root of nonlinear equation

i am working on an electical concepts design problem for a emergency exit door release mechanism. I solved the circuit and found that i(t)=0.25*e^(-t/8c)*cos(t/8c)

i want i(t)=0.1 at t=3 sec

i got the code but it doesnt work

%ft=0.25*exp(-3/c)*cos(3/(8*c))-0.1;

%c= fzero(ft,3);

but i cant get result for c

can someone help me?

Answer 1

[Torsten]

ft = @( c ) 0.25* ...

instead of

ft = 0.25* ...

[Lidaou Ali]

i tried and changed it as you suggested but it still wont give me the answer

[Torsten]

Start with c=3 instead of c=1:

fzero(ft,3)

[Lidaou Ali]

thank you. I am getting different answers for different c values

[Torsten]

If you plot the function, you'll see that it has two roots:

c=-0.238

c=3.297

[Lidaou Ali]

that is correct. i believe i missed an "8"

the right equation is ft=@(c)0.25*exp(-3/(8*c))*cos(3/(8*c))