# Can an integral that is a variable of itself be solved?

1. Oct 7, 2011

### kmarinas86

Under the assumption that the voltage is $V_f\left(1-e^{-\frac{t}{RC}}\right)$, where $V_f$ is the final voltage, how would I determine the relationship between current $I$ and time $t$?

$$I = \int_0^T \frac{V_f\left(1-e^{-\frac{t}{RC}}\right) - RI}{L} \,dt \,$$

$L$ the magnetic inductance, $R$ the resistance, and $C$ the capacitance, are constants.

How would I plot current $I$ as a function of time $t$? (The only variables here are $I$ and $t$.) Let's assume initial conditions of $I=0$ and $t=0$. My problem here is that the variable I am trying to calculate is a variable inside the integral that is used in deriving the variable itself! How are such problems handled? Any help is appreciated!

2. Oct 7, 2011

3. Oct 7, 2011

### SammyS

Staff Emeritus
So, I take it that you mean that $\displaystyle I(T\,) = \int_0^T \frac{V_f\left(1-e^{-\frac{t}{RC}}\right) - RI(t)}{L} \,dt\,.$

Differentiate both sides w.r.t T and solve for I'(T). See if you can integrate the result to get I(τ).

4. Dec 1, 2011

### Simon Bridge

OP has a related question in this other post.

note:
$I=\int RI.dt$ - differentiate both sides gives

$dI/I = R.dt$

... what was the problem?