Resistor in a Circuit with varying resistance.

[Jonnyto]

Homework Statement

In the circuit of Fig. 23, ℰ, R₁, and R₂ have constant values but R can be varied. Find an expression for R that results in the maximum heating in that resistor.

Homework Equations

V=IR
resistors in parallel add up in inverse,
resistors in series add up directly,

The Attempt at a Solution

Okay so the way I undertook this problem was considering as R₂ and R to be parallel, so I found the equivalent capacitance of those two. I got that new resistance and and then I considered R₁ and then the new parallel system of parallels to be resistors in series. Later I used Kirchoff's Loop rule to isolate the new capacitance. ε-iR_eq=0
I used basic algebra to separate the R from all other values. Was this correct? I left the current, i, in the expression. The current is not actually part of the given.

 

[Sunil Simha]

I used basic algebra to separate the R from all other values. Was this correct? I left the current, i, in the expression. The current is not actually part of the given.

You have obtained current through the required resistor as a function of R right? Plug that in the equation of the power dissipated by the resistor to obtain the power P as a function of R.

Now have you used calculus to find the minima or maxima of a function before?

 

[Jonnyto]

Okay how would I solve it using power? Do I set up the power as a differential dw/dt?
However an issue is that i(the current is not isolated) and it is mixed within all the other values. I can only isolate either the varying resistance or the current or at least I think so.

 

[gneill]

Homework Statement

In the circuit of Fig. 23, ℰ, R₁, and R₂ have constant values but R can be varied. Find an expression for R that results in the maximum heating in that resistor.

Homework Equations

V=IR
resistors in parallel add up in inverse,
resistors in series add up directly,

The Attempt at a Solution

Okay so the way I undertook this problem was considering as R₂ and R to be parallel, so I found the equivalent capacitance of those two. I got that new resistance and and then I considered R₁ and then the new parallel system of parallels to be resistors in series. Later I used Kirchoff's Loop rule to isolate the new capacitance. ε-iR_eq=0
I used basic algebra to separate the R from all other values. Was this correct? I left the current, i, in the expression. The current is not actually part of the given.

Um, there' are no capacitors in the given circuit, so surely you're not finding equivalent capacitance. Equivalent resistance perhaps, but not capacitance.

Have you found an expression for the current through resistor R? If so, what did you find?

 

[Sunil Simha]

Okay how would I solve it using power? Do I set up the power as a differential dw/dt?
However an issue is that i(the current is not isolated) and it is mixed within all the other values. I can only isolate either the varying resistance or the current or at least I think so.

Don't write P as dw/dt. Just write power as i²R. In this expression write i in terms of R. You don't need the actual value of i to solve this. (remember, i is the current through the unknown resistor.)

I just want to clarify whether they have taught you to use calculus to find the maxima and minima of a function.