Question

enes 240 1.doc Compatibility Made -Saved to this PC ict O Search View Help Mailings Review References Layout Problem 1 A simple electric circuit consisting of a resistor, following figure. The charge on the capacitor q(t) capacitor, and function of time can inductor is depicted in the be computed a an as a as Switch o 1 -Rt(2L) Cos - t) 2L q(t)=goe Battery V Capacitor Inductor LC Resistor time, qo initial charge, R the resistance, L inductance, and C MATLAB to generate a plot of this function from time t 0 to 0.8, given that qo 10 R = 50, L 5 Where capacitance. Use t= = and C = 104. 1) Use C 10-3 instead and plot the function on the figure with different linetvpes same 2) Use arrows and text to indicate different functions. 3) Add appropriate titles and axis labels to the plot. Print the plot with grid on, and a listing of the script.

Answer 1

MATLAB code is given below in bold letters.

clc;
close all;
clear all;

% define the time vector
t = 0:0.01:0.8;

% define the parameters
R = 50;
L = 5;
C = 1e-4;

q0 = 1; % initial chage

% define the charge
q = q0 * exp(-R*t/(2*L)).*cos(sqrt(1/(L*C)-(R/(2*L))^2)*t);

% now plot the charge as a function of time
figure;
plot(t,q,'linewidth',2);grid on;
xlabel('time');ylabel('C');title('charge');
hold on;

% compute with C = 10^-3
C = 10^-3;
q = q0 * exp(-R*t/(2*L)).*cos(sqrt(1/(L*C)-(R/(2*L))^2)*t);

% now plot the charge as a function of time on the same figure
plot(t,q,'--r','linewidth',2);grid on;
xlabel('time');ylabel('C');
legend('C = 10^-^4','C = 10^-^3');

charge C 104 0.8 C 10-3 0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 0 0.1 0.2 0.3 0.4 0.5 0.6 07 0.8 time - -