How to find Pmax when you know voltage and current?

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SUMMARY

The maximum power level (PMAX) delivered to the flashlight bulb is calculated using the current function i(t) = 2(1-e^(-10t)) A and the voltage function v(t) = 12e^(-10t) V. By deriving the power function p(t) = i(t) * v(t) and setting its derivative to zero, the maximum power is found at t = 0.0693147181 seconds, resulting in PMAX = 6 Watts. This analysis confirms the application of differential calculus in determining maximum power in electrical circuits.

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  • Knowledge of differential calculus for finding maxima
  • Familiarity with exponential functions and their derivatives
  • Basic principles of power calculation in electrical systems
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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
v(t) = 12e^(-10t) V

The Attempt at a Solution


I know p(t) = i(t)*v(t) but that is where my knowledge of this stops
 
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Hi Justin26. Can you use differential calculus?
 
Yes i can. Just unsure on where to begin
 
Start with your expression for power, and find its derivative...
 
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.
 
Better check that derivative.

You are looking for a maximum, so equate your derivative to 0.
 
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
 

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