Modulated AC Source

Electricity and Magnetism Level 2

A modulated A C AC voltage source supplies an R L RL circuit as shown. At time t = 0 t = 0 , there is no current in the inductor. How much energy is dissipated in the resistor between t = 0 t = 0 and t = 10 t = 10 ?

Details and Assumptions:
1) V S ( t ) = 10 sin ⁡ ( 2 π t ) sin ⁡ ( 120 π t ) V_S(t) = 10 \sin(2 \pi t) \sin(120 \pi t)
2) R = 1 R = 1
3) L = 1 120 π L = \frac{1}{120 \pi}


The answer is 125.02.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Nov 9, 2020

Computed the solution using a short script of code. Commented version attached below. The plot of current as a function of time reminds me of the phenomenon of beats. 

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clear all
clc

% Time step and Time vector initialisation:
dt   = 1e-7;
tf   = 10;
t    = 0:dt:tf;

% Circuit parameters:
L    = 1/(120*pi);
R    = 1;

% Initial current:
I(1) = 0;

% Initial resistor heat dissipation:
H    = 0;

for k = 1:length(t)-1

    % Source voltage:
    Vs(k)  = 10*sin(2*pi*t(k))*sin(120*pi*t(k));

    % Heat dissipation integral computation:
    H      = H + I(k)^2*R*dt;

    % Circuit equation for current derivative:
    dI     = (Vs(k) - I(k)*R)/L;

    % Integrating for current as a function of time:
    I(k+1) = I(k) + dt*dI;

end

Answer = H;

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