# maximum real power problem

Determine the value of resistance $$\ R\$$ so that true power of loads is maximum (biggest possible).

Picture of a circuit that came with task:

I solved it this way:

I know that $$\ P = I^2 * R_{real} \$$ and as $$\ I = \frac{U}{\sqrt{(R + 3)^2 + 4^2}} \$$ so $$\ P = \frac{U^2*(R + 3)}{(R + 3)^2 + 4^2}\$$, when I derived that, I get $$\ (3+R)^2 + 4 - 2(3+R)^2 = 0 => (3 + R)^2 - 4 = 0 => 9 + 6R + R^2 - 4 = 0 \$$ which gives me $$\ R_1 = -5, R_2 = -1\$$. The answer that I should've got is R = 5, I don't know what I've did wrong.

• Not "derive", "differentiate". I get R = +1 for maximum (total) power in 3Ω and R. – Spehro Pefhany Jul 29 at 19:49

If the objective is to maximize the power dissipated on the load $$\R\$$. Use,

$$\P_R = \frac{U^2 \cdot R}{(R + 3)^2 + 4^2} = \frac{U^2 \cdot R}{ R^2+6R+9+16} = \frac{U^2}{ R+6+25/R}\$$

Then, differentiate the expression to find its maximum, that is, when $$\R+6+25/R\$$ reaches its minimum.

If you are actually looking for maximizing the power dissipated on both resistors, $$\R\$$ and $$\ 3 \Omega \$$, use,

$$\P_{total} = \frac{U^2 \cdot (R+3)}{(R + 3)^2 + 4^2} = \frac{U^2}{\frac{(R + 3)^2 + 16}{R+3}} = \frac{U^2}{R + 3 + \frac{16}{R+3}}\$$

Then, find when $$\R + 3 + \frac{16}{R+3}\$$ is at its smallest value.

• Just as a note to the OP: Take the derivative and set that equal to zero, in order to find a maximum or minimum. (It's your expectation that you'll find a maximum, of course.) – jonk Jul 29 at 19:34

Well, first of all we can write the input impedance as follows:

$$\underline{\text{Z}}_{\space\text{in}}=3+\text{R}+4\text{j}\tag1$$

The input current is given by:

$$\underline{\text{I}}_{\space\text{in}}=\frac{\hat{\text{V}}\cdot\exp\left(\phi\cdot\text{j}\right)}{3+\text{R}+4\text{j}}\tag2$$

The real power of the load is given by:

$$\text{P}_\text{load}=\text{P}_\text{R}=\text{V}_{\text{R/rms}}\cdot\text{I}_{\text{R/rms}}\cdot\cos\left(\varphi_\text{R}\right)=\text{R}\cdot\text{I}_{\text{R/rms}}^2\cdot\cos\left(\varphi_\text{R}\right)=$$ $$\text{R}\cdot\left(\frac{1}{\sqrt{2}}\cdot\left|\frac{\hat{\text{V}}\cdot\exp\left(\phi\cdot\text{j}\right)}{3+\text{R}+4\text{j}}\right|\right)^2\cdot1=\frac{\text{R}}{2}\cdot\left(\frac{\hat{\text{V}}}{\sqrt{\left(3+\text{R}\right)^2+4^2}}\right)^2=$$ $$\frac{\text{R}}{2}\cdot\frac{\hat{\text{V}}^2}{\left(3+\text{R}\right)^2+4^2}=\frac{\hat{\text{V}}^2}{2}\cdot\frac{\text{R}}{\left(3+\text{R}\right)^2+16}\tag3$$

Now, solve:

$$\frac{\partial\text{P}_\text{load}}{\partial\text{R}}=0\space\Longrightarrow\space\text{R}=5\tag4$$

And the real power of the load is there equal to:

$$\hat{\text{P}}_\text{load}=\frac{\hat{\text{V}}^2}{2}\cdot\frac{5}{\left(3+5\right)^2+16}=\frac{\hat{\text{V}}^2}{32}\tag5$$