NPN transistor

Hello guys! hope all is well. I have a question regarding a basic NPN transistor. ( just started studying about transistors).
In the pic below the lamp is supposed to glow when the switch is closed because the base current flows and allows current to flow from the emitter to the collector.

My question is why would the electrons wait and bother themselves to go through the collector? i would think they would go through the base and take the "easy" quick way back to the positive terminal through the base itself with the slightest base current! ( i know it doesnt happen that way ) . I would like to know why arent they taking the easy way out ( we all know current takes the path of least resistance )

thank u

The OP is using carrier current (i.e., electron current) and not charge current (i.e., conventional current). While unfortunate, that's the way it is.

The notion that "current takes the path of least resistance" is one of the more heavily abused sayings in electronics. If you have a 12V car battery and you connect a light that has a resistance of 12Ω across it and also connect a light that has a resistance of 6Ω across it, does ALL of the current go through the 6Ω light since it is the "path of least resistance"? Of course not. Both lights work just fine. There is more current going through the lower resistance path, but there is still current going through the higher resistance path as well. In fact, in this case, the current in one path is unaffected by whether the other path is even available (e.g., disconnect one light and the other will still have the same current flowing in it).

As for the transistor, it might help for you to think of it along these lines. Imagine I build a water valve that is, itself, operated by water. So I have three ports. The supply port (think of the emitter of an NPN) is connected to the water supply (the faucet, let's say). Then there is a small port (think of the base) that has a needle valve on it that you can turn to adjust how much water flows out of it. Even if the needle valve is full open, though, the port is so small that very little water will flow out of it. But, inside the unit is a bigger valve that lets water flow from the supply port to the output port (think of the collector) but this valve is controlled by the water flowing out the small port control. The more water that is flowing out the small port, the wider open the bigger valve gets. Conceptually, this description is actually pretty close to how many hydraulic servovalves work.

The OP is using electron flow.

Future posters, please stay on topic.

MrChips said:

The OP is using electron flow.

Future posters, please stay on topic.

Click to expand...

And here I thought I had. Silly me.

Wow! People do make a mountain out of a mole hill!
WBahn is right, but for me misses the essential point.

Nathan asks, "... why would the electrons wait and bother themselves to go through the collector? i would think they would go through the base and take the "easy" quick way back to the positive terminal through the base ..."

Well, if you look at the circuit, the base has a 10k resistor and the collector has a 1k resistor. So which is the easier path? The collector of course.
Therefore, as WBahn indicates, roughly 10x as much current goes via the collector as the base.

You should take account of the resistance of the other elements and the fact that the collector current has to go through the LED, which drops a couple of Volts. So the collector current will be only about 8 times as much as the base current.

As far as the transistor is concerned, you can almost ignore it!
Both the base/emitter and collector/emitter junctions are forward biased diodes and just causes a Voltage drop of .6 to .7 V (maybe less for the collector.)

Once you've taken account of these more or less fixed Voltage drops, the remaining Voltages are simply applied to the two resistors - and as you know (&WB reminds you) when two resistors are in parallel, some current goes through each, in inverse proportion to their resistances.