electric-circuitselectric-currentelectrical-resistancevoltage
My question is
Why voltage drops to zero after passing through a single resistance circuit?
Do charges slow down when passing through resistance and if they do then why isn't it effecting current?
How do charges carry on their motion even after voltage drops to zero?
Electrons move because they are in a region of space with a non-zero electric field. They don't accelerate to high speed in a wire because they keep bumping into things; a kind of friction which dissipates energy much like the friction you are used to that explains why resistors get hot. In effect their speed depends on the strength of the local electric field and the nature of the material they move in.
When you hook up a circuit of some kind to a voltage source (battery, generator, wall-wart, whatever), the electric field already present between the terminal of your power supply causes some electrons in the wires to move around. In the course of doing that the electric field gets re-arranged to point along the wires and through the components and so on. There is rather a lot of shuffling that goes on immediately after the power supply is hooked up and I am going to mostly ignore it to focus on what happens when a (short-term) steady-state is established.
Observe the circuit in its operating state: the electric field points along the wires and though components in various way. In some places that field $E$ is weak and in some places it is strong, and in some places the electrons flow fast and in others slow, but there are two rules that must be obeyed:1
The current (number of electrons passing a point) is the same throughout the circuit. This follows because I have restricted out consideration to time when things aren't changing, and if more were passing point A than point B (a little further down the circuit) electrons would be piling up in the space in between them.
The total voltage change around the circuit has to be zero. This is because voltage is a function and can have only one value at any point in space, so if I follow any path that comes back to itself the changes has to equal zero when it returns.2
These rules are written in terms of voltage and current, but previously I was talking about electric fields, so what's the relationship between them?
The current comes into play in the form of Ohm's Law: $V = I R$.
The potential change in a section of circuit with length $d$ and constant electric field $E$ is $\Delta V = E d$, so we can write the voltage rule as $ 0 = V_{ps} - \sum V_i = V_{ps} - \sum_i E_i d_i$ where $V_{ps}$ represents the voltage gain of the power supply.3 Rearranges this gives us $$ V_{ps} = \sum_i V_i = \sum_i E_i d_i \,.$$
One last thing before we're ready to answer the question: the electric field in the wires is usually assumed to be very small compared to the electric field in other things like resistors. Therefore, we can ignore the $Ed$ contributions from the wires in working the maths. This isn't true for very long wires or for very fine wires under low voltages, but we're going with it anyway.
Consider a very simple circuit with a switch in it. A resistor (numbered 1) is connected directly to the battery and to the input of the switch. From the switch the current gets back to the battery either directly or through a second resistor (numbered 2).
The circuit starts with the switch set so that only one resistor is involved. When we hook it up, the fields rearrange themselves such that we have very weak fields in the wires and a very strong field in the resistor: $V_1 = E_1 d_1 = V_{ps}$. To make the current rule work, we have a lot of electrons moving slowly in the wires and a few electrons moving very quickly in the resistor (think of car flowing).
$t = 0$ The original state of the circuit has a field in resistor 1 $E_1 = V_{ps}/d_1$ and very weak fields everywhere in the wires. There is no build up of electrons anywhere and the current flow is steady throughout.
$t = 0 + \epsilon$ The switch has changed state, but the electric fields have not re-arranged yet, so there is zero field in resistor 2. Electrons continue to move through resistor 1 at the same rate as before, when they get through it there is no field to move them through resistor 2. They begin to pile up between resistors 1 and 2. As they do that they begin to reduce the field in resistor 1 and increase it in resistor 2.
$t = 0 + (2\epsilon)$ Now there is a little field in resistor 2 and a little less in resistor 1. Current has started to flow through resistor 2 but there is still less than is flowing through resistor 1. More charge is building up between them and that is driving the field in 2 up more and the field in one down more.
$t = 0 + (\text{several }\epsilon)$ The field in resistor 2 has risen and the field in resistor 1 has dropped until they are almost matched. Current flow through the 2 resistors is almost the same with only a small amount more coming through resistor 1. The charge between them has almost, but not quite stopped changing and that means the fields in them are also almost fixed.
$t = 0 + (\text{many }\epsilon)$ The field in resistor 2 has risen high enough that it's current matches that in resistor 1. This represents the new current of the circuit as a whole and is lower than the original current.
What we learn from this consideration is that any time the flow of electrons is faster through one part of the circuit than another, electrons will accumulate in such a way as to re-distribute the electric field in the circuit so that the flow is more even than before, and that this process happens continually until the flow becomes uniform throughout the circuit. The strength of the electric field is also related to the voltage change over each component and will be adjusted until the total is equal to the supplied voltage.
1 Rules written in a form that applies only to series circuits. For a more complete version, look up Kirchoff's Laws.
2 This is true when you neglect magnetic induction.
3 I am assuming that there is only one power supply. The full treatment of Kirchoff's laws can relax this restriction.
No, they do not all have the same voltage drop. If they were in series, however, they would.
By Ohm's Law, the voltage drop is proportional to the current flowing through a resistor. (So in several series resistors with the same resistance, the drop across each one is the same, since the current across each one is the same). However, because B and C are in parallel, the current is split between those two, meaning that the current in B and C is different from that passing through A.
However, since B and C have the same resistance, you know that the current from A is split 50/50 between them, so since each one gets half of the current, the voltage drop across each one is half of the voltage drop across A.
To find the current, first find the equivalent resistance using parallel and series resistor simplification techniques. Condense the two parallel 10 ohm resistors into a single resistor, and then combine this resistor with the series 10 ohm resistor. This tells you the total (or "equivalent") resistance of the circuit. Since you know the voltage produced and the resistance of the circuit, you can use Ohm's Law to find current.
Best Answer
The ideal circuit, as presented in introductory courses, simplifies the physics in order that the student not be distracted by too many details:
Resistance is present in the wires and connectors, as well as the internal resistance within the battery. Their is a slight, measurable, drop in voltage all along the path of the circuit. Thus the charges in the circuit will always be subject to a voltage.
Since the voltage varies from one point to the next, the potential energy has a spatial gradient; this is the origin of the electric field which is felt by the charges at each point along the circuit. Where the voltage drop is greater, the electric field is stronger.
In reality the voltage does not drop immediately to zero; however, in the ideal case the resistance of the wires is ignored - they aren't really there. In real circuits the wires and other connectors may play an important role in the design.
There is more to say about the physics of an electrical current: within any metal there exists typically one or two free electrons per atom, depending on the metal. These "free electrons" are shielded by the net positive charge of the atomic cores, and move "freely" throughout the metal. Drude called this the electron sea (1900), and this model is still studied in introductory condensed matter, though it has several defects which require quantum theory to repair.
Drude's "electron sea" model
The "electron sea" thus consists of randomly moving electrons which are scattered off of defects and crystal boundaries. Drude assumed a statistical distribution of speeds similar to that which was so successful in the kinetic theory of gasses. The average thermal speed of a free electron in copper is about 1% of the speed of light; OTOH, the drift velocity of a typical current in the same wire is millimeters per second. See Hyperphysics-Microscopic Electric Current for calculations.
The physics tells us that though the current travels very slowly, the changes in the electric field travel at nearly the speed of light. This is usually covered in an upper level undergraduate course in electromagnetic field theory.