Determining the optimal resistance of a variable resistor

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To determine the optimal resistance of a variable resistor, the relationship between current (I), standard resistor resistance (R), and variable resistor resistance (R_v) is crucial. The power dissipated by the variable resistor is given by P = I^2 R_v, where I is dependent on both R_v and R. To maximize power P, it is necessary to express it in terms of a single variable by substituting I into the power equation. This approach allows for differentiation to find critical points and identify the optimal resistance for R_v. Ultimately, the goal is to maximize power dissipation while managing the interdependence of I and R_v.
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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