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Maximizing dissipation rate in a simple circuit

  1. Aug 31, 2006 #1
    QUESTION:
    R1 = 2 ohm
    R2 = 5 ohm
    What value of R3 maximizes the dissipation rate in resistance 3? (battery is ideal -> no internal resistance)

    DIAGRAM:
    _______________
    |.........|...........|
    R1.......R2.........R3
    |.........|...........|
    _ +......|...........|
    - -.......|...........|
    |.........|...........|
    _______________

    (I hope the diagram is clear... please ignore the dots... they are just there as placeholders... below R1 is the battery... the + side is on the top, the - side is on the bottom)

    MY TAKE ON THE PROBLEM:

    We know that the dissipation rate for a resistor is P = i^2*R.
    In order to maximize the dissipation rate, I figured the best way would be to find an equation relating i3 to R3 (and then subbing it in for i3 in order to get only R3s on the right side)... and then simply using calculus to maximize the rate in terms of resistance.

    But therein lies the problem... could not come up with such an equation... I tried using Kirchhoff's loop rule to get these:

    -i3R3 + E - i1R1 = 0 (clockwise traversal through the BIG loop)
    -i3R3 + i2R2 = 0 (clockwise traversal through the RIGHT loop)
    -i2R2 + E - i1R1 = 0 (clockwise traversal throught the LEFT loop)

    needless to say, I had assumed that i1 (going through R1) goes upwards, and that r2 and r3 (through R2 and R3, respectively) are pointing downwards.
    E is the emf of the battery

    Also, I know that i1 = i2 + i3 (junction rule)

    the PROBLEM:
    As I have mentioned, I am unable to find such an equation...

    I need some pointers... if you could PLEASE tell me whether at least I am moving in the right direction...

    I am all out of ideas...

    Thank you :smile:
     
  2. jcsd
  3. Aug 31, 2006 #2
    find total resistance, then find the current I1, through R1.
    then find the voltage drop across R2 and R3
    voltage drop=V - R1*I1
    (I1 should be a function of R3)
    then relate that to power.
    Now you should have Power as a function of R3 and V, take the derivative... and it should be good.
     
  4. Aug 31, 2006 #3

    lightgrav

    User Avatar
    Homework Helper

    that is, your second equation shows that R2 is || R3 ,
    those are in series with R1.
     
  5. Aug 31, 2006 #4
    @ tim_lou

    How can I find i1? The voltage of the battery is unknown... also, the equivalent resistance will contain one unknown (term R3)... also, i1 would, in itself, be unknown.

    There are 3 unknowns and 1 equation... I do not see how your way is possible :confused:

    @lightgrav

    Yes... R2 and R3 are in parallel... and R1 and the equivalent of R2 and R3 are in series... but what next? :frown:
     
  6. Aug 31, 2006 #5

    lightgrav

    User Avatar
    Homework Helper

    Pretend you know the battery Voltage ...
    after all, it will not change as R3 varies

    R3 MUST be a variable, so you can take a deivitive w.r.to it!

    R_2||3 = R2 R3 / (R2+R3) ... R_tot = R1 + R_2||3 ... I1 = V/R_tot
     
  7. Sep 1, 2006 #6
    R1 = 2 ohm
    R2 = 5 ohm
    Voltage does not change with respect to R3...(it's a constant)
    I1 will be in terms of R3
    [tex]I_1=\frac{V}{R_{total}}[/tex]
    Find R_total using R1, R2 and R3

    V will be gone when you take the derivative and set the derivative to zero (finding maximum), R1 and R2 are numbers.

    Power is maxed when the derivative of the Power function is zero (well, check the concavity and sign change to confirm). then you can find R3.

    try to work it out... if you need more hints, just ask
     
    Last edited: Sep 1, 2006
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