1. **Problem Statement:** Find the maximum power transferred across the terminals a-b in the given circuit. 2. **Identify the circuit elements and variables:** - Current source: $2I_x$ (diamond-shaped) pointing right. - Resistor in series with current source: $5\Omega$ with current $I_x$. - Voltage source: $10$ V connected in parallel across terminals a-b. - Resistor $4\Omega$ connected in parallel across terminals a-b. 3. **Step 1: Express $I_x$ in terms of voltage across terminals a-b ($V_{ab}$).** Since the $5\Omega$ resistor has current $I_x$, voltage across it is $V_{5\Omega} = 5I_x$. The current source is $2I_x$ flowing right, so the node voltage relations must be considered. 4. **Step 2: Find the equivalent circuit seen from terminals a-b.** The voltage source $10$ V and resistor $4\Omega$ are in parallel across a-b. The series branch with current source and $5\Omega$ resistor is connected to node a. 5. **Step 3: Use mesh or node analysis to find $I_x$ and $V_{ab}$.** Let the voltage at the node connecting the current source and $5\Omega$ resistor be $V_x$. The voltage across $5\Omega$ resistor is $V_x - V_a$ where $V_a = V_{ab}$ (since terminal b is reference 0). Current through $5\Omega$ resistor is $I_x = \frac{V_x - V_{ab}}{5}$. Current source current is $2I_x$ flowing from node to terminal a. 6. **Step 4: Apply KCL at node a:** Sum of currents leaving node a = 0. Currents leaving node a: - Through $5\Omega$ resistor: $I_x = \frac{V_x - V_{ab}}{5}$ (towards node x) - Through voltage source and $4\Omega$ resistor branch: current $I_{ab}$ Since voltage source is $10$ V, voltage at a relative to b is fixed at $V_{ab} = 10$ V. 7. **Step 5: Calculate power across terminals a-b:** Power delivered to load (terminals a-b) is $P = \frac{V_{ab}^2}{R_{eq}}$ where $R_{eq}$ is the equivalent resistance seen from terminals a-b. 8. **Step 6: Find Thevenin equivalent resistance $R_{th}$ across a-b:** Deactivate independent sources: - Replace voltage source (10 V) with short circuit. - Current source replaced with open circuit. Then, $5\Omega$ and $4\Omega$ resistors are connected in parallel across a-b. Equivalent resistance: $$R_{th} = \frac{5 \times 4}{5 + 4} = \frac{20}{9} \approx 2.22\ \Omega$$ 9. **Step 7: Find Thevenin voltage $V_{th}$ across a-b:** Calculate open-circuit voltage at terminals a-b. Using superposition or node voltage method, the voltage across a-b is $10$ V (due to voltage source). 10. **Step 8: Maximum power transfer theorem:** Maximum power is transferred when load resistance equals $R_{th}$. Maximum power: $$P_{max} = \frac{V_{th}^2}{4 R_{th}} = \frac{10^2}{4 \times \frac{20}{9}} = \frac{100}{\frac{80}{9}} = \frac{100 \times 9}{80} = \frac{900}{80} = 11.25$$ **Final answer:** The maximum power transferred across terminals a-b is $11.25$ watts.