# Thread: determining current in LR circuit

1. ## determining current in LR circuit

A 30-volt electromotive force is applied to an-LR series circuit in which the inductance is 0.1 henry and the resistance is 50 ohms. Find the current i(t)
if i(0)=0 Determine the current as t--> infinity.

V=IR
I=V/R,

but i don't really understand what the question wants. can anyone help?

2. Originally Posted by bleu90
A 30-volt electromotive force is applied to an-LR series circuit in which the inductance is 0.1 henry and the resistance is 50 ohms. Find the current i(t)
if i(0)=0 Determine the current as t--> infinity.

V=IR
I=V/R,

but i don't really understand what the question wants. can anyone help?
Since we are dealing with an LR circuit, the proper differential equation set up is $L\frac{\,dI}{\,dt}+RI=30\implies 0.1\frac{\,dI}{\,dt}+50I=30$

Solve this linear DE, and then apply the initial condition $I(0)=0$ to find C, once you have solved the DE for I(t).

Then take $\lim_{t\to\infty}I(t)$ to answer the last part of the question.

--Chris

3. after dividing the equation by 0.1 and getting the integrating factor,

integrating e^500t*y = 300e^500t

i'm getting confused here. how can i proceed?

4. Originally Posted by bleu90
after dividing the equation by 0.1 and getting the integrating factor,

integrating e^500t*y = 300e^500t

i'm getting confused here. how can i proceed?
Note that once you multiply through by the integrating factor, you get $\frac{d}{\,dt}\bigg[e^{500 t}I\bigg]=300e^{500t}$

Integrating both sides, you get $e^{500 t}I=\tfrac{3}{5}e^{500 t}+C\implies I(t)=\tfrac{3}{5}+Ce^{-500t}$

Now apply the initial condition $I(0)=0$ to find C:

$0=\tfrac{3}{5}+C\implies C=-\tfrac{3}{5}$

Therefore, $I(t)=\tfrac{3}{5}\left(1-e^{-500t}\right)$

Now find $\lim_{t\to\infty}I(t)$

Does this make sense now?

--Chris

5. i got my final answer as 15. is this correct?

6. Originally Posted by bleu90
i got my final answer as 15. is this correct?
$\lim_{t\to\infty}\tfrac{3}{5}\left(1-e^{-500t}\right)=\tfrac{3}{5}\lim_{t\to\infty} \left(1-e^{-500t}\right)=\tfrac{3}{5}\cdot\left(1\right)=\colo r{red}\boxed{\tfrac{3}{5}}$

--Chris