How does a resitance affect electric current in a circuit...??(series)

Question

Consider the following circuit:

Let the resistance of the lamp be $R_1$, and the potential difference across it be $V$. Let the current that flows through the circuit be $I$ in this situation. Say you add another lamp to the circuit (in series with $R_1$) such that its resistance is $R_2 > R_1$. Now the current in the circuit is $I' < I$.

What I want to know is: the moment you add the new lamp and complete the circuit the current in the circuit has reduced but how does this happen (I would like to know more than the just the fact that the resistance has increased so the current reduces I want to know what happens physically)?? The electric field is the only thing that is established instantaneously. So does the new resistance somehow affect the magnitude of the electric field thereby reducing the current (does it have to do something with permitivity of the new resistor, or for that matter of fact does the permitivity of a resistor affect the electric field in any way) or is it that the number of charges that flow through the circuit are the same as in the first case the only difference being the number of charges moving through a given cross section in unit time is less as comapred to before because the drift velocity reduces...??

Answer 1

The resistance of a homogeneous cylindrical wire can be expressed as $$R=\rho\frac{l}{A}$$ (Wikipedia refers to this expression as Pouillet's law.) Here $\rho$ is resistivity, which is determined only by the material properties, but not by its shape, $l$ is the length of the wire, and $A$ is its cross-sectional area. Putting two resistors in series can be thought of as increasing the length of the wire, i.e., increasing the resistance.

The role of $A$ in the above formula is that the bigger is the cross-section, the more electrons can simultaneously pass through it, the bigger is the current (which is by definition the charge which passes a cross-section per unit time.) Connecting resistors in parallel can be understood as increasing the effective cross-section of the wire.

The role of $l$ is a bit more subtle. I warn beforehand that what follows is a simplistic description of electrical resistance (a simplified version of Drude model, in fact): electrons are accelerated by the electric field within the conductor, but collide with the lattice (vir impurities and phonon emission) and lose their energy. If the electric field is $E$ and the average time between the collisions is $\tau$, the characteristic speed of electrons is $$ v_{ave}=\frac{eE\tau}{m}.$$ The number of electrons passing through a cross-section per unit time is then $$ nAv_{ave}=\frac{enAE\tau}{m}, $$ where $n$ is the electron density (concentration) in the conductor. The current is obtained by multiplying the above number by the electron charge, whereas the electric field is the potential difference divided by the length of the conductor, so that we have $$ I=\frac{e^2nAV\tau}{ml}= \frac{e^2n\tau}{m}\times{A}{l}\times V=\frac{V}{R},\\ R=\rho\frac{l}{A},\sigma=\frac{e^2n\tau}{m}=\frac{1}{\rho} $$ where $\sigma$ is called conductivity (the inverse of the resistivity).

We now see where factor $l$ results from: given the same potential difference, the voltage drop per unit wire length is smaller in a longer wire, which results in smaller velocity acquired by electrons between collision events, and hence smaller current.