Determining the optimal resistance of a variable resistor

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SUMMARY

The discussion focuses on maximizing the power dissipated by a variable resistor, denoted as ##P = I^2 R_v##, where ##I = \frac{V}{(R_v + R)}##. The challenge arises from the interdependence of current (##I##) and variable resistance (##R_v##), complicating the differentiation process to find critical points. The solution involves expressing power (##P##) solely in terms of one variable, allowing for the application of calculus to determine the optimal resistance value for the variable resistor.

PREREQUISITES
  • Understanding of Ohm's Law and its application in electrical circuits
  • Familiarity with the concepts of power dissipation in resistors
  • Basic knowledge of calculus, specifically differentiation
  • Ability to manipulate algebraic equations to express variables in terms of others
NEXT STEPS
  • Learn how to express power in terms of a single variable using substitution techniques
  • Study the principles of optimization in calculus, focusing on finding critical points
  • Explore the impact of varying resistance on circuit performance using simulation tools
  • Investigate the applications of variable resistors in real-world electronic devices
USEFUL FOR

Electrical engineers, physics students, and hobbyists interested in optimizing circuit designs and understanding power dissipation in variable resistors.

xAly
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Homework Statement
A simple circuit consists of a supply with a voltage of ##60V## as well as a variable resistor (which can be varied between ##1## and ##25 \Omega##) and resistor (##5\Omega##) connected in series. Determine the resistance of the variable resistor such that the power dissipated by it is maximised.
Relevant Equations
##P = I^2 R, R = \frac{V}{I}##
Let ##R## denote resistance of standard resistor and ##R_v## the resistance of the variable resistor. I know that ##I = \frac{V}{(R_v + R)}##. Now I also know that ##P = I^2 R_v##represents the power dissipated by the variable resistor and that I need to maximise ##P##. The problem I am having is that both and ##I## and ##R_v## are dependent on each other, a decrease in ##R_v## leads to an increase in ##I## and vice versa. Therefore I don't see how to maximise ##P##, usually I would differentiate to find the critical points but I am not sure which one is the independent variable here.
 
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xAly said:
##I = \frac{V}{(R_v + R)}\,\,\,\,\,## ##P = I^2 R_v##
Try to use your equation for ##I## to express ##P## in terms of a single variable.
 

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