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I am a beginner electronics designer, but I have never been able to grasp "when a capacitor discharges".

What causes a capacitor to discharge? Is there a way to calculate a capacitance value to discharge with certain voltage and current values over a specific amount of time?

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    \$\begingroup\$ What causes a pressurized air tank to decide to expel air rather than take it in? \$\endgroup\$ – DKNguyen May 20 at 13:43
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    \$\begingroup\$ Luckily I haven't encountered an electronic component which decided to do anything on it's own. Although it might seem that way if things go awry. \$\endgroup\$ – Arsenal May 20 at 14:03
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    \$\begingroup\$ A capacitor discharges when it has the capability to do that.- that means: To drive a discharging current through a closed current loop. \$\endgroup\$ – LvW May 20 at 17:18
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    \$\begingroup\$ see the answer below c * (dv/dt), thats it. it discharges when the voltage connected to it is less than the voltage stored. current moves while the voltage are different (and there is a closed path to move the current through of course). capacitors c * dv/dt inductors l * di/dt, thats it. \$\endgroup\$ – old_timer May 21 at 1:52
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    \$\begingroup\$ @Cookiebuster it may help if you clarify your background, e.g. a high school student or first year college student isn't a "beginner designer" yet, and this would impact how people explain this. For example, have you taken any courses that teach about differential equations yet? \$\endgroup\$ – Paul May 21 at 15:23
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Capacitors oppose changes of voltage.

If you have a positive voltage X across the plates, and apply voltage Y: the capacitor will charge if Y > X and discharge if X > Y.

calculate a capacitance value to discharge with certain voltage and current values over a specific amount of time

There's the classic RC circuit analysis which should be taught right at the beginning of learning electronics.

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schematic

simulate this circuit – Schematic created using CircuitLab

A charged capacitor is always trying to discharge itself and will do so whenever the circuit it is part of permits current to flow between its two plates.

An analogy would be a stretched rubber band. It will spring back to its relaxed state whenever it is released from whatever is keeping it stretched.

More specifically, a capacitor discharges whenever the voltage in the circuit the capacitor is part of has a smaller magnitude than the voltage stored on the capacitor. So in the circuit above if the voltage across the capacitor is greater than the voltage of the voltage source, Vs, the capacitor will discharge through the resistor, R, until the voltage across the capacitor equals the voltage supplied by Vs. At that point current will stop flowing through R, as there is no voltage difference across it.

If Vs is greater than the capacitor voltage then current will flow through R charging the capacitor until the capacitor voltage is equal to Vs. Again, current will then stop flowing.

The answer by pjc50 links to the equations needed to calculate how long this takes.

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Is there a way to calculate a capacitance value to discharge with certain voltage and current values over a specific amount of time?

$$\text{current taken from the capacitor} = \text{capacitance}\cdot \dfrac{\text{change in voltage}}{\text{change in time}}$$

Or

$$I = C\cdot\dfrac{dv}{dt}$$

Re-arrange as suits your needs.

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The capacitor discharge when the voltage drops from the main voltage level which it connected to like it connected between (5v and GND ) if voltage drops to 4.1v then the capacitor discharge some of its stored charge ,the drop in voltage may caused by many effects like increase in a load current due to internal resistance of non-ideal source .

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