Let’s find the **Thevenin** and **Norton** equivalents for the circuit at terminals \( a \)--\( b \). --- ## **Step 1: Find \( V_{th} \) (Open-Circuit Voltage)** Let \( V_{ab} = V_{th} \). Let the current through \( 150\,\Omega \) be \( I \). ### **Label \( v_x \):** \[ v_x = \text{Voltage across 100}\,\Omega \] ### **Node at top of 100, 750, 56, and controlled source:** Let’s call this node \( V_x \) (with respect to \( b \)), so \( V_x = V_{ab} \). #### **KCL at node \( V_x \):** \[ I_{150} + 0.5v_x = I_{750} + I_{56} \] But let’s express all currents in terms of node voltages. \[ \text{From left:} \] \[ \frac{10 - V_x}{150} + 0.5v_x = \frac{V_x}{750} + \frac{V_x}{56} \] But \( v_x \) is the voltage across \( 100\,\Omega \) resistor. Since its left side is connected to \( 10\,V \), and right side to \( V_x \): \[ v_x = 10 - V_x \] Plug into KCL: \[ \frac{10 - V_x}{150} + 0.5(10 - V_x) = \frac{V_x}{750} + \frac{V_x}{56} \] Expand: \[ \frac{10 - V_x}{150} + 5 - 0.5V_x = \frac{V_x}{750} + \frac{V_x}{56} \] Multiply both sides by common denominator (let’s use 2100, the LCM of 150, 750, and 56 is 2100): - \( \frac{10 - V_x}{150} \) × 14 = \( \frac{140 - 14V_x}{2100} \) - \( 5 - 0.5V_x \) × 2100 = \( 10500 - 1050V_x \) - \( \frac{V_x}{750} \) × 2.8 = \( \frac{2.8V_x}{2100} \) - \( \frac{V_x}{56} \) × 37.5 = \( \frac{37.5V_x}{2100} \) But let's clear denominators by multiplying the entire equation by 2100: \[ 14(10 - V_x) + 2100 \times 5 - 2100 \times 0.5 V_x = 2.8 V_x + 37.5 V_x \] \[ 140 - 14 V_x + 10500 - 1050 V_x = 40.3 V_x \] Combine terms: \[ (140 + 10500) - (14 V_x + 1050 V_x) = 40.3 V_x \] \[ 10640 - 1064 V_x = 40.3 V_x \] \[ 10640 = 1104.3 V_x \] \[ V_x = \frac{10640}{1104.3} \approx 9.64~V \] **So,** \[ \boxed{V_{th} = 9.64~V} \] --- ## **Step 2: Find \( R_{th} \) (Thevenin Resistance)** **Zero the independent voltage source (replace with short circuit):** Redraw the circuit: - 150 Ω, 100 Ω, dependent source (0.5\( v_x \)), 750 Ω, 56 Ω. Let’s apply a test voltage \( V_{test} \) at terminals \( ab \) and find \( I_{test} \). Let node \( a \) be at \( V_{test} \) (relative to \( b \)). Let the node between the 150 Ω and 100 Ω (let’s call it node N) be at voltage \( V \). ### **KCL at node N:** The current through 150 Ω is \( \frac{0 - V}{150} \), through 100 Ω is \( \frac{V - V_{test}}{100} \), and the controlled source injects \( 0.5(V - V_{test}) \) into node a. At node a (V_{test}): \[ \text{Sum of currents leaving:} \] \[ \frac{V_{test} - V}{100} - 0.5(V - V_{test}) + \frac{V_{test}}{750} + \frac{V_{test}}{56} = I_{test} \] But let's solve for \( R_{th} = \frac{V_{test}}{I_{test}} \). **From the left mesh:** \[ \frac{0 - V}{150} = \frac{V - V_{test}}{100} \] \[ - V/150 = (V - V_{test})/100 \] \[ - V/150 \times 100 = V - V_{test} \] \[ -(2/3) V = V - V_{test} \] \[ -(2/3) V - V = - V_{test} \] \[ -(5/3) V = - V_{test} \] \[ V_{test} = \frac{5}{3} V \] \[ V = \frac{3}{5} V_{test} \] Now, \( v_x = V - V_{test} = \frac{3}{5}V_{test} - V_{test} = -\frac{2}{5} V_{test} \). The dependent source injects \( 0.5 v_x = 0.5 \left(-\frac{2}{5} V_{test}\right) = -\frac{1}{5} V_{test} \) into node a. Now, sum currents leaving node a (\( V_{test} \)): \[ \text{Through 100 Ω: } \frac{V_{test} - V}{100} \] \[ \text{Through 750 Ω: } \frac{V_{test}}{750} \] \[ \text{Through 56 Ω: } \frac{V_{test}}{56} \] \[ \text{Through dependent source: } -\frac{1}{5} V_{test} \] Sum: \[ I_{test} = \frac{V_{test} - V}{100} + \frac{V_{test}}{750} + \frac{V_{test}}{56} - \frac{1}{5} V_{test} \] But \( V = \frac{3}{5} V_{test} \): \[ I_{test} = \frac{V_{test} - \frac{3}{5} V_{test}}{100} + \frac{V_{test}}{750} + \frac{V_{test}}{56} - \frac{1}{5} V_{test} \] \[ = \frac{\frac{2}{5} V_{test}}{100} + \frac{V_{test}}{750} + \frac{V_{test}}{56} - \frac{1}{5} V_{test} \] \[ = \frac{2 V_{test}}{500} + \frac{V_{test}}{750} + \frac{V_{test}}{56} - \frac{V_{test}}{5} \] Let’s find a common denominator (the LCM of 500, 750, 56, 5): - \( 500 = 2^2 \times 5^3 \) - \( 750 = 2 \times 3 \times 5^3 \) - \( 56 = 2^3 \times 7 \) - \( 5 \) LCM is \( 2^3 \times 3 \times 5^3 \times 7 = 8 \times 3 \times 125 \times 7 = 24 \times 125 \times 7 = 3000 \times 7 = 21,000 \) So: - \( \frac{2}{500} = \frac{2 \times 42}{21,000} = \frac{84}{21,000} \) - \( \frac{1}{750} = \frac{1 \times 28}{21,000} = \frac{28}{21,000} \) - \( \frac{1}{56} = \frac{1 \times 375}{21,000} = \frac{375}{21,000} \) - \( -\frac{1}{5} = -\frac{1 \times 4,200}{21,000} = -\frac{4,200}{21,000} \) Sum: \[ I_{test} = \left( \frac{84 + 28 + 375 - 4,200}{21,000} \right) V_{test} \] \[ = \left( \frac{487 - 4,200}{21,000} \right) V_{test} \] \[ = \left( \frac{-3,713}{21,000} \right) V_{test} \] So, \[ R_{th} = \frac{V_{test}}{I_{test}} = \frac{V_{test}}{ -\frac{3,713}{21,000} V_{test}} = -\frac{21,000}{3,713} \approx -5.66\, \Omega \] Negative resistance means that the circuit is **active** (because of the dependent source)! This can happen in circuits with dependent sources. --- ### **Summary:** #### **Thevenin Equivalent:** \[ \boxed{ V_{th} = 9.64\,V, \quad R_{th} = -5.66\,\Omega } \] #### **Norton Equivalent:** \[ I_N = \frac{V_{th}}{R_{th}} = \frac{9.64}{-5.66} \approx -1.70\,A \] \[ \boxed{ I_N = -1.70\,A, \quad R_N = -5.66\,\Omega } \] --- ### **Final Answers:** - **Thevenin:** \( 9.64\,V \) in series with \( -5.66\,\Omega \) - **Norton:** \( -1.70\,A \) in parallel with \( -5.66\,\Omega \) If you'd like a neat summary diagram, let me know!