Why is there higher voltage drop across larger resistors?

Question

By simple application of $V = IR$, there must be higher voltage drops across larger resistance. However, $V$ is the integral of $E$ multiplied by displacement. Thus, the electric field must be larger at places with higher resistance.

My question is how does the battery "know" where these high resistance places are and create higher electric fields there. It seems that if there is no wire connected, the battery will produce some electric field. Then when wires are connected, the electric fields will shift such that higher resistance portions will have higher electric fields. Why is this?

Answer 1

The battery doesn't "know" (or "see") the specifics inside the circuit. Instead the mobile charges in the circuit rearrange themselves to create the electric fields.

If you assume that the electric field inside a resistor is too low initially, then that means that charges passing through it will have too little energy. They will slow down. By slowing down, more charges will arrive at the resistor than pass through. This excess of charges creates a stronger electric field. The field will increase in strength until exactly as many charges pass through as arrive at the resistor.