Why is current same in series but potential divides? [duplicate]

Question

I have studied that electric current is constant along a series configuration of resistors, across which voltage divides.

I make some confusion, because I've always thought that electric current flows because of voltage; if voltage decreases, then the current should change as well. But this is not true. I'd like to ask the reason why the current doesn't change.

Answer 1

A series configuration of resistors with resistance $R_i$ with voltage generator with voltage difference $E$ in DC, the circuit has only one closed path, and it is governed by

• Kirchhoff current law: at each node of the circuit, the balance of electric charges, gives you that "the current entering a node is equal to the current leaving that node", and thus the electric current $I$ is constant in the whole circuit.

• Kirchhoff voltage law: the sum of voltage difference along the closed path is zero,

  $ 0 = E - \sum_i \Delta V_i$

  The voltage drop across the $i^{th}$ resistor reads $\Delta V_i = R_i I$, and thus

  $ 0 = E - I \sum_i R_i = E - I R^{equiv, series}$.

Thus:

• electric current reads $ I = \frac{E}{R^{equiv, series}}$;
• voltage drop across $i^{th}$ resistor reads $\Delta V_i = R_i I = \frac{R_i}{R^{equiv,series}} E = \frac{R_i}{\sum_k R_k} E$

Answer 2

if voltage decreases, then the current should change as well...I'd like to ask the reason why the current doesn't change.

The current will change.

Suppose you have a circuit consisting of a single resistor connected across the terminals of a battery. Now suppose you add a second resistor in series with the first one. The resistor will always obey Ohm's Law, $I=V/R$. Putting a second resistor in series means that the two resistors will split the battery voltage (as explained by @basics' answer.) The original resistor will see less voltage than it saw before. Since it sees less voltage, and since it obeys Ohm's Law, we can conclude that the current must be less than it was before.