Finding Maximum Current Entering A Terminal

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madhatter500
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Homework Statement



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

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

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

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[itex]^{-t\alpha}[/itex]. 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?
 
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Let me know if my mathematics is incorrect, but
[itex]\frac{dq^{2}}{d^{2}t}=e^{-t\alpha}-\alpha te^{-t\alpha}=e^{-t\alpha}(1-\alpha t)=0[/itex]
[itex]e^{-t\alpha}\neq0[/itex], therefore [itex]1-\alpha t=0[/itex]

and [itex]t=\frac{1}{\alpha}[/itex]

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/[itex]\alpha^{2}[/itex] - (t/[itex]\alpha[/itex] + 1/[itex]\alpha^{2}[/itex])e[itex]^{-t\alpha}[/itex]

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

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[itex]^{-t\alpha}[/itex]. 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
[itex]\frac{dq^{2}}{d^{2}t}=e^{-t\alpha}-\alpha te^{-t\alpha}=e^{-t\alpha}(1-\alpha t)=0[/itex]
[itex]e^{-t\alpha}\neq0[/itex], therefore [itex]1-\alpha t=0[/itex]

and [itex]t=\frac{1}{\alpha}[/itex]

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!