Current through a single inductor

[modulation]

Homework Statement

Ok, so this isn't a question of how to solve, but rather why the solution works.

Say you have a circuit with 2 inductors in parallel, a 5H and a 20H, in parallel to the inductors is a resistive network with a resistance of 8Ω, and a switch, which has been closed long enough for the inductors to reach DC steady state. The power source is not shown, but initially there is a current through each inductor, 8A through the 5H, and 4A through the 20H. When the switch is opened, the problem asks to find the voltage and current for the inductors, as well as the current through each individual inductor.

Homework Equations

So the solution for the voltage is v(t)=96(e)^{-2t}
, and current for both inductors is i(t)=12(e)^{-2t}

The solution to the current through the individual 5H inductor is
i_1(t)= \frac{1}{5} \int_{0}^{t} 96(e)^{-2x}dx -8

and for the 20H inductor is
i_2(t)= \frac{1}{20} \int_{0}^{t} 96(e)^{-2x}dx -4

The Attempt at a Solution

So what I do not understand is, how does the integral of the voltage, divided by the inductance, subtract the initial current, give the current through an individual inductor over time??
Really curious as to how this is possible mathematically and why??
Am I missing something?

 

[uart]

That follows fundamentally from the V/I characteristics of an inductor.

\frac{di}{dt} = v/L

so

i(t) = \frac{1}{L} \int_0^t v(\lambda) \, d\lambda + i(0)

BTW. The sign of the voltage in your example is technically inconsistent with the direction you're using for current, hence the reason I used "+ i(0)" instead of the minus you had.

 

[modulation]

So even though the voltage was derived from a combined inductance, you can simply divide by the individual inductance to get the current?
<br /> \frac{di}{dt} = v/L<br />

It is just odd , that first off the combined inductance is the product/sum, and then the voltage is the i(t)*R.
Strange that division can undo this. I have searched in my textbook for a more detailed proof of this but no luck.

 

[uart]

So even though the voltage was derived from a combined inductance, you can simply divide by the individual inductance to get the current?
<br /> \frac{di}{dt} = v/L<br />

It is just odd , that first off the combined inductance is the product/sum, and then the voltage is the i(t)*R.
Strange that division can undo this. I have searched in my textbook for a more detailed proof of this but no luck.

Sometimes it's just easier to lump some components together before solving a circuit. This doesn't mean that the basic circuit laws don't still hold for each individual element however.

There are times when you do run into problems here, but this only when the voltages you need to analyze the individual elements get "swallowed up" in the lumping together process. This sometimes happens with Thevenin equivalents for example, but it's not an issue in this case.