0
$\begingroup$

Shouldn't the current passing through a resistor be lesser than that which passes through a circuit?

My understanding is that since Current = Charges/Time. If there exists a resistance to the flow of charges, then that must mean the charges slow down, meaning that more time is required to pass through a point. So, the current should then decrease. But, since this opposition to the flow of charges doesn't exist in the ENTIRE circuit, it should really only decrease the current in the resistor, right?

Am I confusing the resistance of the resistor with the resistance of the wire? Any help will be appreciated! Thanks a bunch for lending me some of your time :D

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

Shouldn't the current passing through a resistor be lesser than that which passes through a circuit?

No. If it were less, charge would accumulate at the point the current enters the resistor.

Improve this answer
$\endgroup$
14
  • $\begingroup$ Current = Charges/time. The time factor increases to pass through a certain point. So, the current should decrease while passing THROUGH the resistor, but the current at the input and the output of the resistor, will, however, remain the same. $\endgroup$
    – ihateelectricalphysics
    Sep 2, 2020 at 19:25
  • $\begingroup$ It doesn’t work that way. If the current is 3 amperes, it means 3 coulombs pass each cross-section per second, whether the cross-section is in the wire or in the resistor. $\endgroup$
    – G. Smith
    Sep 2, 2020 at 19:28
  • $\begingroup$ Doesn't a resistor provide a resistance to the flow of charge? Won't that slow down the charge? Won't that reduce current? $\endgroup$
    – ihateelectricalphysics
    Sep 2, 2020 at 19:32
  • $\begingroup$ It reduces the current in the entire circuit compared to what it would be if the resistor weren’t there. $\endgroup$
    – G. Smith
    Sep 2, 2020 at 19:35
  • 2
    $\begingroup$ The analogy is fine. But isn’t it obvious that you can’t have 3 gallons per second passing through the wide pipe and 2 gallons per second passing through the narrow pipe? Where would the other one gallon per second go? I’m done because comments are not for extended discussions. $\endgroup$
    – G. Smith
    Sep 2, 2020 at 19:52
2
$\begingroup$

maybe you compare it with the flow of water in a pipe instead of the batterie take a pump. The resistor is a very narrow part of the pipe compared to the rest of the water circuit. Now since this narrow resistor is in the circuit less water per second can flow thru than without so the narrow part determines the water current , but its less in all the parts of the tubes. In the narrow part it will flow faster, since the same amount must pass the a smaller area, but the current here liters/second is everywhere the same, since where els should the water coming to the narrow part stay if not flown the, an the water leaving the narrow will flow back to the pump. Same with electric current, the resistor regulates how mach charge can pass an area per second, the speed of the charge in the resistor will be higher thats why the resistor gets hot, but all the charges which go in area must get out .

Improve this answer
$\endgroup$
3
  • $\begingroup$ So, the resistor only regulates how many charges will pass through the RESISTOR ITSELF and not through the circuit, right? $\endgroup$
    – ihateelectricalphysics
    Sep 2, 2020 at 20:22
  • $\begingroup$ No, I tried ti emphasize, that the resistor regulates how many charges go thru any part of the circuit. $\endgroup$
    – trula
    Sep 2, 2020 at 21:01
  • $\begingroup$ I think I'm starting to understand it even more now. The water analogy is confusing me but I think I've cleared it in the comment of Mr. G. Smith's answer. $\endgroup$
    – ihateelectricalphysics
    Sep 2, 2020 at 21:03

Your Answer

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

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