to about 63.2% after τ, and essentially fully charged (99.3%) after about 5τ. When the voltage source is replaced with a short circuit, with the capacitor fully charged, the voltage across the capacitor drops exponentially with t from V towards 0.
For brevity, the notation omits to always specify the unit (ohm or farad) explicitly and instead relies on implicit knowledge raised from the usage of specific letters either only for resistors or for capacitors, [nb 1] the case used (uppercase letters are typically used for resistors, lowercase letters for capacitors), [nb 2] a part's appearance, and the context.
The energy (measured in joules) stored in a capacitor is equal to the work required to push the charges into the capacitor, i.e. to charge it. Consider a capacitor of capacitance C, holding a charge +q on one plate and −q on the other.
The diode and the switch are simplified as either a short circuit when they are on or by an open circuit when they are off. When in the off-state, the capacitor C is charged by the input source through the inductor L 1. When in the on-state, the capacitor C transfers the energy to the output capacitor through the inductance L 2.
A discharged or partially charged capacitor appears as a short circuit to the source when the source voltage is higher than the potential of the capacitor. A fully discharged capacitor will take approximately 5 RC time periods to fully charge; during the charging period, instantaneous current can exceed steady-state current by a substantial ...
One of the capacitors is charged with a voltage of , the other is uncharged. When the switch is closed, some of the charge = on the first capacitor flows into the second, reducing the voltage on the first and increasing the voltage on the second. When a steady state is reached and the current goes to zero, the voltage on the two capacitors must ...
Since S1 is a short while closed, and the instantaneous voltage V L1 is approximately V IN, the voltage V L2 is approximately −V C1. Therefore, D1 is opened and the capacitor C1 supplies the energy to increase the magnitude of the current in I L2 and thus increase the energy stored in L2. I L is supplied by C2. The easiest way to visualize ...