Question

how do you prove that the solution to L*di/dt+R*i(t)=v(t)=1 is i(t)=1/R(1-e^-R/L*t)?

Answer 1

Are you familiar with the equation of the this type \[\frac{ dn }{ dt } = -N \lambda \]
(The radioactivity equation)

Answer 2

@Abhishek619

Answer 3

i see.. im not sure if the radioactive equation applies as the initial differential equation relates to current (i) in an electrical circuit with voltage (v), resistance (R) and inductance (L), with respect to time (t).

Answer 4

well, you can always use this. i'll show you how to do this. you just need to identify if your differential equation is in this form or not.

Answer 5

\[L \frac{ di }{ dt }=v-iR\]
here, L, v and R are constants.
\[L \frac{ di }{ dt }=-R(i-\frac{ v }{ R })\]
\[\frac{ di }{ dt }=-\frac{ R }{ L }(i-\frac{ v }{ R }) \]
try to integrate on both sides.
\[\frac{ di }{ (i-\frac{ v }{ R }) }=-\frac{ R }{ L } dt\]
now integrate using the proper limits.

Answer 6

\[\int\limits_{0}^{i}\frac{ di }{ (i-\frac{ v }{ R }) }=-\frac{ R }{ L } \int\limits_{0}^{t}dt.\]
can you integrate now?

Answer 7

i see, thankyou very much :)

Answer 8

you're welcome! :)

Answer 9

to make it more easier, whenever you find such equation you can use the solution of the radioactivity equation.
\[\frac{ dN }{ dt }=- \lambda N\]
the solution for this differential equation is
\[N=N_{o}e ^{- \lambda t}\]
No is the amount of N at time t=0.
try to think on this and work out.

Answer 10

the solution to the radioactivity differential equation also comes from the above how we've integrated the above integration.

Answer 11

this is Kirchoff's law for RL circuits?

Answer 12

@oldrin.bataku yes! :)

Answer 13

cool