Nonlinear Resistor

Electricity and Magnetism Level 3

The DC circuit pictured above contains a nonlinear resistor, whose voltage / resistance characteristic is shown.

In this circuit, what resistance value does the nonlinear resistor take in DC steady-state?

Details:
- R 0 = 2 Ω R_0 = 2 \, \Omega
- α = 1 V − 1 \alpha = 1 \, V^{-1}
- Assume that the non-linear resistor obeys Ohm's Law. R = V N I = R 0 e − α V N R = \frac{V_N}{I} = R_0 e^{-\alpha V_N} , where I I is the current


The answer is 0.43709.

Steven Chase by Steven Chase , 37, USA

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1 solution

Steven Chase
Feb 10, 2018

Current in the circuit:

I = 5 1 + 2 e − V N = 5 − V N 1 5 = ( 1 + 2 e − V N ) ( 5 − V N ) I = \frac{5}{1 + 2\, e^{-V_N}} = \frac{5 - V_N}{1} \\ 5 = (1 + 2\, e^{-V_N})(5 - V_N)

Solve this numerically for V N V_N and then plug back in to get the nonlinear resistance value. It turns out to be about 0.43709 Ω 0.43709 \, \Omega

1 Helpful 1 Interesting 1 Brilliant 0 Confused

I just stumbled upon this problem and I got it wrong. My thoughts on the concept of this problem are as such:

  • When I think of a nonlinear resistance, the first thought that comes to mind is that Ohm's law is invalid.

  • I like to think of nonlinear resistance as: R = d V d I R = \frac{dV}{dI} . If I apply the same argument here, I get an expression for V N V_N as a function of I I .

In this problem, I observe that you implicitly assumed that Ohm's law is valid. Why does this assumption hold true?

Karan Chatrath - 6 months ago

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I simply conceived of it as having those properties. I probably should have stated that it obey's Ohm's Law. In hindsight, I don't know if anything like this exists in reality. I think what you described is similar to a varistor. The varistor appears to have a fixed V-I characteristic. So I suppose for a varistor, you could sweep V over a range (guess and check), and for each V, there would be a corresponding pre-defined I. And then you could calculate Ohm's Law for the standard resistor (1 ohm) to see if that expression yields the same current as the V-I characteristic did. If so, that is the operating voltage. That would be a good problem.

Steven Chase - 6 months ago

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'Varsitor' sounds like an interesting read. I'll take a look and see if I can cook up a question on it, sometiime.

Karan Chatrath - 6 months ago

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