Maximizing dissipation rate in a simple circuit

In summary, the problem involves finding the value of R3 that maximizes the dissipation rate in resistance 3. This can be done by finding an equation relating i3 to R3 and using calculus to maximize the rate in terms of resistance. However, finding such an equation is proving to be a challenge. One possible approach is to find the total resistance using R1, R2, and R3, then use it to find the current I1. Next, the voltage drop across R2 and R3 can be found and related to power. Taking the derivative of the power function and setting it to zero can help find the maximum power and therefore, the value of R3 that maximizes the dissipation rate. However,
  • #1
eisbrecher
3
0
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:
 
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  • #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.
 
  • #3
that is, your second equation shows that R2 is || R3 ,
those are in series with R1.
 
  • #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:
 
  • #5
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
 
  • #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:

1. How can I increase the dissipation rate in a simple circuit?

The dissipation rate in a simple circuit can be increased by using components with higher resistance values, increasing the voltage or current in the circuit, or by adding resistors in parallel.

2. What factors affect the dissipation rate in a simple circuit?

The main factors that affect the dissipation rate in a simple circuit are the resistance of the components, the amount of current flowing through the circuit, and the voltage applied to the circuit.

3. How does increasing the dissipation rate affect a simple circuit?

Increasing the dissipation rate in a simple circuit can result in higher temperatures and potentially damage the components if they are not able to handle the increased heat. It can also affect the overall performance of the circuit by causing fluctuations in voltage and current.

4. Is it always beneficial to maximize the dissipation rate in a simple circuit?

No, it is not always beneficial to maximize the dissipation rate in a simple circuit. While increasing the dissipation rate can improve the performance of the circuit, it can also lead to overheating and damage if not done carefully. It is important to consider the limitations of the components and the overall goal of the circuit before attempting to maximize the dissipation rate.

5. Are there any safety concerns when trying to maximize the dissipation rate in a simple circuit?

Yes, there are safety concerns when trying to maximize the dissipation rate in a simple circuit. Increasing the dissipation rate can result in higher temperatures and potentially cause burns or damage to the components. It is important to use caution and follow proper safety measures when working with circuits and electricity.

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