Optimizing a Circuit

If the current through $R_1$ is $I_1$ , and the current through $R_2$ is $I_2$ , then the current through $R_3$ must be $I_1+I_2$ , and we have $P_1 \; = \; I_1^2 \hspace{1cm} P_2 \;=\; 2I_2^2 \hspace{1cm} P_L \; = \; 3(I_1+I_2)^2$ Now $P_1 + P_2 \; = \; I_1^2 + 2I_2^2 \; = \; \tfrac13\big[(I_1-2I_2)^2 + 2(I_1+I_2)^2\big] \; \ge \; \tfrac23(I_1+I_2)^2 \; = \; \tfrac29P_L$ with equality when $I_1 =2I_2$ . This current balance can be achieved by choosing $V_1 = V_2$ . Thus $P_1 + P_2 + P_L \; \ge \; \tfrac{11}{9}P_L \; = \; \tfrac{110}{9} \; = \; \boxed{12.2}$

We've seen some nice formal solutions already (see Mark H and Chew-Seong). I will also present an "intuitive physical approach". This solution is not strictly legitimate (a priori), but it works, and it is borne out by the formal solutions.

1) Suppose we'll acheive maximum efficiency with a source that doesn't dissipate any power internally while unloaded.

2) Requirement (1) mandates that $V1 = V2$ , so as to have zero circulating current in the unloaded case.

3) If $V1 = V2$ , the equivalent source can be represented as a Thevenin equivalent consisting of a source $V$ in series with a resistance of $\frac{R_1 R_2}{R_1 + R_2}$ . Call this equivalent resistance $R_{TH}$ . $R_{TH} = \frac{2}{3}$ .

4) When the Thevenin equivalent is connected to the load, the load voltage $V_L = V \times \frac{R_L}{R_L + R_{TH}}$ (because of the voltage divider principle). This works out to $V_{L} = \frac{9}{11} V$ .

5) The load power can be expresed as $\frac{V_L^{2}}{R_L} = 10$ . Since $R_L = 3$ , $V_L = \sqrt{30} = \frac{9}{11} V$

6) $V = \frac{11 \sqrt{30}}{9}$

7) The total power then works out to $\frac{V^{2}}{R_{TH} + R_L} = \frac{121 \times 30}{81 \times (\frac{2}{3} + 3) } \approx 12.2$

Let the current through $R_1$ , $R_2$ and $R_L$ due to $V_1$ alone be $I_{11}$ , $I_{21}$ and $I_{L1}$ respectively and those due to $V_2$ be $I_{12}$ , $I_{22}$ and $I_{L2}$ . Then the resultant current $I_1 = I_{11} - I_{12}$ , $I_2 = I_{21} - I_{22}$ and $I_L = I_{L1} + I_{L2}$

And we have:

$\begin{aligned} I_{11} & = \frac {V_1}{R_1+R_2||R_L} = \frac {V_1}{1+2||3} = \frac {V_1}{1+\frac {2\times 3}{2+3}} = \frac {5V_1}{11} \end{aligned}$

By current division, $I_{21} = \dfrac 35 I_{11} = \dfrac {3V_1}{11}$ and $I_{L1} = \dfrac 25 I_{11} = \dfrac {2V_1}{11}$ .

Similarly, $I_{22} = \dfrac {V_2}{R_2+R_1||R_L} = \dfrac {V_2}{2+1||3} = \dfrac {V_2}{2+\frac {1\times 3}{1+3}} = \dfrac {4V_2}{11}$

By current division, $I_{12} = \dfrac 34 I_{22} = \dfrac {3V_2}{11}$ and $I_{L2} = \dfrac 14 I_{22} = \dfrac {V_2}{11}$ .

Now, we have:

$\begin{aligned} P_1+P_2+P_L & = I_1^2R_1 + I_2^2R_2 + I_R^2R_L \\ & = 1(I_{11} - I_{12})^2 + 2(I_{21} - I_{22})^2 + 3(I_{L1} + I_{L2})^2 \\ & = \left(\frac {5V_1}{11}-\frac {3V_2}{11}\right)^2 + 2\left(\frac {3V_1}{11}-\frac {4V_2}{11}\right)^2 + 3\left(\frac {2V_1}{11} + \frac {V_2}{11}\right)^2 \\ & = \frac {(25V_1^2-30V_1V_2+9V_2^2) + (18V_1^2-48V_1V_2+32V_2^2) + (12V_1^2+12V_1V_2+3V_2^2)}{121} \\ & = \frac {55V_1^2-66V_1V_2+44V_2^2}{121} \end{aligned}$

It is given that $P_L = \dfrac {12V_1^2+12V_1V_2+3V_2^2}{121} = 10$ watts and we need to find the minimum of $P_1+P_2+P_L$ we can use Lagrange multipliers as follows.

