Maybe it's a stupid question but as the title says, in DC analysis how does the current from Vcc flows?
Where is the – and the + of the voltage source? And how does the current flows in AC analysis when we also have Vi involved?
Electronic – How does the current flow from VCC
circuit analysis
Related Solutions
why can't electrons also flow from the negative terminal in battery 2, through the right bulb, and into the positive terminal of battery 1?
Because the only way to get to the negative terminal of battery 2 is to come from the body of the battery, and the only way to get there is from the positive terminal of the battery. And the only way to get there is through the switch in the friend's house, which is open so no current can flow through it.
Batteries don't create electrons from nothing. They "pump" them from one terminal to the other via a chemical reaction, causing current to flow (through the battery) from a low potential to a high potential.
Current only flows in complete circuit means the current has to flow through the battery just as much as it has to flow through every other circuit element.
As an aside, it's also why it's silly to say that current always flows from high potential to low --- the battery is a trivial example of when it's the other way around. And it's also silly to say that all current is a flow of electrons in the opposite direction from conventional current --- the battery is a trivial and everyday example of when ionic currents should be considered.
Both Martin and immibis have it correct. Here's a slightly more complete circuit, showing how I1 is absorbed by the op-amp output, then flows to the op-amps' negative supply voltage (the op-amp's positive supply voltage is not shown, but is still required).
You cannot accurately measure I1 at the -ve power supply, because there is additional quiescent current to bias the op-amp's internal transistors.
Best Answer
Whenever you see Vcc or +9V or any other voltage flag, it's really just a shortcut way of saying "this is a voltage source relative to ground." That means I can redraw your circuit as:
simulate this circuit – Schematic created using CircuitLab
So the + side of the voltage source is connected to the "V" in your circuit, and the minus side is ground. Vi works the same way. That first capacitor will effectively be a short to AC (if the capacitor is large enough). So current will flow in from Vi in to the base of Q1, which will cause more current to flow through RC and the base of Q2, which will cause even more current to flow through RC and RE. It's a Darlington Pair circuit.