# How current flows in a circuit as per Kirchhoff's law if the voltage dropped = voltage applied

According to the Kirchhoff's law the voltage drop is always equal to the voltage applied, then how come the current even flows in the last part of the circuit?

For example if in a circuit which has a 9 volts battery, there is a resistance which drops the current from 9 volts to 0 volts(i.e., 9 ohms), then if we look at atomic level then the electrons flowing in after the resistor will repel the positive terminal because of their net charge.

I know that I am misunderstanding something but if you know the correct theory then please answer.

• Electrons are attracted to the positive terminal. Current doesn't drop. Current isn't measured in volts. Commented Nov 29, 2020 at 11:32
• One way of looking at it : divide that resistor into 9 equal parts. Each 1 ohm resistor has 1 Volt across it : the current in each is the same 1 Amp no matter whether the resistor goes from 0V to 1V or 8V to 9V.
– user16324
Commented Nov 29, 2020 at 13:38
• See this answer. You write, "if we look at atomic level then the electrons flowing in," and develop some conclusions from assumptions you believe are true. It's better to visit the physics and see what really is true. So read the above.
– jonk
Commented Nov 30, 2020 at 5:16