# Engineering Use a MATLAB Workspace to get the unknown values in an AC circuit

#### Fatima Hasan

Homework Statement
The question is attached below.
Homework Equations
-
The plot of part b is attached below.
I got a warning on part a , I don't know where is my mistake .
Part A
Matlab:
clc;
clear all;
%parta
f=800;
t=0:10e-6:0.005;
xc=10^(-6);
xl=40e-3;
for i=1:(0.005)/(10e-6)
w=2*f*pi;
zeq=xl*j*w+50;
is(i)=2*(1600*t(i)*pi);
c=1/(xc*j*w);
i1(i)=is(i)*c/(c+zeq); %apply CDR
vo(i)=i1(i)*zeq;
end
figure (1)
subplot(2,1,1) , plot (t,vo)
subplot (2,1,2) , plot (t,is)
%partb
f=10:10:1000;
xl=40e-3;
xc=1e-6;
r=50;
for i=1:100 %(1000-10)/10+1=100
w(i)=2*f(i)*pi;
zeq(i)=xl*j*w(i)+r;
c(i)=1/(xc*j*w(i));
l(i)=xl*w(i)*j;
zeq(i)=r+l(i);
zt(i)=zeq(i)+c(i);
is(i)=2*(cosd(0)+sind(0)*j); %convert to recantgular form
vo(i)=(is(i)*c(i)/zt(i))*zeq(i); %Apply CDR to find the current and multiply it by Zeq to get the voltage
end
logscale1=20*log(abs(vo));
subplot(2,1,1) , plot(f,logscale1)
subplot(2,1,2) , plot(f,logscale2)

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#### BvU

Homework Helper
I don't have Matlab, but Octave gives an error in line 11: the index $i$ in array $is$ is invalid because $i$ is not an integer...

#### RPinPA

Homework Helper
I don't have Matlab, but Octave gives an error in line 11: the index $i$ in array $is$ is invalid because $i$ is not an integer...
That sounds likely just looking at the code. Actually providing the text of the message would be useful, we can't tell much from "I got a warning somewhere in my code".

I know that 0.005/10e-6 looks like it should be an integer, but floating point arithmetic in computers is not exact. Many numbers that are terminating decimals in base 10 are repeating decimals in base 2.

I would try changing the limits of the FOR loop to "for i=1:round(0.005/10e-6)"

#### Fatima Hasan

That sounds likely just looking at the code. Actually providing the text of the message would be useful, we can't tell much from "I got a warning somewhere in my code".

I know that 0.005/10e-6 looks like it should be an integer, but floating point arithmetic in computers is not exact. Many numbers that are terminating decimals in base 10 are repeating decimals in base 2.

I would try changing the limits of the FOR loop to "for i=1:round(0.005/10e-6)"
I resolve it again .
Plots are attached below.
Matlab:
clc;
clear all;
%parta
f=800;
xl=40e-3;
xc=1e-6;
r=50;
t=0:1e-6:5e-3
is=2*cos(1600*pi*t);
w=2*pi*f;
l=xl*j*w;
c=1/(xc*j*w);
z=l+r;
zeq=(z*c)/(c+z);
vo=is*zeq;  %v=total current*total Z
figure (1)
subplot(2,1,1) , plot (t,vo)
subplot(2,1,2) , plot (t,is)
%part b
f=10:10:1000;
xl=40e-3;
xc=1e-6;
r=50;
for i=1:100 %(1000-10)/10+1=100
w=2*pi*f(i);
c=1/(xc*j*w);
l=xl*w*j;
z=l+r;
zeq=(z*c)/(z+c);
is=2*(cosd(0)+sind(0)*j); %convert to rectangular form
vo(i)=is*zeq;
vom(i)=abs(vo(i));
end
figure (2)
semilogx(f,vom)
figure (3)
semilogx(f,voa)

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Last edited:

#### Fatima Hasan

I missed (i) , it should be voa(i)=rad2deg(angle(vo(i)));

"Use a MATLAB Workspace to get the unknown values in an AC circuit"

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