AC with Variable RL (2-19-2021)

Set source current due to $V_1$ be $I_1$ and that due to $V_2$ be $I_2$ . Then:

$L_1 \dot{I}_1 + R(I_1+I_2) = V_1 \ \dots \ (1)$ $L_2 \dot{I}_2 + R(I_1+I_2) = V_2 \ \dots \ (2)$ $I_1(0)=I_2(0) = 0$

We are asked to compute:

$H = \int_{0}^{0.1} R(I_1 + I_2)^2 \ dt$

This problem has been handled numerically, details of which are shown in the code below.

As for energy analysis, multiply (1) by $I_1 \ dt$ and (2) by $I_2 \ dt$ and add:

$V_1I_1 \ dt + V_2I_2 \ dt = L_1I_1 \ dI_1 + L_2I_2 \ dI_2 + R(I_1 + I_2)^2 \ dt$ $\int_{0}^{t}V_1I_1 \ dt + \int_{0}^{t}V_2I_2 \ dt = \frac{L_1I_1^2}{2} + \frac{L_2I_2^2}{2} + \int_{0}^{t} R(I_1 + I_2)^2 \ dt$

The above equation clearly tells us that the energy provided by both sources is partly stored in the inductors and partly dissipated in the resistor. Since energy is being dissipated in the resistor, the energy of the circuit is not conserved. If energy was conserved, the energy provided by the source would not be dissipated in any way by the circuit.

As for power dependence on resistance:

$P = R(I_1 + I_2)^2 = (I_1 + I_2)^2 + \sin(\omega_R t) (I_1 + I_2)^2$

Naively speaking, I would say the power is a linear combination of a component with constant resistance and that with variable resistance. This is obviously not true, because $I_1$ and $I_2$ both depend on the resistance as can be seen from the governing equations. So essentially, the dependence of power dissipation on the resistance is nonlinear. It is a relation that I do not fully understand at this stage.

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clear all
clc

% Parameters of circuit:
w           = 120*pi;
wr          = 20*pi;
L1          = 5/w;
L2          = 3/w;

% Time step and end time definitions:
dt          = 1e-5;
tf          = 0.1;
t           = 0:dt:tf;

% Initial conditions:
I1(1)       = 0;
I2(1)       = 0;

% Heat dissipation in resistance initialisation:
H           = 0;

for k = 1:length(t)-1

  % Definitions of source voltages and time-varying resistance:
  T         = t(k);
  V1        = 15*sin(w*T + pi/6);
  V2        = 10*sin(w*T);
  R         = 1 + sin(wr*T);

  % Governing equations of circuit:
  dI1       = (V1 - R*(I1(k) + I2(k)))/L1;
  dI2       = (V2 - R*(I1(k) + I2(k)))/L2;

  % Explicit Euler numerical integration:
  I1(k+1)   = I1(k) + dt*dI1;
  I2(k+1)   = I2(k) + dt*dI2;

  % Computing heat dissipation integral:
  H         = H + (I1(k+1) + I2(k+1))^2*R*dt;
end

ANSWER = H
% ANSWER = 1.2167

Below is a plot of power dissipated at the resistor, the resistor value and the current. The power lost at the resistor follows both the resistor value and the current - the higher the resistor value the more power is dissipated, as long as there is current flowing through the resistor. The resistor value is set externally. There must be some power input into the system to drive the current, since energy is given away at the resistor.