$\begin{aligned} F(V_1,V_2,\lambda) & = 55V_1^2-66V_1V_2+44V_2^2 - \lambda (12V_1^2+12V_1V_2+3V_2^2-1210) \\ \frac {\partial F}{\partial V_1} & = 110V_1 - 66V_2 - \lambda (24V_1 + 12V_2) \\ \frac {\partial F}{\partial V_2} & = - 66V_1 + 88V_2 - \lambda (12V_1+6V_2) \\ \frac {\partial F}{\partial \lambda} & = 12V_1^2+12V_1V_2+3V_2^2-1210 \end{aligned}$

\(\begin{array} {} \dfrac {\partial F}{\partial V_1} = 0 & \implies 55V_1 - 33V_2 = \lambda (12V_1+6V_2) &...(1) \\ \dfrac {\partial F}{\partial V_2} = 0 & \implies - 66V_1 + 88V_2 = \lambda (12V_1+6V_2) &...(2) \\ (1)-(2): & \implies 121V_1 - 121V_2 = 0 \\ & \implies V_1 = V_2 \\ \dfrac {\partial F}{\partial \lambda} = 0 & \implies 12V_1^2+12V_1^2+3V_1^2 = 1210 \\ & \implies 27V_1^2 = 1210 \end{array} \)

Therefore, the minimum $P_1+P_2+P_L$ is as below.

$\begin{aligned} P_1+P_2+P_L & = \frac {55V_1^2-66V_1V_2+44V_2^2}{121} = \frac {33 V_1^2}{121} = \frac {33\times 1210}{27 \times 121} \approx \boxed{12.2} \end{aligned}$

$\text{Let the respective currents voltages be } I_1,\ I_2,\ I_L\ \ and\ V_L.\\ I_L=\sqrt{\dfrac{10} 3}=I_1+I_2.\ \ \ \implies\ I_2=\sqrt{\dfrac{10} 3} - I_1.\\$ $\text{Power loss in internal resistance }=\ I_1^2*1+(\frac {10} 3 -2*\sqrt{\frac{10} 3}*I_1+I_1^2)*2.\\$ $Differentiating \ and\ equating\ to\ zero\ for\ minimum, 6I_1^2 - 4*\sqrt{\dfrac{10} 3}= 0.\\$ $\implies\ I_1=\dfrac 2 3 *\sqrt{\dfrac{10} 3}.\\ P_1+P_2+P_L=I_1^2*1 + I_2^2*2 + 10 \\ =\dfrac 4 9 *\dfrac{10}3 +\left (\sqrt{\dfrac{10} 3} - \dfrac 2 3 *\sqrt{\dfrac{10} 3} \right ) ^2 + 10=\ \dfrac{40}{27} + \dfrac{20}{27} + 10\ =\Large\ \ \ \color{#D61F06}{12.222}$

$\ \ OR\ \$ A better logical solution is as under.
$Let\ I_1,\ I_2,\ I_L,\ be\ respective \ currents,\ and\ V_L\ load\ voltage\ between\ AB.\\ The\ ends\ of\ R_1\ and\ R_2\ meet\ at\ A,\ \ so\ are\ the\ -tives\ of\ V_1\ and\ V_2\ meet\ at\ B.\\ \therefore\ for\ minimum\ power\ loss,\ current\ flow\ at\ no\ load\ is\ avoided\ by\ making \ V_1 = V_2.\\ So\ pd\ between\ + tive\ terminals\ and\ A,\ must\ be\ equale. \\ \implies\ \ \ I_1*R_1=I_2*R_2,\ \ \implies\ I_1 = 2*I_2.\\ \therefore\ I_1 + I_2 = \frac 3 2*I_1=I_L=\ \sqrt{\dfrac{10} 3}\ \ \ \implies\ I_1=\frac 2 3*\sqrt{\dfrac{10} 3}.\\ P_1+P_2+P_L=I_1^2*1 + I_2^2*2 + 10 = \dfrac 4 9 *\dfrac{10}3 +\dfrac 1 9 *\dfrac{10}3*2 + 10 \ \ \ =\Large\ \ \ \color{#D61F06}{12.222}$

Let $i_{L}$ be the current passing through $R_{L}$ .

