Does Ohm's Law Apply to a Circuit with Current but No Voltage?

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Ohm's Law applies even in circuits where current flows but voltage appears to be zero, as the ideal connecting wires have negligible resistance. The confusion arises from the mathematical interpretation of V=IR, leading to the indeterminate form 0/0 when both voltage and current are zero. In practical scenarios, there is always some resistance in the wires, resulting in a small voltage drop. The distinction between "indeterminate" and "undefined" is clarified, with indeterminate forms requiring further analysis to determine their values. Understanding these concepts is essential for accurately applying Ohm's Law in real-world circuits.
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in a circuit like the one in the attached picture, the voltage between two points in between two resistors should be 0.

But there is current flowing through the circuit.

So what’s going on here? Does ohms law not apply in this situation for some reason?
 

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Add the two points where you are taking measurements to the schematic.
 
Ohm's law still applies. In the idealized case, the connecting wires between the resistors have zero resistance thus essentially zero voltage drop is required to produce a current. In reality, there is some resistance and thus some small voltage drop between two adjacent points on the conducting wires.

(Assuming I understand your question.)
 
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One does not normally apply Ohms Law to an ideal wire because you get what I assume is confusing you which is V=IR leading to 0 = I*0 or I = 0/0 and the problem w/ that is that 0/0 is undefined.
 
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0/0 is indeterminate.
 
cabraham said:
0/0 is indeterminate.
In the context of a limit, 0/0 is an "indeterminate form". That is if one has two functions, f() and g() and if ##\lim_{x \to c}f(x) = 0## and ##\lim_{x \to c}g(x) = 0## then one cannot determine from that information alone whether ##\lim_{x \to c} \frac{f(x)}{g(x)}## exists or, if it does, what value it takes.

As a standalone formula, ##\frac{0}{0}## is simply undefined.
 
cabraham said:
0/0 is indeterminate.
What's the difference between indeterminate and undefined?
 
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phinds said:
What's the difference between indeterminate and undefined?
"Indeterminate" is used in the context of limits, as @jbriggs444 already said, and means that some work is required to determine whether they represent numbers. Some indeterminate forms are ##[\frac 0 0]##, ##[\frac \infty \infty]##, ##[\infty - \infty]##, and ##[1^\infty]##. Most textbooks write these forms in brackets to emphasize that they are "forms" rather than actual numbers.
The following limits are examples of the first three types I listed:
##\lim_{x \to 1}\frac{x^2 - 1}{x - 1}##
##\lim_{t \to \infty}\frac{t^2 + 2}{t^3 - 1}##
##\lim_{y \to \infty}y^2 - y^3##
These are called indeterminate forms because it's not obvious at first glance that they represent a number. By taking a limit and subsequent algebraic or other operations, one can determine that a limit actually exists or not.

As for undefined -- the division of any number by zero is undefined, as is taking the square root (or fourth root or any even root) of a negative number is undefined (if we're dealing with the real-valued square root function). In general, attempting to evaluate a function at a number not in its domain is undefined.
 
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