# Homework Help: Kirchhoff's Rule

1. Feb 14, 2013

### StarMan1234

1. The problem statement, all variables and given/known data

In the figure R1 = 2.22 Ω, R2 = 5.08 Ω, and the battery is ideal. What value of R3 maximizes the dissipation rate in resistance 3?

R2 and R3 are wired in parallel to each other. R1 is in series with the combination of R2 and R3. Emf is not given.

2. Relevant equations
E-I1R1-I2R2=0 (first loop)
E-I1R1-I3R3=0 (second loop)
1/R= sum of 1/R for parallel resistors
R= R1+... RN for series resistors
P=IV=I^2*R=V^2/R

3. The attempt at a solution
I began by solving the system of equations for the loops. So, I2R2=I3R3. I know the sum of the currents for resistors 2 and 3 is equal to the current through resistor 1. I know I need to take the derivative of an expression for power and set it equal to zero to maximize the dissipation rate. I am having a lot of trouble formulating the power expression however. I understand voltage is equivalent in parallel resistors and current is equivalent in parallel resistors, but I am not sure which is the proper way to form the power function.

2. Feb 14, 2013

### voko

Find the resistance of the entire group (it will be a function of R3), let's call it R(R3). Then the current through the entire group is I = V/R(R3).

The current J through R3 is I minus the current through R1R2.

The dissipated power is J2R3.

3. Feb 15, 2013

### ehild

You can do what voko suggested, writing up the resultant resistance, but the problem can be solved also by Kirchhoff's laws.
You can assume any value for the emf. Let it be E=10 V. You have shown that the voltage across both R3 and R2 is the same, call it U. How can you write up I2 and I3 with U and the resistances?

You have shown also the equation for the first loop:E-I1R1-I2R2=0 . With U, it is E-I1R1-U=0. Isolate I1.

All currents expressed with U, substitute them into the equation for the currents I1=I2+I3, and isolate U as function of R3.

The power dissipated on R3 can be written as P=U2/R3. It is function of R3. Minimise.

ehild

Last edited: Feb 15, 2013
4. Feb 15, 2013

### jaumzaum

5. Feb 15, 2013

### StarMan1234

Hey ehild,

I isolated I1.
I1= (E-U)/R1

I2=U/R2
I3=U/R3
I1=(U/R2)+(U/R3)
I don't know how solve this equation for U as a function of R3. I'm really stuck.

But if you can help me to do that, then I think I can use that equation by plugging it into P=U^2/R3, taking the derivative and setting it equal to zero. Can anyone help with this?

6. Feb 16, 2013

### jaumzaum

Let A and B be the terminals of R3. Take out R3. Take out battery. What is the equivalent resistance? That's the resistance that maximizes dissipation in R3. Simple like that

R2//R1 = 5.08*2.22/7.3 = 1.54 Ω

7. Feb 16, 2013

### ehild

Plug in (E-U)/R1 for I1: (E-U)/R1=(U/R2)+(U/R3). E is just a constant. If it is easier for you, add it a numerical value. Isolate U.

ehild

8. Feb 16, 2013

### ehild

jaumzaum: The rule of this Forum is that we do not give out full solution. Using Thevenin equivalent is the easiest method, but it is not sure that the OP learnt about it, or he knows that the optimum load is equal to the internal resistance of the source.

ehild

9. Feb 16, 2013

### StarMan1234

I have U= (E/R1)/((1/R1)+(1/R2)+(1/R3))

But what I don't understand is how to solve it in terms of R3 like you said. This equation just looks like a mess.

10. Feb 16, 2013

### ehild

Plug in the numerical data and call 1/R3=x. Find the derivative of xU2

ehild

11. Feb 16, 2013

### jaumzaum

Sorry.
I just found useful the OP compared both solutions to see how the theorem can simplify a lot the exercise itself. This problem is asking to be solved by thevenín :)

12. Feb 16, 2013

### ehild

Well, it is useful, but the rules do not allow it. You can guide the OP.
I haven't seen the picture in the original post.

ehild

13. Feb 16, 2013

### micromass

It is not allowed to give solutions. Please don't do this ever again.