Why doesn't current decrease on the other side of a resistor? (1 Viewer)

Delirium

New Member
Joined
Jul 29, 2006
Messages
13
Gender
Male
HSC
2007
Ohm's law states: I = V / R . Jacaranda Preliminary HSC Physics states that measuring current on both sides of a resistor will give the same reading. But if you go by Ohm's law, increasing the variable R will decrease the value of the fraction V / R (k/x -> 0 as x -> infinity - basic maths). Therefore decreasing I (current). What's the deal here?

And how does a resistor reduce potential difference at a higher rate than when charge flows through a wire with very low resistance?

Please don't give me a textbook answer for these questions, as I have heard way too many already. They all seem to hide behind big, all-encompassing words. Please try and give me an explanation using "fundamental concepts" (i.e., in terms of the motion of the electrons, etc.).

This problem has been stressing me out for quite some time now, and anybody I have asked up until now has given me a textbook answer.
 
Last edited:

airie

airie <3 avatars :)
Joined
Nov 4, 2005
Messages
1,143
Location
in my nest :)
Gender
Female
HSC
2007
You're doing it as the exact opposite of the actual process. The voltage is what decreases as resistance increases, thus the current stays contant.

I think you're confusing voltage with current here. Remember current is a measure of the rate of charge flow in a circuit, which has to be constant in a series circuit (or a buildup of charge would occur at the position of the resistor), while voltage is the electric potential energy of the electrons, which is lost as electrons pass through the resistor, mostly as heat. See here, in I = V / R, V is what changes with R, while I is constant.

When you talk about "potential difference", it is just the difference between the higher voltage end (or what you'd call the "positive terminal") of the power supply and its lower voltage end. "Inside" the power supply (if we could call it like that :p) the electrons are raised to a higher potential, which is gradually decreased in the external circuit by the various components. Now rearrange Ohm's Law, you get V=IR. Since I is constant, V is directly proportional to R, therefore a resistor, with a higher resistance, reduces such potential difference to a greater extent than a low-resistance wire does.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top