Tricky circuit with variable resistor

[zeralda21]

Homework Statement

An electric circuit comprises two series resistors A and B, the first of which has a given resistance R_A and the second variable resistance R_B, over which a constant DC voltage U is added. How shall the resistance R_B be selected so that the power P developed in resistor B will be at its maximum?

All components may be assumed to be idealized.

Homework Equations

Ohms law $$U=RI$$ and
$$P=UI=RI^2$$

The Attempt at a Solution

My strategy is to calculate the current in the circuit and consequently the power in B with $$P_B=R_B*I^2$$ which shall be maximized.

The total resistance in the circuit is $$R=R_A+R_B$$ and the current is therefore $$I = U/(R_A+R_B)$$ and the power is hence: $$P_B=R_B*I^2=\frac{R_{B}U^2}{(R_A+R_B)^2}$$. How can I proceed and maximize this ?

 

[barryj]

Your power equation is correct. Solving max problems like this is normally a calculus problem. Without calculus, I suggest that you try assigning a value to Ra and do a plot of power on you calculator and find the max.

 

[gneill]

How do you go about finding the maximums and minimums of any function?

 

[barryj]

One of the first things you learn in calculus is how to take the derivative of a function. The derivative of a function is another function that gives the slope of the original function at any point. So, if one sets the derivative to zero and solve for the variable, you can find the minimums and maximums of the function. Consider a function y = x^2 - 4x + 5. The derivative is 2x-4. Setting 2x-4 = 0 gives x = 2 and that is where the minimum is. Calculus is cool!

 

[zeralda21]

Alright! Yes I know about derivatives and just solved this problem by introducing a dimensionless quantity $$x=\frac{R_A}{R_B}$$ and maximizing the function $$f(x)=\frac{x}{(1+x)^2}$$