Circuit with Nonlinear Resistor

Electricity and Magnetism Level 3

An inductor of inductance $L$ and a nonlinear resistor are connected in series to a DC voltage supply ( $V_S$ ). The voltage-current characteristic of the resistor is given by:

$\frac{dV_N}{dI} = R_o \mathrm{e}^{-\alpha V_N}$

Here, $V_N$ is the voltage across the resistor at any time $t$ . At time $t=0$ , the inductor has no energy and consequently, there is no voltage across the resistor. Find the steady state value of current flowing through the resistor.

Note:

The voltage-current relationship is defined for the resistor and it clearly does not obey Ohm's law.
$\alpha=1$
$V_S = 2$
$R_o = 2$
$L = 1$
$\mathrm{e} \approx 2.71828$ is Euler's number.
Assume all quantities to be defined in SI units.

Bonus:

When $\alpha = 0$ , does the result meet your expectation?

Inspiration

The answer is 3.1945.

1 solution

Steven Chase
Dec 11, 2020

My original problem defines the resistance in an absolute sense, whereas this problem defines it in an incremental sense. Re-arrange to solve for the incremental current in terms of the incremental voltage.

$dI = \frac{1}{R_0} e^{\alpha V_N} d V_N$

The voltage and current associated with the nonlinear resistor both start at zero. The voltage ends at $V_S$ in steady state, because without a changing current, there can be no voltage across the inductor. Integrate $dI$ to get the final current.

$I_{final} = \int_0^{V_S} \frac{1}{R_0} e^{\alpha V_N} d V_N = \frac{e^{\alpha V_S} - 1}{\alpha R_0}$

Plugging in numbers yields:

$I_{final} = \frac{e^2 - 1}{2}$

The final value of the current is undefined for $\alpha = 0$ , but taking the limit of the $I_{final}$ expression as $\alpha$ goes to zero results in:

$\lim_{\alpha \to 0} I_{final} = \frac{V_S}{R_0}$

This makes sense, since for $\alpha = 0$ , the nonlinear resistor simply behaves as a standard resistor.

Circuit with Nonlinear Resistor

The answer is 3.1945.

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