Question

Calculate equivalent resistance using MATLAB

Answer 1

r=0;
k=0;
type=input('Type of connection (series or parallel): ','s');
if strcmpi('series',type)
while k>=0
k=input('Resistor rating (enter 0 to stop): ');
r=r+k;
if k==0
break;
end
end
disp(['Equivalent resistance: ',num2str(r)])
end

if strcmpi('parallel',type)
while k>=0
k=input('Resistor rating (enter 0 to stop): ');
r=1/(1/r+1/k);
if k==0
break;
end
end
disp(['Equivalent resistance: ',num2str(r)])
end

Answer 2

the series part works fine but not the parallel part, what am I doing wrong?

Answer 3

reminder please, \(\frac1R = \sum_i \frac1{r_i} \) ?

but what exactly doesn't work?

Answer 4

initiallly your r =0

Answer 5

i see, so how do i go about solving it?

Answer 6

you must ask the value of one resistance before going into the loop.

Answer 7

ok, let me try it out first

Answer 8

alternately you assume some initail resistance and adjust the same before displaying the final result...

Answer 9

I wouldn't start with a default value like 1 or .. 2 ? because that would fix one 'resistor rating' to that value; I don't think he wants that.

Answer 10

@reemii
correct ..

Answer 11

if strcmpi('parallel',type)
r=input('Resistor rating: ');
while k>=0
k=input('Resistor rating (enter 0 to stop): ');
r=1/(1/r+1/k);
if k==0
break;
end
end
disp(['Equivalent resistance: ',num2str(r)])
end

I tried it like this but didnt work

Answer 12

tell us what values you enter, and what the output is. we'll understand better what happens

Answer 13

no matter what value, or how many values i enter, the output remains 0

Answer 14

I get it.
you must make the test (k>0) before the line r=1/(1/r+1/k); because you must break before doing anything forbidden like dividing by zero or a negative value.
I think that r is always evolving fine, until you decide to "break". then you have this 1/0 division and 1/(1/r + infinity) always gives you ZERO.

Answer 15

And how do I make such a test? Another while loop?

Answer 16

No, instead of
———
k=input('Resistor rating (enter 0 to stop): ');
r=1/(1/r+1/k);
if k==0
break;
end
———
test the value of k immediately like this:
———
while k>0 %% this could be simply "while true", don't you think?
k=input('Resistor rating (enter 0 to stop): ');
if k <= 0
break
end
r=1/(1/r+1/k);
end % of while
———

Answer 17

before the loop you might also make this test:
———
if r < 0
disp('you dont even want to supply one valid value? too bad...');
else
while .... % now the loop
...
end

Answer 18

I mean, "if r <= 0"

Answer 19

Thank you! I get it now