DP 1-2 The voltage across a circuit element is v(t) = 20 (1–e^–8t)V when t ≥ 0 and v(t) = 0 when t < 0. The current in this element is i(t) = 30e^–8t mA when t ≥ 0 and i(t) = 0 when t < 0. The element current and voltage adhere to the passive convention. Specify the power that this device must be able to absorb safely.

Hint: Use MATLAB, or a similar program, to plot the power.

P is in mW when t is in s. Converting to W the equation becomes

To find the maximum power we can find the maximum of curve or we can take the first derivative and set it equal to zero.

Take reciprocal of equation

Take ln of both sides

Now plug into P(t)

Max Power that can be absorbed is 0.150 W

MATLAB Plots and Code follows:

%Program to plot out Current vs time for Homework Problem 1.DP1-2.

%Version 2018-12-24 DW Donovan

clear all;

t = [0:0.001:2]';

P = 0.6*(exp(-8*t)-exp(-16*t));

tt1 = 'PH 320 Homework Problem DP1-2';

tt2 = 'Power vs Time for a Circuit Element';

ttn = 'D.W. Donovan -- ';

tnl = '\newline';

ttf = [tt1 tnl tt2 tnl ttn date];

xl = 'Time, t, (s)';

yl = 'Power, P, (W)';

sp = .04;

axxmin = min(t)-sp;

axxmax = max(t) + sp;

axymin = min(P) - sp;

axymax = max(P) + sp;

figure

hold on

plot(t, P,'k-','LineWidth',5)

title (ttf,'FontSize', 16)

xlabel(xl, 'FontSize', 16)

ylabel(yl, 'FontSize', 16)

axis([axxmin axxmax axymin axymax])

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This page last updated on December 26, 2018