DC Series Circuit (Voltage Drop )?

Question

9VDC with three resistors are connected in a series.
I understand each will have a voltage drop.
I also understand the sum of all voltage drops will be equal to the source voltage.

Does this mean, the resistors are consuming all of the voltage?
That there is no voltage left, coming out of the final resistor?
If voltage is the pressure which current travels through a circuit, then is that the same as saying no current is returning to the battery ( coming out of the final resistor) ?
I am just not quite sure what voltage drop means, I know how to figure it and all but not sure what it really means is happening.
If anyone can explain I really appreciate it and I thank you kindly.

Answer 1

Voltage drop is the difference in voltage at the resistor from the point where the current gets in to the point where the current gets out. In your case voltage drop in first resistor is, say V1, the second resistor V2, and third V3. When they are connected to the battery their sum is equal the battery voltage called electromagnetic strength EMS. Thus the EMS is the driving force that forces the current to flow through the resistors. Resistors do not have the counter force. They are passive elements. The EMS causes the accumulation of the electric charge at the entrance vs. exit which manifests as the voltage drop. In summary, the EMS is the cause and the voltage drop the consequence.

Answer 2

#1 - The current is always the same in a DC series circuit, the load does not drop current. The voltage drop will vary at each load point, depending on where measured from each load. If you measured the voltage drop between ground and the grounded leg of the final resistor, it would read 0. But, if you measured it from the + side of the same resistor, to ground, it would have a reading. A better understanding of Ohms law and some more exercises like this will help, especially a lab or two where you actually measure Vdrops on a breadboard.

Answer 3

Voltage is not 'consumed'. Power is consumed and it is the product of voltage and amperage.
Take the total resistance and divide the voltage by that that gives you the number of amps flowing. Then you can calculate the voltage drop across each resistor by R*A. Now you can calculate the power dissipated across all three resistors V*A. (you can also do that for each resistor separately).
There is voltage drop always across the whole load. i.e. the whole voltage is applied over the whole load, with the amperage fixed by the combination of total resistance and voltage. Then calculating the amperage across each load and the voltage that will take, you can calculate the power dissipated by each individual part of the load.