I'm new to electronics and have made some simple projects using a LM7805. I'm using the following schematics:
I don't understand why this circuit works. Why doesn't it short circuit? I thought electricity always takes the "easiest" path which in my opinion would be the following:
Why does the electricity even bother going through the 7805 to begin with while it can just fill up the 0.33uf capacitor and then go right through it back to the battery?
Let's have a look inside a capacitor to see what prevent the short circuit.
A capacitor consists of two conducting plates. And there's an isolating plate between the two conducting plates.
How is current able to pass through the the isolating plate?
When there's a change in electrons (charge) on one conducting plate, a change in the charge of the other conducting plate occurs. The two conducting plates affect each other because the isolating plate is very thin. This way current can pass through a capacitor. I consider it as a virtual current.
When you apply an AC signal, a short circuit will happen because there's a rapid change in the charge of the plates.
When the frequency is decreased, lower current can pass. Capacitor is considered as a resistor and this resistance is called "capacitive reactance".
When you apply a DC voltage (zero frequency):
At the beginning, the capacitor will charge because applying DC voltage itself is considered a change in charge.
After completing charging, current will stop flowing because there's no change in the charges.
Capacitors do not present a short circuit path. The Capacitor will charge then stop when full, acting as a parallel voltage source when the Input voltage source drops for whatever reason.
And electricity does not just take the easiest path, or path of least resistance. Electricity will flow through all parallel paths. You may see a greater current through one path due to its lower resistance, but it will go through other paths too.
Your thinking is actually 100% correct!
The "electricity" (current) does "fill up" the 0.33uF capacitor before flowing to the 7505 regulator.
However, the capacitor "fills up" in a fraction of a second. After that, flow to the capacitor essentially stops and current flows to the regulator only.
Another way to view it is that the capacitor goes from a "short circuit" when it's fully discharged to an "open circuit" when it's fully charged.
Note - in practice, things are slightly more convoluted due to parasitic effects: https://en.wikipedia.org/wiki/Parasitic_element_(electrical_networks)
When you apply a voltage to the 7805 the input capacitor is charged and then the DC current through this capacitor stops. At the same time the other way for the current is,through the 7805 to the connected load at the output and a little bit to the ground connection.
So you are right. There would be a current through the capacitor but only during the charging of the capacitor. This current is limited because the capacitor has an internal resistance in series with the actual capacitor. Once the capacitor has internally the same voltage level as the input voltage the charging stops.
