0
\$\begingroup\$

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.

\$\endgroup\$
3
  • \$\begingroup\$ Electrons are attracted to the positive terminal. Current doesn't drop. Current isn't measured in volts. \$\endgroup\$
    – Andy aka
    Commented Nov 29, 2020 at 11:32
  • \$\begingroup\$ 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. \$\endgroup\$
    – user16324
    Commented Nov 29, 2020 at 13:38
  • \$\begingroup\$ 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. \$\endgroup\$
    – jonk
    Commented Nov 30, 2020 at 5:16

1 Answer 1

0
\$\begingroup\$

The situation you’re describing is the ‘steady state’, where the 9V from the battery exactly matches the 9V across the load. But there’s only 9V across the load because a current is flowing. The current won’t suddenly stop flowing because that would create an imbalance between the 9V from the battery and the less-than-9V across the load, which is obviously impossible.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.