1. **Problem Statement:** Calculate the power drawn by the 6Ω resistor using Thevenin's theorem. 2. **Thevenin's Theorem:** Thevenin's theorem states that any linear electrical network with voltage and current sources and resistors can be replaced at terminals A-B by an equivalent voltage source $V_{th}$ in series with a resistance $R_{th}$. 3. **Process for Application:** - Remove the load resistor (6Ω) from the circuit. - Calculate the open-circuit voltage across the terminals where the load was connected; this is $V_{th}$. - Calculate the equivalent resistance seen from these terminals with all independent voltage sources replaced by short circuits and current sources by open circuits; this is $R_{th}$. - Reconnect the load resistor to the Thevenin equivalent circuit. - Calculate the current through the load resistor using Ohm's law. - Calculate power using $P = I^2 R$. 4. **Step 1: Remove 6Ω resistor and find $V_{th}$:** - Thevenin voltage $V_{th}$ is the voltage across the terminals where 6Ω was connected. - Analyze the remaining circuit to find $V_{th}$. 5. **Step 2: Find $R_{th}$:** - Turn off independent sources: voltage sources become short circuits, current sources become open circuits. - Calculate equivalent resistance seen from the terminals. 6. **Step 3: Calculate current through 6Ω resistor:** $$I = \frac{V_{th}}{R_{th} + 6}$$ 7. **Step 4: Calculate power drawn by 6Ω resistor:** $$P = I^2 \times 6$$ **Note:** Without exact circuit values for the voltage and current sources in the branches, numerical calculation cannot be completed here. **Slug:** "thevenin power" **Subject:** "electrical engineering" **Desmos:** {"latex":"","features":{"intercepts":true,"extrema":true}} **q_count:** 1