Inductance: given V waveform, find I

[courteous]

Homework Statement

An inductance of 3 mH has the following voltage waveform: for $$0<t<2\text{ }ms$$, $$v=15\text{ }V$$; for $$2ms<t<4ms$$, $$v=0$$; and for $$4\text{ }ms<t<6\text{ }ms$$, $$v=-30\text{ }V$$. Assuming $$i(0)=0$$, find the current at time (a) 1 ms, (b) 4 ms, (c) 5 ms.

Ans. (a) 5 A; (b) 10 A; (c) 0

Homework Equations

$$u(t)=L\frac{di(t)}{dt} \Rightarrow i(t)=\frac{1}{L}\int v(t)dt$$

The Attempt at a Solution

Even for (a) case (for which I somehow get the "correct" solution ), I'm not sure about the physical background: how do I use integral in $$i(t)$$ equation if I have to find currents at particular instances for t = {1, 4 and 5}?!

[PLAIN]http://img848.imageshack.us/img848/9894/dsc00990k.jpg

(a) $$i(t=1)=\frac{15}{L}\int dt=5t=5\text{ }A$$

(b) What is $$v(t=4)$$ ... 0 V or -30 V?

(c) As in (a) case, $$i(t=5)=\frac{-30}{L}\int dt=-10t=-50\text{ }A \neq 0\text{ }A$$

 

[chaoseverlasting]

As the voltage is constant between those two intervals, (t=0, t=2), your integral will give you $$i(t)=\frac{1}{L}Vt + C$$.

Here the constant is the value of the current at time t=0. It is given that i(0)=C=0.

As L=3mH, V=15v and t=1ms, substituting these values you get your answer.

Second part or your answer requires the current at t=4ms.

Before we get to t=4, what was the state at t=2?

Using the above realation, the current at t=2ms was 10A. This was your initial current, and thus C=10A in this case (I'm referring to the integration constant).

Using the same equation, as V=0, i(t=2,t=4)=C=10 A (it is constant between t=2ms and t=4ms).

Similarly, can you work out the last part?

 

[courteous]

So the last part can be isolated? So that the initial current at t=4 is 10 A ... and t=4 can be thought of as t=0? Then t=5 is equivalent to t=1:
$$i(t=5)=\frac{1}{3}(-30)(1)+10=0\text{ }A$$

I also found it easier to reverse $$i(t)$$ into $$v(t)$$ and look at derivatives of current.

[PLAIN]http://img683.imageshack.us/img683/341/dsc00991ep.jpg

Thank you! I really got some intuition behind it (if ever so slightly).

 

[chaoseverlasting]

So the last part can be isolated? So that the initial current at t=4 is 10 A ... and t=4 can be thought of as t=0? Then t=5 is equivalent to t=1:
$$i(t=5)=\frac{1}{3}(-30)(1)+10=0\text{ }A$$

No! What that means is that if we started analyzing the circuit at t=4, the value of current that we would have to deal with is 10A. So, the 'initial condition' at t=4 is i(t=4)=10A

Mathematically, the substitution that you've used will work. What it really means is that at t=5, the voltage waveform has been applied for 1 second. In that sense you could say that t=5 implies that the voltage has been applied for t=1 seconds.

I also found it easier to reverse $$i(t)$$ into $$v(t)$$ and look at derivatives of current.
.

It's the same principle that you have to apply in either case. Use whatever you find easier to apply, but understand both.