Inductor Voltage Maximum and Maximum Voltage Calculation

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Discussion Overview

The discussion centers around a homework problem involving the calculation of the maximum voltage across a 25mH inductor given a specific current function. Participants explore the differentiation of the current function to find the time at which the voltage is maximized and the corresponding maximum voltage value.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states the need to differentiate the current function and set the derivative to zero to find the time of maximum voltage.
  • Another participant questions the correctness of the calculated time of 0.01339 s, suggesting it does not appear to be a zero of the function.
  • A participant acknowledges the error in their previous calculation and attempts to solve the equation 2cos400t - sin400t = 0, leading to an angle calculation.
  • There is a realization that the angle calculated does not directly correspond to time, prompting a correction in the variable used.
  • A final correction is made, converting the angle back to time, resulting in a new time of approximately 0.0027679 s.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the maximum voltage calculation, as the discussion focuses primarily on the time calculation and the correctness of derivatives and algebraic manipulations. Multiple competing views on the correct approach and calculations are present.

Contextual Notes

There are unresolved aspects regarding the correctness of the derivative and the algebraic steps taken, as well as the dependence on the correct interpretation of the angle in relation to time.

Zarathuztra
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Homework Statement


The current in a 25mH inductor is known to be - 10A for t <= 0 and (-10cos400t - 5sin400t)e^-200t A for t >= 0. Assume the passive sign convention.

Part A: At what instant of time is the voltage across the inductor maximum?
Part B: What is the maximum voltage?

Homework Equations


V=L*di/dt


The Attempt at a Solution



I took the derivative of (-10cos400t - 5sin400t)e^-200t, then took the derivative again, set to zero, and found the value of t at that point.

5000*e^(-200t)*sin(400t)

1000000*e^(-200t)(2cos400t - sin400t) = 0 @ t = 0.01339 s

Where am I going wrong? Am I missing something?
 
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t = 0.01339 s doesn't look like a zero of that function to me...
 
your right, that's way off... looking at it again.

I need 2cos400t - sin400t to equal 0...

2cosx = sinx
2 = tanx
arctan(2) = x

x = 1.107s

this didn't work.. is my derivative correct? I'm pretty sure my algebra and trig were correct.
 
Last edited:
Zarathuztra said:
your right, that's way off... looking at it again.

I need 2cos400t - sin400t to equal 0...

2cosx = sinx
2 = tanx
arctan(2) = x

x = 1.107s

this didn't work.. is my derivative correct? I'm pretty sure my algebra and trig were correct.

You've solved for the angle, x. But the angle is not t...
 
hahaha, what a silly mistake, I used a variable in place of 400t and didn't sub the variable out... x/400 = t so, 1.107/400 = t = .0027679s
 
Yes, that looks much better :smile:
 

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