Non-linear Resistance and Ohm's Law

[Chewy0087]

hey there, my question is just regarding non-linear resistors, basically if you had a situation whereby the current through a non-linear resistor is given by

$$I = 0.5 V^2$$

and you know the current through it (and thereby it's voltage), so for arguments sake current = 2A and therefore voltage = 2V. at this point, would the "resistance" of the resistor still obey Ohms law? so would therefore be 1 ohm (at this current)? or would it simply be the case that it has no "resistance" at all? or it's "resistance" is the fact that I = 0.5 V^2...

thanks in advance, it's quite a simple problem but i just thought of it

 

[Mapes]

Its resistance would be $1\,\Omega$, but it wouldn't follow Ohm's Law, which states that the ratio $V/R$ is constant.

 

[Dale]

hey there, my question is just regarding non-linear resistors, basically if you had a situation whereby the current through a non-linear resistor is given by

$$I = 0.5 V^2$$

and you know the current through it (and thereby it's voltage), so for arguments sake current = 2A and therefore voltage = 2V. at this point, would the "resistance" of the resistor still obey Ohms law? so would therefore be 1 ohm (at this current)? or would it simply be the case that it has no "resistance" at all? or it's "resistance" is the fact that I = 0.5 V^2...

thanks in advance, it's quite a simple problem but i just thought of it

It would not follow Ohm's law, but in any sufficiently small neighborhood around some specific voltage you could model it as an ideal resistor that does follow Ohm's law in series with an ideal voltage source.

 

[Chewy0087]

hmm I see, thanks a lot for clearing this up for me, but it begs a question then,

if the question was such that it asked (with my earlier model) something along the lines of, given a 9V battery, what fixed resistance resistor must be placed in series with it in order for the current in the circuit to be 2A would that mean that;

given the current resistance of the non linear resistor the resistance of the fixed one must satisfy $$(\frac{1}{1+R})* 9=2$$

or would you simply say that in order for that to occur the current would be 2A and 7V must pass through the linear resistor giving it a resistance of 14 ohms?

i think the answer is the former, but it makes it confusing

 

[Dale]

You would solve:
$$\sqrt{2 I} + R I = V$$
Where V is the battery voltage, I is the current, and R is the unknown resistance.

 

[Chewy0087]

You would solve:
$$\sqrt{2 I} + R I = V$$
Where V is the battery voltage, I is the current, and R is the unknown resistance.

hmm okay, so you totally discount the "current resistance" of the non-linear resistance? because clearly the first way i outlined doesn't yield the correct result

thanks for the help, much obliged