Deriving RL circuit (integration factor)

**imagemania** · June 3rd 2011, 02:43 AM

I'm having trouble with this, here's my attempt:
$V=V_{R}+V_{L}$
$= IR + L \frac{dI(t)}{dt}$

Now introduce f(t) for the integratino factor, divide by L to get the first order on its own

$f(t)\frac{V}{L} = f(t)\frac{IR}{L} + f(t)\frac{dI(t)}{dt}$
$f '(t) = \frac{Rf(t)}{L}$
$f(t) = e^{\int \frac{R}{L} dt}$
So far this seems somewhat logical, as the final answer has the minus factor in it.

Continuing:
$e^{\frac{Rt}{L}}\frac{V}{L} = \frac{d}{dt} [e^{\frac{Rt}{L}} I]$
Integrating this gets me back to
$V = IR$
as theres nothing i can do with the exponentials

The answer i should be getting is:
$I = I_{0} [1-e^{-\frac{Rt}{L}}]$

Hope you can help me out

**mr fantastic** · June 3rd 2011, 03:23 AM

Originally Posted by imagemania

I'm having trouble with this, here's my attempt:
$V=V_{R}+V_{L}$
$= IR + L \frac{dI(t)}{dt}$

Now introduce f(t) for the integratino factor, divide by L to get the first order on its own

$f(t)\frac{V}{L} = f(t)\frac{IR}{L} + f(t)\frac{dI(t)}{dt}$
$f '(t) = \frac{Rf(t)}{L}$
$f(t) = e^{\int \frac{R}{L} dt}$
So far this seems somewhat logical, as the final answer has the minus factor in it.

Continuing:
$e^{\frac{Rt}{L}}\frac{V}{L} = \frac{d}{dt} [e^{\frac{Rt}{L}} I]$
Integrating this gets me back to
$V = IR$
as theres nothing i can do with the exponentials

The answer i should be getting is:
$I = I_{0} [1-e^{-\frac{Rt}{L}}]$

Hope you can help me out

What about the constant of integration? You have completely ignored it.

It would help if the original question was posted. Assuming a series RL circuit and assuming V does not depend on time:

5. Application of ODEs: Series RL Circuit

**imagemania** · June 3rd 2011, 07:03 AM

It's not a question it's just something I am trying to derive.

As for the constant, the first integral should have a constant (though generally in first order integrating factor questions it's saved till the end). In this case it woudl still cancle anyway from what i can see.
As for the second integral, it would be between 0 and t (i.e. no constant).

So following that:
i would get
$\frac{V}{L}e^{\frac{Rt}{L}+c} = \frac{d}{dt}[e^{\frac{Rt}{L}+c}I]$

$\frac{V}{R}e^{\frac{Rt}{L}+c} = Ie^{\frac{Rt}{L}+c}$

**topsquark** · June 3rd 2011, 11:30 AM

Originally Posted by imagemania

It's not a question it's just something I am trying to derive.

As for the constant, the first integral should have a constant (though generally in first order integrating factor questions it's saved till the end). In this case it woudl still cancle anyway from what i can see.
As for the second integral, it would be between 0 and t (i.e. no constant).

So following that:
i would get
$\frac{V}{L}e^{\frac{Rt}{L}+c} = \frac{d}{dt}[e^{\frac{Rt}{L}+c}I]$

$\frac{V}{R}e^{\frac{Rt}{L}+c} = Ie^{\frac{Rt}{L}+c}$

Mr. F was not talking about the constant for that integration.
$V = IR + L \frac{dI}{dt}$

Your integrating factor is correct, so let's jump to
$\frac{d}{dt} \left ( e^{(R/L)t}I \right ) = e^{(R/L)t}$

Integrating:
$e^{(R/L)t}I + C = \frac{V}{R}e^{(R/L)t}$

$I = -Ce^{-(R/L)t} + \frac{V}{R}$

Now define I0 = V/R and use the condition that I(t = 0) = 0 A.

-Dan

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