$R_{L}i_{L}^2=10$

$i_{L}=\sqrt{\frac{10}{3}}$

Let $i_{1}$ and $i_{2}$ be the currents passing through $R_{1}$ and $R_{2}$ , respectively.

Since $V_{1}$ and $V_{2}$ are variable quantities we can create a sistem of equations and find $i_{1}$ and $i_{2}$ in function of $V_{1}$ and $V_{2}$ so $i_{1}$ and $i_{2}$ can assume any value.

So, to find the minimum value of $P_{1}+P_{2}$ we can use the Cauchy-Schwarz inequality:

$(i_{1}+i_{2})^2\le(i_{1}^2+2i_{2}^2)(1+\frac{1}{2})$

$(\sqrt{\frac{10}{3}})^2\le\frac{3}{2}(P_{1}+P_{2})$

$\frac{20}{9}\le{P_{1}+P_{2}}$

So, the minimum vale of $P_{1}+P_{2}+P_{L}$ is $10+\frac{20}{9}=12.22$

Another way using $V_1,\:V_2$ instead of the currents to optimize the dissipated power. First calculate all currents using loop analysis. Let $I_1,\:I_2$ be the currents through $R_1,\:R_2$ flowing upwards: $\begin{gathered} \begin{pmatrix} 4&3\\ 3&5 \end{pmatrix}\si{\ohm}\cdot\begin{pmatrix} I_1\\I_2 \end{pmatrix}=\begin{pmatrix} R_1+R_L&R_L\\ R_L&R_2+R_L \end{pmatrix}\cdot\begin{pmatrix} I_1\\I_2 \end{pmatrix}\overset{!}{=}\begin{pmatrix} V_1\\V_2 \end{pmatrix}\quad\Rightarrow\quad\begin{pmatrix} I_1\\I_2 \end{pmatrix}=\frac{\SI{1}{\siemens}}{11}\begin{pmatrix} 5V_1-3V_2\\ -3V_1+4V_2 \end{pmatrix}\\ I_L=I_1+I_2=\frac{\SI{1}{\siemens}}{11}(2V_1+V_2) \end{gathered}$

Use $P_L\overset{!}{=}\SI{10}{\watt}$ : $\displaystyle\begin{aligned} P_L=R_LI_L^2=\frac{\SI{3}{\siemens}}{11^2}(2V_1+V_2)^2\overset{!}{=}\SI{10}{\watt}\quad\Rightarrow\quad V_2=-2V_1\pm 11\sqrt{\frac{10}{3}}\si{\volt}&&(*) \end{aligned}$

Calculate the dissipated power in all resistors. Complete the square in the third step: $\displaystyle\begin{aligned} P_L+P_1+P_2&=P_L+R_1I_1^2+R_2I_2^2=\SI{10}{\watt}+\frac{\SI{1}{\siemens}}{11^2}(5V_1-3V_2)^2+\frac{\SI{2}{\siemens}}{11^2}(-3V_1+4V_2)^2&&\\ &\underset{(*)}{=}\SI{10}{\watt}+\frac{\SI{\cancel{11^2}}{\siemens}}{\cancel{11^2}}\left( 3V_1^2\mp 22\sqrt{\frac{10}{3}}V_1\si{\volt}+\SI{\frac{410}{3}}{\volt^2} \right)\\ &=\SI{10}{\watt}+\SI{3}{\siemens}\left( \left( V_1\mp\frac{11}{3}\sqrt{\frac{10}{3}}\si{\volt} \right)^2+\SI{\frac{410}{9}}{\volt^2}-\SI{\frac{1210}{27}}{\volt^2} \right)\\ &\geq\SI{10}{\watt}+\SI{3}{\siemens}\left( \SI{\frac{410}{9}}{\volt^2}-\SI{\frac{1210}{27}}{\volt^2} \right)=\SI{\frac{110}{9}}{\watt}\approx\boxed{\SI{12.2}{\watt}} \end{aligned}$

Rem.: The voltages $V_1,\:V_2$ for my solution are $\displaystyle V_1=\pm\frac{11}{3}\sqrt{\frac{10}{3}}\si{\volt}\underset{(*)}{=}V_2$