## Problem Restatement- **Source (phasor** \( V_s = 11\angle(-60^\circ) \) volts - **Goal:** Maximize real power delivered to the load by appropriately choosing the load impedance. - **Find:** The **maximum real power** in the load, in **watts**. --- ## Step 1: Maximum Power Transfer Theorem (AC) For **maximum real power** transfer in AC circuits: - The **load impedance** \( Z_L \) should be the **complex conjugate** of the source (Thevenin) impedance \( Z_{th} \): \[ Z_L = Z_{th}^* \] - The **maximum power** delivered to the load is: \[ P_{max} = \frac{|V_{th}|^2}{4R_{th}} \] where \( R_{th} \) is the real part of \( Z_{th} \). --- ## Step 2: Missing Information You **did not specify the source (Thevenin) impedance**. To proceed, we **must know** the source impedance \( Z_{th} \), or at least its real part \( R_{th} \). ### Common Case: If Source Impedance is Purely Real, \( R_{th} \) If \( Z_{th} = R_{th} \) (pure resistance), then: \[ P_{max} = \frac{|V_s|^2}{4R_{th}} \] If you know \( R_{th} \), substitute it here. --- ## Step 3: Plug in Values - \( |V_s| = 11 \) volts So, **if** \( R_{th} \) is known: \[ P_{max} = \frac{121}{4R_{th}} \] *(Units: watts)* --- ## Step 4: Final Answer (General Form) ### **Maximum Real Power in the Load:** \[ \boxed{ P_{max} = \frac{121}{4R_{th}} \quad \text{watts} } \] - \( R_{th} \): Real part of the source (Thevenin) impedance, in ohms. --- > **If you provide the value of the source impedance, I can give a numerical answer.** > Without \( R_{th} \), this is the general expression. --- ###  *Illustration of a simple AC source and load circuit (not the exact circuit, for illustration purposes only)* --- **Summary:** The maximum real power delivered to the load is \(\frac{121}{4R_{th}}\) watts, where \(R_{th}\) is the real part of the source impedance. Please supply the source impedance for a numerical answer.

\[ \boxed{P_{max} = \frac{121}{4R_{th}} \text{ watts}} \]