Well the derivative is right. I even double checked it online. That being said thanks for your help. I really appreciated!
Final answer broken down is t = 0.0693147181 and PMAX = 6
Alright I get -240e^-10t+480e^-20t after i derive it but that does not give me the PMAX of the equation when the only t that is given is t=0 current starts to flow.
Homework Statement
After t=0, the current entering the positive terminal of a flashlight bulb is given by:
i(t) = 2(1-e^(-10t)) A
and the voltage across the bulb is v(t) = 12e^(-10t) V.
Determine the maximum power level delivered to the flashlight.
Homework Equations
i(t) = 2(1-e^(-10t)) A...