Question

Okay, this question had me and my friend confused today.

In a certain electrical circuit, the current "i" satisfies the differential equation:-

L (di / dt) = E - Ri

(a) Solve the equation to show that E = Ri + Ae^(-Ri / L)

Answer 1

The solution you have written down isn't quite right. But in any case, this solution is easily solved by what is called separation of variables:

\[ L \frac{di}{dt} = E - Ri \ \ \Longrightarrow \ L \frac{di}{E-Ri} = dt\]

Now integrate...

\[ \int L \frac{di}{E-Ri} = \int dt \]
Hence ... you take it from here

Answer 2

Oh sorry, the solution should be "E = Ri + Ae^(-Rt / L)" :)

Would L in this case be taken over to the other side to get

\[\int\limits \frac{ di }{ E - Ri } = \int\limits \frac{ dt }{ L }\]

From here I ended up with

\[\ln |E - Ri| = \frac{ t }{ L }\]

Correct me if im wrong ;o

Answer 3

\[\ln |E = Ri| = \frac{ t }{ L } + C\]

I mean, can't miss out the C ;)

Answer 4

Not quite. You have a -R in front of the i, so when you integrate, we had better see that appear somewhere as constant.

Answer 5

Oh I see :) thanks.