Explain what is meant by skin depth of a conductor and explain how it is related to attenuation constant.

The impedance of a series RL circuit is given by $Z = R + jX_L$.

$$Z = 12 \, \Omega + j12 \, \Omega$$

To convert this to polar form, we find the magnitude $|Z|$ and phase angle $\theta_Z$.

$$|Z| = \sqrt{R^2 + X_L^2} = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \, \Omega$$ $$\theta_Z = \arctan\left(\frac{X_L}{R}\right) = \arctan\left(\frac{12}{12}\right) = \arctan(1) = 45^\circ$$

So, the impedance is $Z = 12\sqrt{2} \angle 45^\circ \, \Omega$.

Using Ohm's Law for AC circuits, $I = \frac{V}{Z}$.
Given $V = 52 \angle 30^\circ \, \text{V}$.

$$I = \frac{52 \angle 30^\circ \, \text{V}}{12\sqrt{2} \angle 45^\circ \, \Omega}$$ $$|I| = \frac{52}{12\sqrt{2}} = \frac{13}{3\sqrt{2}} = \frac{13\sqrt{2}}{6} \, \text{A}$$ $$\theta_I = 30^\circ - 45^\circ = -15^\circ$$

So, the current is $I = \frac{13\sqrt{2}}{6} \angle -15^\circ \, \text{A}$.

Active power can be calculated using the formula $P = |I|^2 R$, where $|I|$ is the magnitude of the current and $R$ is the resistance.

$$P = \left(\frac{13\sqrt{2}}{6}\right)^2 \times 12 \, \Omega$$ $$P = \left(\frac{169 \times 2}{36}\right) \times 12$$ $$P = \left(\frac{169}{18}\right) \times 12$$ $$P = \frac{169 \times 12}{18} = \frac{169 \times 2}{3} = \frac{338}{3} \, \text{W}$$ $$P \approx 112.67 \, \text{W}$$

The active power in the circuit is $\text{112.67 W}$.