Electrical Circuit: Finding Max/Min Voltage in 5 min

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SUMMARY

The discussion focuses on determining the maximum and minimum voltage in an electrical circuit described by the equation V = 12 - 9t + 2t² over a 5-minute interval. The minimum voltage is calculated to be 1.875 V occurring at 2.25 minutes, while the maximum voltage is 17 V at 5 minutes. Participants emphasize the importance of understanding the vertex of the parabola and the derivative to find these critical points accurately.

PREREQUISITES
  • Understanding of quadratic functions and their properties
  • Knowledge of calculus, specifically derivatives
  • Familiarity with graphing parabolas
  • Basic electrical circuit concepts
NEXT STEPS
  • Study how to find the vertex of a quadratic function
  • Learn about calculating derivatives in calculus
  • Explore applications of parabolas in electrical engineering
  • Research optimization techniques in mathematical modeling
USEFUL FOR

Students and professionals in electrical engineering, mathematics educators, and anyone interested in optimizing voltage calculations in electrical circuits.

Koyuki
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The problem is:
"In an electrical circuit, the voltage, V volts, as a function of time, t minutes, is given by V = 12 - 9t + 2t^2. Determine the greatest and least values of voltage during the first 5 min. When do these values occur?"

The answers given are exact.
Least: 1.875 V @ 2.25 min
Most: 17 V @ 5 min

Well I know how to get the max, but the min?

Thanks!
 
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Koyuki said:
...
Well I know how to get the max, but the min?
...
The graph is a parabola. You can find the vertex if you remember how, or you can find where the derivative is 0.
 

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