I just started with small projects with capacitors, like astable multivibrator and I do not understand where the capacitor discharges. The theory always show the capacitor charged by the battery, then the battery is taken away and the capacitor is discharged in a circuit with only a bulb or a resistor.

But in circuits like astable multivibrator, what happens during discharging, where does the current created by the capacitor during discharging go? Back Into the battery?

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  • \$\begingroup\$ There is no one answer. You have to show a specific circuit. Capacitors can be discharged through resistors, transistors, inductors, or even just shorted. If the transistor terminals are connected to each other when the voltage is not zero, the current will be limited by series resistance of the capacitor, and whatever resistance is in wires or traces connecting the terminals together. If a capacitor has a voltage across it, it will always discharge if there is a pathway. When it discharges, current flows from one terminal to the other, through whatever external path is available. \$\endgroup\$ – mkeith Jan 21 '15 at 19:25
  • \$\begingroup\$ Just look at the astable multivibrator circuit, it is a classic circuit \$\endgroup\$ – Sunny Jan 21 '15 at 20:15
  • \$\begingroup\$ If you are looking at a circuit post a link. How do I know whether whatever "classic" multivibrator circuit I pull up will be the same as yours? Have the same reference designators? Help us help you. \$\endgroup\$ – mkeith Jan 21 '15 at 20:27
  • \$\begingroup\$ See the circuit above for example \$\endgroup\$ – Sunny Jan 21 '15 at 21:45
  • \$\begingroup\$ I have not seen this circuit before. C1 charges through L1 and R1 when TR1 is off. It discharges through TR1 when TR1 is on. Note that when C1 is charging, the charge current also flows through the base of TR2, which should keep TR2 on. The circuit is symmetric. C2 discharges through TR2 when it is on, etc. \$\endgroup\$ – mkeith Jan 22 '15 at 4:12

If capacitor is connected to a closed circuit it will either charge or discharge, depending on the voltage "forced" on it (the equilibrium or steady state voltage, that is reached when there is no current flow through the capacitor). If this voltage is higher than the capacitors \$Q/C\$, such as with case when it is connected to a power supply, it will charge to reach this voltage. If there is no voltage source, or it's voltage lower than \$Q/C\$, the capacitor will discharge to reach this voltage as well. In case of passive elements only this voltage would be \$0\$. So the capacitor will act as a power source, and the current will "go" into capacitor itself (in order to equalize the charges on the capacitor plates) through the passive components while the energy is dissipated as heat and/or light and/or other types of energy, depending on components.

  • \$\begingroup\$ The current will go through passive elements, but there is the battery on the way, so is the current going back to battery? \$\endgroup\$ – Sunny Jan 21 '15 at 17:57
  • \$\begingroup\$ Well, if the capacitor has a higher voltage than the battery, it will eventually charge the battery (if it is chargeable, of course). \$\endgroup\$ – Eugene Sh. Jan 21 '15 at 18:00
  • \$\begingroup\$ It cannot have higher voltage, because it was charged by the battery \$\endgroup\$ – Sunny Jan 21 '15 at 18:25
  • \$\begingroup\$ Then there is no way the capacitor will discharge through the battery. \$\endgroup\$ – Eugene Sh. Jan 21 '15 at 18:26
  • \$\begingroup\$ Through what is it then discharging? \$\endgroup\$ – Sunny Jan 21 '15 at 18:52

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