Circuit With Nonlinear Resistor - 2

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. Let the heat dissipated by the resistor at $t=10$ be $H$ . Enter your answer as $\lfloor 10H \rfloor$ .

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 and $\lfloor . \rfloor$ represents the Floor function.
Assume all quantities to be defined in SI units.

Bonus:

Is your result in agreement with the energy conservation principle? Also, when $\alpha = 0$ , does the result match expectations?

The answer is 425.

1 solution

Steven Chase
Dec 12, 2020

This was a fun one. Start with the voltage loop equation for the circuit.

$V_N = V_S - L \dot{I}$

From this, we can derive the expression for the time derivative of the current.

$\dot{I} = \frac{V_S - V_N}{L}$

Then apply the chain rule to the expression for the nonlinear resistor.

$\frac{d V_N}{dI} = \frac{d V_N}{dt} \frac{dt}{dI} = R_0 e^{-\alpha V_N} \\ \dot{V}_N = R_0 e^{-\alpha V_N} \dot{I}$

To summarize:

$\dot{I} = \frac{V_S - V_N}{L} \\ \dot{V}_N = R_0 e^{-\alpha V_N} \dot{I}$

The expressions are sufficient to perform numerical integration. In addition, the powers associated with the source, resistor, and inductor are:

$P_S = V_S I \\ P_R = V_N I \\ P_L = (L \dot{I}) I$

In the above expressions, $P_S$ uses a source convention and $P_R$ and $P_L$ use load conventions. When integrated, we expect the source energy to equal the resistor and inductor energies combined.

$E_S = E_R + E_L$

The plots below show $V_N$ and $I$ , as well as the energies for $\alpha = 1$ . It is clear that energy conservation is satisfied.

The next set of plots shows the results for $\alpha = 0$ . Since it is just a standard resistor in this case, we expect there to be a direct proportionality between $V_N$ and $I$ . The circuit time constant in this case is $\tau = \frac{L}{R_0} = 0.5$ , so the resistor should have the full source voltage across it after about $t = 5 \tau = 2.5$ . These expectations are met.

Simulation Code:

import math

dt = 10.0**(-5.0)

alpha = 1.0
VS = 2.0
R0 = 2.0
L = 1.0

###################################

t = 0.0
count = 0

VN = 0.0
I = 0.0

Id = (VS-VN)/L
VNd = R0*math.exp(-alpha*VN)*Id

ES = 0.0
ER = 0.0
EL = 0.0

print "t VN I ES ER EL (ES-(ER+EL))"

while t <= 10.0:

    VN = VN + VNd*dt
    I = I + Id*dt

    Id = (VS-VN)/L
    VNd = R0*math.exp(-alpha*VN)*Id

    PS = VS*I
    PR = VN*I
    PL = (L*Id)*I

    ES = ES + PS*dt
    ER = ER + PR*dt
    EL = EL + PL*dt

    t = t + dt
    count = count + 1

    if count % 1000 == 0:

        print t,VN,I,ES,ER,EL,(ES-(ER+EL))

###################################

print ""
print ""

print dt
print t
print ER
print math.floor(10.0*ER)

Nice solution @Steven Chase , but would you mind sharing how you solved the system of differential equations for I and VN? Or you just computed it with the programme above?

Veselin Dimov - 5 months ago

My solution combines a state space with numerical integration. The state space represents the time rate of change of the state variable in terms of the state variable $x$ and a forcing function $u$ .

$\dot{x} = f(x,u)$

Then the numerical integration looks like:

$x_k = x_{k-1} + \dot{x}_{k-1} \Delta t \\ x_k = x_{k-1} + f(x_{k-1},u_{k-1}) \Delta t$

Steven Chase - 5 months ago

Circuit With Nonlinear Resistor - 2

The answer is 425.

1 solution

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