How can I maximize the power consumed in this resistor?

Question

^{simulate this circuit – Schematic created using CircuitLab}

I have gotten really stuck on this question. The question is:

1. Determine R3 so that maximum power is consumed in R3.
2. What is that maximum power? Answer in microwatts.

I am not sure where to begin. I have begun my naming the R3 as R(Load) and calculating I(Short) from Node A to Node B by just putting a wire there instead of R3, but I am not able to get the voltages in the nodes. I am really confused and hoping for some guidance.

Answer 1

This is a standard network reduction task: note that you get maximum power delivery to a component by matching the internal resistance of a voltage/current source feeding it.

So you go about reforming the network. Replace I(B) paralleled with R2 with a 9V voltage source in series with R2, then add R4 into it. Essentially, you now have two voltage sources connected in parallel to R3, one with an internal resistance of 4kohm, one with 2kohm, making for a total of 1.33kohm, so that is what R3 needs to be in order to maximize power output into it.

It's left as an exercise to the reader to calculate the actual amount of power, but you have to be aware that the current source and the voltage source have opposing polarity, so the resulting voltage/power will not be all that impressive (at least it isn't zero in which case the value of R3 would be actually irrelevant).

Answer 2

It's always a bit messy trying to work with both current and voltage sources so convert I(B) and R2 to an equivalent voltage source V(B) and redraw: -

Then join R2 and R4 together to make 4 kΩ and place on the other limb of V(B).

Now you have two voltage sources commoned to 0 volts and each has a series single resistor. From here you can go several ways but pretty much all ways include source transformation (that's what I did with I(B) making it V(B) btw.

Is this enough of a hint to figure out the rest yourself? Do you understand what the R3 resistance needs to be to get maximum power transfer?

Answer 3

First, I will present a method that uses Mathematica to solve this problem.

Well, we are trying to analyze the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$\begin{cases} \begin{alignat*}{1} \text{I}_\text{a}&=\text{I}_0+\text{I}_4\\ \\ \text{I}_0&=\text{I}_2+\text{I}_3\\ \\ \text{I}_1&=\text{I}_2+\text{I}_4\\ \\ \text{I}_\text{a}&=\text{I}_1+\text{I}_3 \end{alignat*} \end{cases}\tag1$$

When we use and apply Ohm's law, we can write the following set of equations:

$$\begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_1-0}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_2-\text{V}_1}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}_2-0}{\displaystyle\text{R}_3}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}_2-\text{V}_3}{\displaystyle\text{R}_4} \end{alignat*} \end{cases}\tag2$$

We also know that \$\displaystyle\text{V}_1-\text{V}_3=\text{V}_\text{i}\$.

Now, we can solve for \$\displaystyle\text{V}_2-\text{V}_1\$:

$$\text{V}_2-\text{V}_1=\frac{\displaystyle\text{R}_2\left(\text{I}_\text{a}\text{R}_3\text{R}_4-\text{V}_\text{i}\left(\text{R}_1+\text{R}_3\right)\right)}{\displaystyle\text{R}_2\left(\text{R}_1+\text{R}_3\right)+\text{R}_4\left(\text{R}_1+\text{R}_2+\text{R}_3\right)}\tag3$$

Where I used the following Mathematica code to find \$(3)\$:

In[1]:=Clear["Global`*"];
solution = 
  FullSimplify[
   Solve[{Ia == I0 + I4, I0 == I2 + I3, I1 == I2 + I4, Ia == I1 + I3, 
     I1 == (V1 - 0)/R1, I2 == (V2 - V1)/R2, I3 == (V2 - 0)/R3, 
     I4 == (V2 - V3)/R4, V1 - V3 == Vi}, {I0, I1, I2, I3, I4, V1, V2, 
     V3}]];
FullSimplify[(V2 /. solution[[1]]) - (V1 /. solution[[1]])]

Out[1]=(Ia R2 R3 R4 - R2 (R1 + R3) Vi)/(R2 (R1 + R3) + (R1 + R2 + R3) R4)

So, the power in the resistor \$\text{R}_2\$ is given by:

$$\text{P}_{\text{R}_2}=\frac{\displaystyle\left(\text{V}_2-\text{V}_1\right)^2}{\displaystyle\text{R}_2}=\text{R}_2\left(\frac{\displaystyle\text{I}_\text{a}\text{R}_3\text{R}_4-\text{V}_\text{i}\left(\text{R}_1+\text{R}_3\right)}{\displaystyle\text{R}_2\left(\text{R}_1+\text{R}_3\right)+\text{R}_4\left(\text{R}_1+\text{R}_2+\text{R}_3\right)}\right)^2\tag4$$

In order to find the maximum, we can use:

$$\frac{\displaystyle\partial\text{P}_{\text{R}_2}}{\displaystyle\partial\text{R}_2}=0\space\Longleftrightarrow\space\text{R}_2=\frac{\displaystyle\text{R}_4\left(\text{R}_1+\text{R}_3\right)}{\displaystyle\text{R}_1+\text{R}_3+\text{R}_4}\tag5$$

Where I used the following Mathematica code to find \$(5)\$:

In[2]:=FullSimplify[
 Solve[{D[(
     R2 (Ia R3 R4 - (R1 + R3) Vi)^2)/(R2 (R1 + R3) + (R1 + R2 + 
          R3) R4)^2, R2] == 0}, R2]]

Out[2]={{R2 -> ((R1 + R3) R4)/(R1 + R3 + R4)}}

Using your values, we get:

$$\text{R}_2=\frac{\displaystyle2000\left(3000+1000\right)}{\displaystyle3000+1000+2000}=\frac{\displaystyle4000}{\displaystyle3}\approx1333.33\space\Omega\tag6$$

And the power is:

$$\text{P}_{\text{R}_2}=\frac{\displaystyle3}{\displaystyle16000}\approx0.0001875\space\text{W}\tag7$$