Simple Series Powers

Start by defining the voltage as the reference:

$$\mathbf{V} := V_m \phase{\ang{0}}$$

Thus, we can find the current:

$$\frac{1}{\cos{\theta} + j \sin{\theta}} = \frac{1}{\cos{\theta} + j \sin{\theta}} \frac{\cos{\theta} - j \sin{\theta}}{\cos{\theta} - j \sin{\theta}} = \frac{\cos{\theta} - j \sin{\theta}}{\cos^2{\theta} + \sin^2{\theta}} = \cos{\theta} - j \sin{\theta} = \cos{(-\theta)} + j \sin{(-\theta)}$$ $$\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{V_m \phase{\ang{0}}}{\abs{\mathbf{Z}} \phase{\theta}} = \frac{V_m}{\abs{\mathbf{Z}}} \frac{1}{\cos{\theta} + j \sin{\theta}} = \frac{V_m}{\abs{\mathbf{Z}}} \parens{\cos{(-\theta)} + j \sin{(-\theta)}} := I_m \phase{-\theta}$$

From this, we get:

$$\begin{gather*} \mathbf{V} = \mathbf{I} \mathbf{Z} \\ V_m = \parens{R + jX} I_m \parens{\cos{\theta} - j \sin{\theta}} \\ V_m \frac{1}{\cos{\theta} - j \sin{\theta}} = I_m \parens{R + jX} \\ V_m \parens{\cos{\theta} + j \sin{\theta}} = I_m \parens{R + jX} \end{gather*}$$ $$\begin{gather*} \mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} \\ I_m \parens{\cos{\theta} - j \sin{\theta}} = \frac{V_m}{R + jX} = \frac{V_m}{R + jX} \frac{R - jX}{R - jX} = \frac{V_m \parens{R - jX}}{R^2 + X^2} \end{gather*}$$ $$\begin{align*} V_m \cos{\theta} &= I_m R \\ V_m \sin{\theta} &= I_m X \\ I_m \cos{\theta} &= V_m \frac{R}{R^2 + X^2} \\ I_m \sin{\theta} &= V_m \frac{X}{R^2 + X^2} \end{align*}$$

Using the equations for average power and reactive power (noting that the angle $\theta$ works because it is the angle in which current lags behind the voltage), we get our expressions for average and reactive power delivered to the entire load $\mathbf{Z}$:

$$\boxed{ P = \frac{1}{2} V_m I_m \cos{\theta} } \qquad \boxed{ Q = \frac{1}{2} V_m I_m \sin{\theta} }$$

Now, to find the average and reactive power delivered to $R$ and $jX$, we first note that $R$ will only receive average power while $jX$ will only receive reactive power:

$$\begin{align*} P_X &= 0 \\ Q_R &= 0 \end{align*}$$

Due to a common current $\mathbf{I}$ and conservation of AC power, we can also deduce:

$$\begin{align*} P_R &= P_{\mathbf{Z}} = \frac{1}{2} I_m^2 R = I_{\text{rms}}^2 R \\ Q_X &= Q_{\mathbf{Z}} = \frac{1}{2} I_m^2 X = I_{\text{rms}}^2 X \end{align*}$$

However, $R$ and $jX$ experience different voltages $\mathbf{V}_R := V_{Rm} \phase{\theta_R}$ and $\mathbf{V}_X := V_{Xm} \phase{\theta_X}$.

By voltage division:

$$\begin{align*} \mathbf{V}_R &= \frac{R}{R + jX} \mathbf{V} = V_m \frac{R}{\mathbf{Z}} = V_m \frac{R}{\abs{\mathbf{Z}} \phase{\theta}} = V_m \frac{R}{\sqrt{R^2 + X^2} \parens{\cos{\theta} + j \sin{\theta}}} = V_m \frac{R}{\sqrt{R^2 + X^2}} \parens{\cos{(-\theta)} + j \sin{(-\theta)}} \\ \mathbf{V}_X &= \frac{jX}{R + jX} \mathbf{V} = V_m \frac{jX}{\mathbf{Z}} = V_m \frac{jX}{\abs{\mathbf{Z}} \phase{\theta}} = V_m \frac{jX}{\sqrt{R^2 + X^2} \parens{\cos{\theta} + j \sin{\theta}}} = V_m \frac{X}{\sqrt{R^2 + X^2}} j \parens{\cos{(-\theta)} + j \sin{(-\theta)}} \end{align*}$$

We can find the magnitudes, rearrange, then substitute into equations (1) and (2):

Thus, we have found our solutions.