When Is the Voltage Across an Inductor Maximum?

[zealeth]

Homework Statement

The current in a 28mH inductor is known to be −10A for t≤0and (−10cos(400t)−5sin(400t))e^(−200t) A for t≥0. Assume the passive sign convention. At what instant of time is the voltage across the inductor maximum? What is the maximum voltage?

Homework Equations

v(t) = L*di/dt

The Attempt at a Solution

v(t) is at a max when di/dt is max since L is constant.

Taking the derivative of the equation for t≥0, I have:

di/dt = 5000e^(-200t)*sin(400t)

Finding the maximum using my calculator, I found t = 0.0028 ms.

Plugging this back into di/dt, di/dt = 2569.8

v(0.0028) = 0.028 * 2569.8 = 72.3 V.

Now both of these answers are correct, but my question is there a better way to find t than just plugging it into my graphing calculator and using the built in function to find the maximum? Is there a way this can be solved on a basic/scientific calculator? I know I could take the 2nd derivative of i(t) and set it equal to 0 but there is an infinite number of roots for t<0 and a good number of them for t>0. My professor never solved a problem similar to this in class so I'm not sure what his method is.

 

[BvU]

Your ${dI\over dt}$ looks a little strange. Differentiating $I = \left (A \cos(\omega t) + B \sin (\omega t) \right )e^{-\alpha t}$ is differentiating a product. How come you end up with only one sin term ?

Never mind, I can reproduce, sorry. Filling in numbers is useful sometimes...

So now differentiate again and rewrite as $A^\prime \sin(\omega t + \phi)\ \exp(-200t)$.

From the $\exp(-200t)$ that descends monotonically it is obvious the first maximum is the maximum.

 

[NascentOxygen]

I know I could take the 2nd derivative of i(t) and set it equal to 0 but there is an infinite number of roots for t<0 and a good number of them for t>0.

There is no easy out, that's what you'll have to do. As BvU pointed out, you are looking for the
local maximum that lies 0 <t< T/2 because as time goes on the oscillations get smaller (it's a decaying exponential you have there).

I checked your answer; and I agree with it.