MIDTERM_2_1

# MIDTERM_2_1 - %Q(i 1)= Q(i(h(R*C*Q(i(V*h/R%This line shows...

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% Charging a capacitor using Euler's Loop Method to solve the DIFFY-Q % The charge and current as a function of time on a capacitor in an RC % circuit % Charging a capacitor h = 1; % time increments R = 1e6; % the resistance in ohms C= 1e-5; % the capacitance in farads V=10 ; % the voltage of the power supply a = 0; % the starting time b = 60; % the ending time >= 6RC m = (b - a) / h; Q = zeros(1, m+1); Q(1) = 0; t = a:h:b; for i = 1:m Q(i+1)=Q(i)+exp(-t./(R*C)).*(V*h)/R;
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Unformatted text preview: %Q(i+1)= Q(i)-(h/(R*C))*Q(i)+(V*h)/R; %This line shows diff eqn end Qex = C*V*(1- exp(-t/(R*C))); %This is the exact solution from analytical %integration I=[(1/h)*diff(Q),0] format bank disp( [t' Q' Qex'] ) subplot(1,2,1) plot(t, Q,'r') xlabel( 'time in seconds' ), ylabel( 'charge on capacitor in coulombs' ) hold on plot(t, Qex ), hold off subplot(1,2,2) plot(t,I,'k') xlabel( 'time in seconds' ), ylabel( 'current in the circuit in amps' )...
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## This note was uploaded on 09/15/2011 for the course EE 150 at USC.

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