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?
2. Relevant equations
v(t) = L*di/dt
3. 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.