Why does voltage split, if potential changes with distance alone? 
*

*Isn't electric potential a distance-dependent quantity, like gravitational potential?
So why is it that a resistor can bring down voltage in a series circuit?
Shouldn't voltage be changed only by the amount of distance it travels?


*Why does voltage split in a series circuit.
P.S. I am a 10th grader. Needs an intuitive understanding. Please don't use complicated laws, formulas etc.
 A: Treating circuit in terms of resistors, inductances, capacitors, etc. is called lumped-element model. Lumped element model is an approximation that ignores the spatial effects (among other things):

The lumped-element model (also called lumped-parameter model, or lumped-component model) simplifies the description of the behaviour of spatially distributed physical systems, such as electrical circuits, into a topology consisting of discrete entities that approximate the behaviour of the distributed system under certain assumptions. It is useful in electrical systems (including electronics), mechanical multibody systems, heat transfer, acoustics, etc. This may be contrasted to distributed parameter systems or models in which the behaviour is distributed spatially and cannot be considered as localized into discrete entities.
Mathematically speaking, the simplification reduces the state space of the system to a finite dimension, and the partial differential equations (PDEs) of the continuous (infinite-dimensional) time and space model of the physical system into ordinary differential equations (ODEs) with a finite number of parameters.

Treating potential as a distance-dependent quantity is simply beyond of the lumped-element model is suitable for.
A: For simplicity I have chosen identical resistors but the analysis can be extended to situations where this is not so.
Consider a resistor of length $d$, resistance $R$, uniform cross-sectional area and composition,  with a potential difference of $V-0=V$ across it as shown in the top diagram.

An electric field $V/d$ is developed across the resistor and this drives a current through the resistor.
Now add another identical resistor in series.
The potential difference across each resistor is $V-V/2= V/2=V/2-0$, ie half the original value so the electric field in each resistor is now $V/(2d)$ and this now dives a current which is half the value of the current with one resistor.
So the effective resistance of the two resistors in series is equal to twice the resistance of one resistor.
Within the top resistor, if $x$ is the distance from the right hand side of the resistor, $0$, then the potential at position $x$ is $\frac Vd\cdot x$, ie dependent of the position at which the potential is measured.
A: $$R=\frac{\rho L}{A}$$
From this equation, I like to imagine resistors as Either :
"long wires" that are  wrapped up.
Firstly the introduction of a resistor has no bearing on the potential difference across the entire circuit, its just that the introduction of the resistor  changes the potential difference across a certain section of the wire.  Since The PD is constant across the entire circuit, and the E field is constant ( for simplicity), the introduction of a resistor is like adding a new length of wire, The net E field everywhere must decrease to make  V = ED constant across the entire circuit. This means that the potential difference across other parts of the wire is now decreased since the E field has decreased over those regions, and hence the potential difference across the resistor must therefore increase (To make V constant across the entire ciruit
Mathematically consider a circuit of wire length $d$ with a potential $V_{0}$ and section of wire $K$:
$$V_{0} = E_{0} D$$
$$E_{0} = \frac{V_{0}}{D}$$
Potential across the gap:
$$E_{0}K = \frac{V_{0} K }{D}$$
Potential across the rest of the circuit:
$$E_{0}(D-K) = \frac{V_{0}(D-K)}{D}$$
Lets introduce a resistor, Modelled as a wire of length $L$ The potential is equal for both circuits
$$V_{0} = E_{1} [D +L-K]$$
$$E_{1} = \frac{V_{0}}{[D +L-K]}$$
Potential across the gap:
$$E_{1}L = \frac{V_{0} L}{[D +L-K]}$$
Potential across the rest of the circuit
$$E_{1}(D-K) = \frac{V_{0} (D-K)}{[D +(L-K)]}$$
We can see that as L goes to infinity, the potential difference across the resistor approached the potential difference across the battery, and the potential of the rest of the circuit approaches zero. hence, "resistors increase the potential drop, and decrease the potential drop elsewhere"
We can also consider resistors as "Wires with a lower area":
This is fairly more complex to analyse, The area of the wire, influences the net field in the wire due to the changing of the distribution of surface charges, a smaller area wire means that the net field in the wire is actually larger increasing the potential difference across the resistor
Why does the voltage split in series?
Forget the voltage drop in the wires for the minute, Across the terminal of the battery, the is a potential drop of V, This means that regardless of any path we take, the voltage drop across those 2 points is V. Imagine i have a resistor with resistance R, if there is zero voltage drop across the wire then there must be a voltage drop V across the resistor ( as i pick a path from the terminals intersecting with the resistor).
If I add another identicle resistor, there is an increase in length of the wire. The addition of this resistor shouldnt have any impact on the potential difference across the circuit. however, the presence of surface charges will want to make it so that the current is constant by making E constant.
For the same potential difference across both ressitors, if the wire is longer,  the total E field at any point in space is halfed. ( As distance doubles so E must decrease for the PD to remain constant) Thus the potential difference across each resistor is halfed.
More mathematical Approach by considering the resistors as 2 joint wires of a certain length:
$$E = -\nabla V$$
$$-\int_{a}^{b} \vec{E} \cdot \vec{dr} = V(b)-V(a) = V_{batt}$$
The Key idea is that E is constant due to surface charges( if area of wire is constant, which is true for 2 equal resistors)
$$-|\vec{E}|\int_{a}^{b} dr = V_{batt} $$
$$-|\vec{E}|L = V_{batt} $$
for 2 equal resistors of length $\frac{L}{2}$ This becomes
$$-|\vec{E}|\frac{L}{2} + -|\vec{E}|\frac{L}{2} = V_{batt} $$
$$-|\vec{E}|\frac{L}{2} = \frac{V_{batt}}{2} $$
Therefore the potential difference across one of the resistors is therefore $\frac{V_{batt}}{2}$
A: I want to answer your question about voltage. Think of electricity as water. The flow is current. And the distance it falls from a water fall is potential.
If you were working on series. You wound be doubling the height of the water fall.
If you working on parallel, you would be working at the normal height, but opening a second opening for the water to flow.
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