# Can non-ohmic conductors have a constant resistance?

The diagram below shows the relation between a direct current I in a certain conductor and the potential difference, V, across it. When V < 1.8V, the current is negligible.

Which statement about the conductor is correct?

A It does not obey Ohm's law but when V > 1.8V, its resistance is 4.

B It does not obey Ohm's law but when V = 3V, its resistance is 10.

The exercise book's answer key states that the correct answer is A, implying that resistance remains constant at 4 Ohms so long as V > 1.8. However, it is clear that the conductor is non-ohmic as it does not pass through the origin of the I-V graph. Thus, its resistance should not be constant. It the answer not B?

Preceding this question one whose solution seemed the contradict the above. It is as follows:

The graph shows the variation with potential difference V of the current I in an electrical component.

The resistance is measured for the current I(y) and for the current I(x). What is the change in the resistance of the component?

A zero

B (V(x)/I(x))-(V(y)/I(y))

As you guys have mentioned, it is the effective resistance of the component that matters. Therefore, the solution of this question should be A, where the resistance is constant as the graph is linear with a constant gradient. But the same answer key states that it is B. Hence the reason for my confusion.

Is this just a case of the people who wrote the answers being loose and inconsistent with their definition of "resistance"?

• The slope of the line is 1V per 250mA, or 4V/A. That is the same as a 4 Ohm resistor. Oct 25, 2019 at 13:54
• @SolomonSlow But when applying V=RI when V=3V and I=300mA, resistance is given to be 10 ohms. What am I missing here? Oct 25, 2019 at 13:57
• It is clear that, above 1.8V current vs voltage is a straight line - it has a specific resistance there given by the slope of the line. Whether the line goes through the origin is irrelevant. Only the slope matters. Oct 25, 2019 at 13:58
• @JonCuster If resistance is constant, then why does applying V=RI when V=3V and I=300mA yield a resistance of 10 ohms? Oct 25, 2019 at 14:00
• Resistance is only constant above 1.8V as stated in the problem. You are hung up trying to determine the slope by forcing the line through the origin. It does not go through the origin. Determine the slope locally at 3V. Oct 25, 2019 at 14:02

In simple terms Ohm's law is $$V\propto I$$ which means that if a graph of potential difference $$V$$ against the current $$I$$ is drawn it would be a straight line of gradient $$\frac{V-0}{I-0} = \frac VI$$ and since the definition of resistance is $$R=\frac VI$$ in this case the gradient of the line is the resistance.

The graph presented to you has portions which have a constant inverse gradient $$\frac{\Delta V}{\Delta I}$$ and this is called the small signal/incremental/dynamic/differential resistance.

Here the incremental resistance is $$\frac{4-2}{0.55-0.50}=\frac {2}{0.5} = 4\,\Omega$$.

It is a useful parameter if within the range of linearity one wants to know how much the voltage might change for a given change in the current.

Now when the potential difference is $$3\, \rm V$$ the current is $$300\,\rm mA$$ and so the resistance is $$R = \frac V I = \frac{3}{0.3} = 10 \,\Omega$$ but note that at $$4\,\rm V$$ the resistance is $$\frac {4}{0.55} \approx 7.3 \,\Omega$$.
So the resistance of this circuit element decreases as the voltage increases.

• So the term resistance should be re-termed incremental resistance? Then wouldnt the answer B be more accurate given that they asked for the resistance and not the incremental resistance? Oct 25, 2019 at 14:13
• @DavidToh I think that the correct answer is $B$ as no mention is made of incremental etc in option $A$. Oct 25, 2019 at 15:43

Neither A nor B obeys Ohm's law since the ratio of V to I is not constant for all V and I.

For values of V greater than 1.8 Volts, you need to look at the changes in voltage and current, or the ratio $$\frac{\Delta V}{\Delta I}$$ and not $$\frac{V}{I}$$. Choice B is using $$\frac{V}{I}$$.

Hope this helps.

• Thank you for the answer! An edit has been made to the question to add more details as to the reason for my confusion. Oct 25, 2019 at 14:40