0
\$\begingroup\$

What is more accurate definition of Ohm's law and resistance? is it

$$R=\frac {V}{I}$$

or

$$R=\frac{dV}{dI}$$

This is doubt that developed in my mind during a class where professor derived power equation where he used second one for resistance in the derivation. I checked Wikipedia. They showed the first relation as accurate. Of course if the first relation is correct and resistance \$R\$ is constant, then we can use second relation. But what if resistance is not constant?

For a practical problem, suppose my voltage source is current dependent and is given by

$$V=I^2+2I$$

Then how will you find resistance of given circuit at a given value of current \$I\$?

\$\endgroup\$
  • 2
    \$\begingroup\$ If the relation between voltage and current in your circuit is non-linear then most of Ohm's law goes out the window. But not all of it. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 17 '14 at 6:07
  • \$\begingroup\$ so if the relation etween voltage and current is non-linear,what should we have to?which part of ohm's law can be applied?or any other advanced law like ohm's law??? \$\endgroup\$ – jjoyk Jan 17 '14 at 6:13
  • \$\begingroup\$ Ohm's law still stands. You just can't use it for everything. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 17 '14 at 6:13
  • \$\begingroup\$ So what we have to do when relation become non-linear for circuit solving??? \$\endgroup\$ – jjoyk Jan 17 '14 at 6:14
  • 1
    \$\begingroup\$ Just fall back to Maxwell's equations. \$\endgroup\$ – Samuel Jan 17 '14 at 6:18
3
\$\begingroup\$

This answer is probably inherently displeasing to the feeling of natural order for some :-) :

A law of nature is simply a statement of observed results under defined conditions.

Ohms law is essentially a statement that the ratio of the two two variables V & I is typically observed to remain approximately constant as the variables vary.

It is arguably saying the opposite of what it may seem - ie not so much
"R is the ratio between ..." but
it is more "if the ratio between V & I is constant then we call this constant resistance" and "this approximates typical behaviour of a significant proportion of real world products".

At any given moment R IS the ratio between V and I. If this ratio has changed then R has changed. So V/I never changes for specific values of V & I with all other conditions held constant, whereas dV/dI typically does change in the real world.

So R = V/I is an accurate statement
R = dV/dI is usually an approximation and
where it all falls apart it just means that the observation does not apply under thos e conditions.


That's woolier than I'd like but seems to convey what I'm trying to say. I hope :-).

\$\endgroup\$
  • \$\begingroup\$ Much easier than that, Russell. Its just the product rule. \$\endgroup\$ – Scott Seidman Jan 17 '14 at 12:07
  • \$\begingroup\$ @ScottSeidman - Either, no, that does not cover my point or, no, your compact summary fell below the bottom bound of my Mentat capabilities - more spice needed. [[Note to self: Watch for sandworms]]. \$\endgroup\$ – Russell McMahon Jan 17 '14 at 14:15
  • \$\begingroup\$ Just pointing out that mathematically, the ONLY situation where R is not equal to dV/dI are situations in which R is not constant with respect to current. \$\endgroup\$ – Scott Seidman Jan 17 '14 at 17:08
  • 1
    \$\begingroup\$ @ScottSeidman - We agree. My " ...if the ratio between V & I is constant then we call this constant resistance ..."is the flip side of that coin (or perhaps the same side viewed from an only slightly different angle. ) \$\endgroup\$ – Russell McMahon Jan 18 '14 at 1:16
  • \$\begingroup\$ There are many real-world situations in which dV/dI is relatively constant but doesn't equal V/I. In such cases, it is often useful to regard such a system as being equivalent to a resistor (of value dV/dI) in series with a voltage source (of value V-I(dV/dI)). \$\endgroup\$ – supercat Aug 28 '14 at 2:08
8
\$\begingroup\$

If a circuit element is Ohmic, then the voltage across and current through are proportional

$$V \propto I $$

and the constant of proportionality is the resistance \$R\$

$$V = R \cdot I$$

This relationship, Ohm's Law, is obviously a linear relationship and thus the slope of the associated VI curve is just the constant of proportionality \$R\$

$$\frac{dV}{dI} = \frac{V}{I} = R $$

However, for a non-linear circuit element, e.g., \$V = R \cdot (I + \epsilon I^3)\$, the voltage across is not proportional to the current through and thus

$$\frac{V}{I} \ne \frac{dV}{dI}$$

Now we can define the terms static resistance and dynamic (or differential) resistance:

$$R_{static} = \frac{V}{I} $$

$$r_d = \frac{dV}{dI} $$

The static resistance is useful for DC analysis while the dynamic resistance is useful for small-signal analysis (where we linearize the circuit element about the DC operating point).

For more, see the Wikipedia article section static and differential resistance.

\$\endgroup\$
  • \$\begingroup\$ The term ohmic is also sometimes used to junction that is low resistance as well as non-rectifying. Warwick ohmic contact \$\endgroup\$ – M. Enns Mar 27 '17 at 2:30
2
\$\begingroup\$

You're entering the field of superposition and small signal analysis with \$\frac{dV}{dI}\$. It simplifies a complex model in such a way that you can work reasonably accurate with it and with reasonably simple equations.

\$\endgroup\$
1
\$\begingroup\$

I think the important bit you are missing is that when we use the variable \$V\$ for voltage we are already talking about a voltage difference, not an absolute value. It is implied here that we are talking about the difference between the voltage at some point and the voltage at ground (0V by definition). So, Ohm's law is already in the form

$$ R = \frac{dV}{dI}$$

and the two forms are equivalent.

If you start talking about how real resistors or nonlinear elements behave then you need to stop talking about Ohm's law. Ohm's law applies to ideal resistances only.

\$\endgroup\$
  • \$\begingroup\$ I think your argument shows that Ohm's law is in the form $R = \Delta V/\Delta I$ for some macroscopic $\Delta V$; but I believe that Jibin's worry is that this won't return the same answer as a computation with "an infinitesimal $\textup dV$", i.e., the derivative. \$\endgroup\$ – LSpice Feb 5 '14 at 21:16
1
\$\begingroup\$

Start with \$v=ir\$, and differentiate with respect to \$i\$.

$$ \frac{\mathrm{d}v}{\mathrm{d}i} = i \frac{\mathrm{d}r}{\mathrm{d}i} + r \frac{\mathrm{d}i}{\mathrm{d}i} = r + i \frac{\mathrm{d}r}{\mathrm{d}i} $$

If \$r\$ is constant with respect to \$i\$, you get your equation. If not, you need to include the extra term. Calc is your friend!

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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