I'm doing a super simple Arduino led project. I followed a tutorial online that wired up the LEDs like this:
I'm wondering why this capacitor doesn't just short the power supply.
Can anyone provide me with some insight?
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The capacitor is in fact a short circuit, however only temporarily.
When you first turn on the power supply, the capacitor will act like a short circuit during this initial transient phase. There will be a large inrush current as the capacitor charges up (*).
Similarly, if you take a fully discharged capacitor, and connect it to a power supply which is already at some DC voltage, you will get a sudden inrush current into the capacitor which if large enough (e.g. very large capacitor) can be enough to overload the power supply. This is especially problematic if switching a large capacitive load into a circuit.
Once the power supply reaches its steady-state DC voltage, and the capacitor charged up to the same voltage, the capacitor will no longer act as a short circuit on the supply because it cannot conduct a DC current as explained by the other answers.
If however you have some noise or ripple signal, or a voltage transient, the capacitor will again act like a low impedance or short circuit (frequency dependent) for this noise signal or voltage ripple. This is how the capacitor has the effect of smoothing out the power supply.
(*) The power supply may provide some form of inrush current limiting or controlled voltage ramp-up, in which case the current flow will be limited.
There is no conductive path from one terminal to another in a capacitor. The symbol shows this.
Figure 1. Non-polarised, polarised and variable capacitors. The symbol represents two parallel plates separated by an air-gap or a dielectric (insulator).
Figure 2. An unwound electrolytic capacitor showing the two layers of foil and one of the insulation layers. Source: Translators Cafe.
The capacitor can't pass DC current but it can accept charge and if the voltage across it varies current will flow in and out of both terminals. In this manner it is able to pass an AC current through it.
It is possible, at least theoretically, to damage a power supply by connecting a discharged capacitor across its terminals. However, in real life the characteristics of the power supply and the capacitor usually prevent this from happening.
When you first connect the capacitor to the supply a large current will flow to charge the capacitor. The magnitude of this current is limited by the output resistance of the supply and the effective series resistance (ESR) of the capacitor. Neither of these resistances are shown on your cartoon schematic, but they are there nonetheless. As long as these resistances limit the current to a safe level no damage will occur. Another important consideration is that this surge current lasts for only a very short period of time...once the capacitor is charged to the supply voltage you will see relatively smaller currents.
You can imagine the capacitor as a battery. The charge on a capacitor is dictated by voltage and it's capacitance. Since capacitance will be static and charge doesn't teleport, we can see that as charge increases, the voltage increases. The voltage can only go as large as what's feeding it so there's a maximum amount of charges the capacitor can hold, which is again dictated by the voltage.
Its purpose here is to provide a current source for the LEDs. There's a few reasons for this, the top one is that large instantaneous current draws cause a droop in the voltage from the power supply, so the capacitor has a lower voltage across it, which makes it lose charges. So it turns into a current source for the short amount of time of the droopped voltage.
An additional fun question to ask yourself is why you had to add a resistor to the control signal going to the LEDs!
A capacitor is two electrodes separated by an insulating layer. In some cases the insulating layer could be air or vacuum. In other cases it could be a dielectric (such as some type of plastic or ceramic). Unless there is some type of fault, such as dielectric failure, the capacitor is not a short-circuit at DC, by the definition of a capacitor.
DC doesn't travel through a capacitor as long as its charged.
The capacitor on LED strips is there to dampen sudden Amp changes in the system, when your LED Strip suddenly needs more power etc. The thin cables on these PSUs are most often too high inductance to manage these fast, big shifts well.
The capacitor was connected in parallel with the circuit. Current reaches both at the same time and it would be too late for it to protect the circuit. This would have been different if the connection was done in series inline with the circuit. The current would have to reach and pass through the capacitor before it reaches the circuit to protect it.
It IS going to short the power supply. It's also going to cause the capacitor to explode, in all likelihood. The problem is that the capacitor is installed backwards. Electrolytic capacitors are polarized, and that dashed white line on the side of the capacitor is actually a series of arrows pointing to the negative terminal. Polarized capacitors really don't like being installed backwards and will typically "vent" their frustration in rather violent ways.