Optimal Power Transfer Impedance

Let the internal impedance of the AC voltage source be $\widetilde Z_i = (a+jb) \ \Omega$ and $\widetilde Z_L = (A+jB) \ \Omega$ . Then, the load current is given by:

$\begin{aligned} I \angle \phi & = \frac {V\angle 0}{\widetilde Z_i + \widetilde Z_L} \\ & = \frac V{a+b+j(A+B)} \\ & = \frac V{\sqrt{(a+A)^2+(b+B)^2}\angle \tan^{-1} \frac {b+B}{a+A}} \\ \implies \phi & = - \tan^{-1} \frac {b+B}{a+A} \end{aligned}$

The active power transferred to the load is maximum when the power factor $\cos \phi$ is maximum.

$\begin{aligned} \cos \phi & = \cos \left( - \tan^{-1} \frac {b+B}{a+A} \right) \\ & = \cos \left( \tan^{-1} \frac {b+B}{a+A} \right) \\ & = \frac 1{ \sqrt{1 + \left(\frac {\color{#3D99F6}b+B}{a+A} \right)^2}} \end{aligned}$

Note that $\cos \phi$ is maximum when $\color{#3D99F6}B=-b$ and $\cos \phi = 1$ .

When $B=-b$ , then $\widetilde Z_i + \widetilde Z_L = a + A$ . Then the equivalent circuit is the AC voltage source is series with its internal resistance $R_i = a$ and load resistance $R_L= A$ and maximum active transferred to the load is when $R_L = R_i$ or $A=a$ (see Note ).

Therefore, the active power transferred to the load impedance is maximum when $A=a$ and $B=-b$ and when $a=b=3$ , $A+B=3-3 = \boxed{0}$ .

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Note: Consider a voltage source $V$ with internal resistance $r$ in series with a load resistance $R$ . Then the load current is $I = \dfrac V{r+R}$ and the power transferred to $R$ is given by:

$\begin{aligned} P & = V_RI \\ & = \frac {RV}{r+R} \times \frac V{r+R} \\ & = \frac {RV^2}{(r+R)^2} \\ & = \frac {V^2}{\frac {r^2}R+ 2 + R} \end{aligned}$

$P$ is maximum, when $\dfrac {r^2}R + R$ is minimum. By AM-GM inequality $\dfrac {r^2}R + R \ge 2r$ . Equality occurs when $R=r$ . Therefore, maximum power transferred occurs when load resistance equals to internal resistance.