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.

 

[DeeNos]

To find the current i(t) in an inductor, we can use the equation u(t)=L\frac{di(t)}{dt}. This equation relates the voltage across an inductor to the rate of change of the current through it. In this problem, we are given the voltage waveform v(t) and we need to find the current at different time instances.

(a) At t=1 ms, the voltage v(t) is 15 V. Using the equation i(t)=\frac{1}{L}\int v(t)dt, we can integrate the voltage waveform from t=0 to t=1 ms to find the current i(t) at t=1 ms. This gives us i(t=1)=\frac{1}{L}\int_0^1 15 dt=5 A.

(b) At t=4 ms, the voltage v(t) is 0 V. This means that there is no change in the voltage across the inductor at this time. Therefore, the current through the inductor will remain constant and will be equal to the current at t=2 ms. This gives us i(t=4)=10 A.

(c) At t=5 ms, the voltage v(t) is -30 V. Using the same equation as in (a), we can integrate the voltage waveform from t=0 to t=5 ms to find the current i(t) at t=5 ms. This gives us i(t=5)=\frac{1}{L}\int_0^5 (-30) dt=0 A. This means that the current through the inductor has returned to its initial value of 0 A. This is because the voltage waveform repeats itself after every 6 ms, and at t=5 ms, we are at the start of a new cycle.

Overall, we can use the equation i(t)=\frac{1}{L}\int v(t)dt to find the current at any time instance as long as we have the voltage waveform and the initial current value. The physical explanation for this is that the inductor resists changes in current, and the voltage across it is directly proportional to the rate of change of current through it.