Hey everyone,

I'm having trouble understanding how capacitors discharge in series circuits. I understand how it discharges in the schematic that I've attached but I can't see how it can discharge in a series circuit as the stored electrons can't pass through the capacitor to make both sides neutral.

 

In a circuit like that, electricity flows in a circle. If the capacitor was in series, electrons would flow from one plate, around the circle and to the other plate. Just like a battery.

Also maybe it would help you to know that the little electrons are not freely flying around the circuit. Picture a soda straw that is full of water, this is similar to a wire. When you put one drop of water into one side of the straw, one drop comes out the other side, but it's not the same drop. You have to put in a lot of drops for your first drop to make its way all the way out of the straw. Wires are similar. They're full of electrons and when one goes in one end, a different one pops out the other end. You have to put a lot of electrons in one end before those same electrons start coming out the other end. So with the capacitor in series, when one electron leaves a plate, it's most likely a different electron that comes in on the other plate.

 

Hello,

The capacitor is PARALLEL to the battery.
When disconnecting the battery, the led with the resistor will discharge the capacitor.
You will see the led go dim to off.
I do not see a capacitor in series.

Bertus

 

The capacitor remains charged if the battery stays connected. Disconnect the battery and the capacitor charge will reduce as current flows through the resistor and LED.

 

This is the capacitor discharge equation

• VC- VC is the voltage that is across the capacitor after a certain time period has elapsed.
• V0- V0 is the initial voltage across the capacitor before the discharging begins where it's connected in series with a resistor in a closed circuit. In simple terms, this is the voltage that the capacitor initially has before the discharge process begins.
• Time, t- Time, t, is the period of time which has elapsed since the discharge process has begun. t is measured in unit seconds. It is a very important parameter in this equation because it determines how much the capacitor discharges. The more time that has elapsed, the more the capacitor will discharge. Conversely, the less time that has elapsed, the less the capacitor will have discharged.
• Resistance, R- R is the resistance of the resistor to which the capacitor is connected to in the circuit, as shown in the diagram above. This affects the discharging process in that the greater the resistance value, the slower the discharge, while the smaller the resistance value, the quicker the discharge, and, thus, the lower the amount of voltage, VC, across the capacitor.
• Capacitance, C- C is the capacitance of the capacitor in use. C affects the discharging process in that the greater the capacitance, the more charge a capacitor can hold, thus, the longer it takes to discharge, which leads to a greater voltage, VC. Conversely, a smaller capacitance value leads to a quicker discharge, since the capacitor can't hold as much charge, and thus, the lower VC at the end.

 

Mostly caps discharge in a series circuit by reversing the voltage. (AC)
They are also discharged in a series circuit by connecting different nets as needed. (the 555 for example)
Other than that it could be capacitor leakage, or alternate leakage paths back thru the circuit when power is disconnected.

 

I'm having trouble understanding how capacitors discharge in series circuits. I understand how it discharges in the schematic that I've attached but I can't see how it can discharge in a series circuit

Mr Betrus.is right it is the circuit that is confusing you.

 

I'm having trouble understanding how capacitors discharge in series circuits.

As shown, the capacitor will never discharge.

 

This is the capacitor discharge equation
View attachment 208302

• VC- VC is the voltage that is across the capacitor after a certain time period has elapsed.
• V0- V0 is the initial voltage across the capacitor before the discharging begins where it's connected in series with a resistor in a closed circuit. In simple terms, this is the voltage that the capacitor initially has before the discharge process begins.
• Time, t- Time, t, is the period of time which has elapsed since the discharge process has begun. t is measured in unit seconds. It is a very important parameter in this equation because it determines how much the capacitor discharges. The more time that has elapsed, the more the capacitor will discharge. Conversely, the less time that has elapsed, the less the capacitor will have discharged.
• Resistance, R- R is the resistance of the resistor to which the capacitor is connected to in the circuit, as shown in the diagram above. This affects the discharging process in that the greater the resistance value, the slower the discharge, while the smaller the resistance value, the quicker the discharge, and, thus, the lower the amount of voltage, VC, across the capacitor.
• Capacitance, C- C is the capacitance of the capacitor in use. C affects the discharging process in that the greater the capacitance, the more charge a capacitor can hold, thus, the longer it takes to discharge, which leads to a greater voltage, VC. Conversely, a smaller capacitance value leads to a quicker discharge, since the capacitor can't hold as much charge, and thus, the lower VC at the end.

Good explanation, but T equals time in seconds to charge or discharge to 60%. If you invert the answer it gives the frequency more or less. This might be accurate, it's been a while. Ha

 

I didn't want to confuse the kid.

 

As shown, the capacitor will never discharge.

Almost all capacitors will discharge eventually. That is because almost all real capacitors have a leakage resistance in parallel with the actual capacitance. Some capacitors have more, some have less.
In addition, for any connection arrangement that provides the capacitor with a voltage to charge up to, there is a current path. If the charging voltage source is removed and replaced by it's internal resistance, current can flow and the capacitor can discharge.
Of course in some simulators the simulation is incorrect i that the capacitors are all perfect and there is no leakage current. Those simulators do not agree with reality.