Finding Maximum Current Entering A Terminal

  • Thread starter Thread starter madhatter500
  • Start date Start date
  • Tags Tags
    Current Maximum
AI Thread Summary
To find the maximum current entering a terminal, the charge expression q = 1/α² - (t/α + 1/α²)e^(-tα) is differentiated with respect to time, resulting in the current equation. The first derivative is te^(-tα), and to find the maximum, the second derivative is set to zero, leading to the equation e^(-tα)(1 - αt) = 0. This indicates that the maximum occurs at t = 1/α. The discussion highlights the importance of careful differentiation and algebra in solving for maximum current values. The final answer confirms the correct approach to the problem.
madhatter500
Messages
2
Reaction score
0

Homework Statement



The expression for the charge entering an upper terminal of a component is:

q = 1/\alpha^{2} - (t/\alpha + 1/\alpha^{2})e^{-t\alpha}

Find the maximum value of the current entering the terminal if \alpha = 0.03679 s^{-1}

Homework Equations






The Attempt at a Solution



Since they gave us the equation for the charge, to find current we simply take the derivative with respect to time of the given equation. I did this and ended up with te^{-t\alpha}. This seems to check out just fine. I then realized that they want the maximum value that this derivative can be. So then I went ahead and took the second derivative and tried to set it equal to 0 (since that would be a local max or min). This ended up being very messy and it didn't work out. Is there an easier way to find the maximum current?
 
Physics news on Phys.org
Let me know if my mathematics is incorrect, but
\frac{dq^{2}}{d^{2}t}=e^{-t\alpha}-\alpha te^{-t\alpha}=e^{-t\alpha}(1-\alpha t)=0
e^{-t\alpha}\neq0, therefore 1-\alpha t=0

and t=\frac{1}{\alpha}

You can go from there

edit: this is going from your 1st derivative, I haven't checked it myself
 
madhatter500 said:

Homework Statement



The expression for the charge entering an upper terminal of a component is:

q = 1/\alpha^{2} - (t/\alpha + 1/\alpha^{2})e^{-t\alpha}

Find the maximum value of the current entering the terminal if \alpha = 0.03679 s^{-1}

Homework Equations






The Attempt at a Solution



Since they gave us the equation for the charge, to find current we simply take the derivative with respect to time of the given equation. I did this and ended up with te^{-t\alpha}. This seems to check out just fine. I then realized that they want the maximum value that this derivative can be. So then I went ahead and took the second derivative and tried to set it equal to 0 (since that would be a local max or min). This ended up being very messy and it didn't work out. Is there an easier way to find the maximum current?

Not too comfortable with your first derivative. t appears in two places in the original expression. I think you may have overlooked one??
 
MetalManuel said:
Let me know if my mathematics is incorrect, but
\frac{dq^{2}}{d^{2}t}=e^{-t\alpha}-\alpha te^{-t\alpha}=e^{-t\alpha}(1-\alpha t)=0
e^{-t\alpha}\neq0, therefore 1-\alpha t=0

and t=\frac{1}{\alpha}

You can go from there

edit: this is going from your 1st derivative, I haven't checked it myself

Perfect. This gives me the correct answer. Looks like I need to brush up on my algebra! Thank you!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top