Question

To be completed on MATLAB. Thanks!

1. Create variables for the source voltage (5V), LED forward voltage (1.85V), and the LED maximum current (30 milliamps). 2. Input the value of the first resistor. 3. If the value is negative: 1. Print Negative resistance, reversing. 2. Set the resistor's value to its negation (for example - resistance = -resistance 4. Repeat steps 2 and 3 for the other two resistors. 5. Calculate the effective resistance using the parallel resistor formula. 6. Calculate the current through a function based on the two voltages and the effective (total) resistance. 7. Convert the current to milliamps by multiplying by 1000. 8. If the current in milliamps is greater than the diode's maximum current rating: 1. Display the message WARNING: Unsafe current. 9. Otherwise, 1. Display the message Current is safe. 10. Display the current in milliamps.

Answer 1

MATLAB script

clc
clear
close
sv=5; fv=1.85;mc=30;
r1=input('Enter the value of first resistor');
if r1<0;
disp('Negative resistance,reversing')
r1=-r1;
end
r2=input('Enter the value of first resistor');
if r2<0;
disp('Negative resistance,reversing')
r2=-r2;
end
r3=input('Enter the value of first resistor');
if r3<0;
disp('Negative resistance,reversing')
r3=-r3;
end
ref=(r1*r2*r3/(r1*r2+r2*r3+r3*r1));
c=ref*(sv+fv)*1000;
if c>mc
disp('WARNING:Unsafe current')
else
disp('current is safe')
fprintf('current in function is %g\n',c)
end

Result

Enter the value of first resistor 1e-5
Enter the value of first resistor -2e-5
Negative resistance,reversing
Enter the value of first resistor 10e-5
current is safe
current in function is 0.0428125 milliamps