# Maximum Power Transfer Theorem Solution We solve this using **Maximum Power Transfer The**. > Maximum power is delivered to the load when > \[ > R_L = R_{th} > \] So we must first find the **Thevenin equivalent voltage (\(V_{oc}\))** and **Thevenin resistance (\(R_{th}\))** seen at terminals **1–2**. --- ## Step 1: Find Open-Circuit Voltage (\(V_{oc}\)) When the load is disconnected, **no current flows through \(R_3\)**. Therefore the voltage at terminal **1** equals the voltage of the top node. The current source: \[ I = 15\,\text{mA} \] Resistors connected from the node to ground: - \(R_1 = 10\,\text{k}\Omega\) - \(R_2 = 12\,\text{k}\Omega\) These are **in parallel**. \[ R_{eq} = R_1 \parallel R_2 \] \[ R_{eq} = \frac{10k \times 12k}{10k + 12k} \] \[ R_{eq} = \frac{120k}{22} \] \[ R_{eq} = 5.45\,\text{k}\Omega \] Now find the node voltage: \[ V_{oc} = I \times R_{eq} \] \[ V_{oc} = 15\,\text{mA} \times 5.45\,\text{k}\Omega \] \[ V_{oc} = .015 \times 5454 \] \[ V_{oc} \approx 81.8\,\text{V} \] --- ## Step 2: Find Thevenin Resistance (\(R_{th}\)) Deactivate the current source. A **current source becomes an open circuit**. Now the resistance seen from terminals **1–2** is: \(R_3\) in series with \((R_1 \parallel R_2)\) Already calculated: \[ R_1 \parallel R_2 = 5.45\,\text{k}\Omega \] Given: \[ R_3 = 5\,\text{k}\Omega \] \[ R_{th} = R_3 + (R_1 \parallel R_2) \] \[ R_{th} = 5k + 5.45k \] \[ R_{th} = 10.45\,\text{k}\Omega \] --- ## Step 3: Maximum Power Transfer Condition \[ R_L = R_{th} \] \[ R_L = 10.45\,\text{k}\Omega \] --- ## Step 4: Maximum Power Delivered \[ P_{max} = \frac{V_{oc}^2}{4R_{th}} \] \[ P_{max} = \frac{(81.8)^2}{4 \times 10450} \] \[ P_{max} = \frac{6691}{41800} \] \[ P_{max} \approx .16\,\text{W} \] --- # ✅ Final Answers **Thevenin Voltage** \[ V_{oc} \approx 81.8\,\text{V} \] **Thevenin Resistance** \[ R_{th} \approx 10.45\,\text{k}\Omega \] **Load for Maximum Power** \[ R_L = 10.45\,\text{k}\Omega \] **Maximum Power Delivered** \[ P_{max} \approx .16\,\text{W} \quad (160\,\text{mW}) \]