Phase Angle for Max Power

The current $I$ through the circuit is given by:

$\begin{aligned} I & = \frac {1\angle \delta - 1\angle 0^\circ}{\vec Z} \\ & = \frac {\cos \delta + j \sin \delta - 1}{0.05+j1.00} \\ & = \frac {(\cos \delta - 1 + j \sin \delta)(0.05-j1.00)}{(0.05+j1.00)(0.05-j1.00)} \\ & = \frac {0.05(\cos \delta - 1)+\sin \delta + j(0.05\sin \delta - \cos \delta + 1)}{1.0025} \end{aligned}$

The active power consumed by the source on the right is maximum when $\Re (I)$ is maximum. That is when $0.05\cos \delta+ \sin \delta$ is maximum.

$\begin{aligned} f(\delta) & = 0.05\cos \delta + \sin \delta) \\ f'(\delta) & = - 0,05 \sin \delta + \cos \delta & \small \color{#3D99F6} \text{Equating }f'(\delta) = 0 \\ \implies \tan \delta & = \frac 1{0.05} = 20 \\ \implies \delta & \approx 87.14^\circ \end{aligned}$

Note that $f''(87.14^\circ) < 0$ . Therefore the active power consumed by the source on the right is maximum when $\delta \approx \boxed{87.14}^\circ$ .

Voltage on the impedance is the difference between them:

$\large \displaystyle V = 1 \angle \delta - 1 \angle 0$

$\color{#20A900} \boxed{ \large \displaystyle V = [\cos(\delta) - 1] + j \sin(\delta) }$

Current on the impedance is the same current of the whole cicruit:

$\large \displaystyle I = \frac{V}{0.05 + j}$

$\color{#20A900} \boxed{ \large \displaystyle I = \frac{[0.05(\cos(\delta) - 1) + \sin(\delta)] + j[(\cos(\delta) - 1) - \sin(\delta)]}{1.0025} }$

Power on the right source is:

$\large \displaystyle S = 1 \angle 0 \cdot I^*$

$\color{#20A900} \boxed{ \large \displaystyle S = \frac{[0.05(\cos(\delta) - 1) + \sin(\delta)] + j[(\cos(\delta) - 1) - \sin(\delta)]}{1.0025} }$

Active power is:

$\large \displaystyle P = \Re (S)$

$\color{#20A900} \boxed{ \large \displaystyle P = \frac{0.05(\cos(\delta) - 1) + \sin(\delta)}{1.0025} }$

Differentiating $P$ with respect to $\delta$ and making it equal to $0$ :

$\large \displaystyle \frac{dP}{d \delta} = \frac{-0.05 \sin(\delta) + \cos(\delta)}{1.0025} = 0$

$\large \displaystyle \tan(\delta) = 20$

$\color{#20A900} \boxed{ \large \displaystyle \delta = \tan^{-1}(20)}$ or $\color{#20A900} \boxed{ \large \displaystyle \delta = \tan^{-1}(20) + 180^o }$

However, the second derivative of $P$ with respect to $\delta$ :

$\large \displaystyle \frac{d^2 P}{d \delta^2} = \frac{-0.05 \cos(\delta) - \sin(\delta)}{1.0025}$

Is negative on $\tan^{-1}(20)$ (maximum) and positive on $\tan^{-1}(20) + 180^o$ (minimum)

Hence the solution is:

$\color{#3D99F6} \boxed{ \large \displaystyle \delta = \tan^{-1}(20) \approx 87.14^o}$