Let's solve **Step 1**: **Find the Thevenin equivalent to the left of terminals a,b using the test source method.** --- ### **Step 1.1: Remove the Load** Remove the 40 Ω resistor between \( a \) and \( b \) (treat it as the load). --- ### **Step 1.2: Thevenin Resistance (\(R_{th}\)) Using Test Source Method** 1. **Deactivate all independent sources:** - Replace the 4 A current source with an open circuit. 2. **Apply a test source (say \( V_{test} \))** between terminals \( a \) and \( b \). Let the resulting current through the test source be \( I_{test} \). 3. **Find \( R_{th} = \frac{V_{test}}{I_{test}} \).** #### **Redrawn Circuit with Test Source** - 60 Ω (top branch) - 20 Ω series with dependent source (middle branch) - 80 Ω (bottom branch) All are open at the left (since current source is open). The dependent current source is \( 160 i_x \), where \( i_x \) is the current through the 80 Ω resistor. Let's define the current directions: - \( i_1 \): through 20 Ω branch - \( i_2 \): through 80 Ω branch - \( i_x = i_2 \) (current through 80 Ω resistor) - \( I_{test} = i_1 + i_2 \) **Apply KCL at node left of a:** \[ i_1 = \frac{V_{test} - 160i_x}{20} \] \[ i_2 = \frac{V_{test}}{80} \] \[ I_{test} = i_1 + i_2 \] But the dependent source is in series with 20 Ω, so the voltage across 20 Ω + dependent source = \( V_{test} \): Let’s sum voltages from left to right across the 20 Ω and dependent source: \[ V_{test} = 20i_1 + V_{dep} \] But \( V_{dep} \) is the voltage across the dependent current source. Since this is a current source, we need to use KCL at nodes. Alternatively, **use mesh analysis**: #### **Assign Mesh Currents:** - Mesh 1: 20 Ω & dependent source (current \( i_1 \)), flows from left to \( a \) - Mesh 2: 80 Ω (current \( i_2 \)), flows from left to \( a \) The test source connects the two meshes at \( a \) and \( b \). **Apply KCL at node left of a:** \[ i_1 + i_2 = I_{test} \] But let's use node voltage at \( a \) relative to \( b \): Let \( V_a = V_{test} \), \( V_b = 0 \) - Current up through 20 Ω + dependent source: \( i_1 \) - Current up through 80 Ω: \( i_2 = \frac{V_{test}}{80} \) - \( i_1 \) flows from test source, through 20 Ω and dependent source, to node \( a \). But the dependent source is a current source of \( 160i_x \) upwards in the branch, and \( i_x = i_2 \). So, the current through the 20 Ω resistor is \( i_1 \), then the dependent current source adds \( 160i_2 \) upwards. From KCL at node between 20 Ω and dependent source: \[ i_1 = -160i_2 \] But the current through the test voltage source is: \[ I_{test} = i_1 + i_2 \] But with \( i_1 = -160i_2 \): \[ I_{test} = -160i_2 + i_2 = -159i_2 \] So, \[ i_2 = \frac{V_{test}}{80} \] Therefore, \[ I_{test} = -159 \cdot \frac{V_{test}}{80} \] So, \[ R_{th} = \frac{V_{test}}{I_{test}} = \frac{V_{test}}{-159 \cdot \frac{V_{test}}{80}} = -\frac{80}{159} \Omega \] **Negative resistance indicates the dependent source is providing power (active network).** --- ### **Step 1.3: Find Thevenin Voltage (\(V_{th}\))** 1. **Reconnect the 4A current source.** 2. **Open terminals a-b (remove load).** 3. **Find open-circuit voltage \(V_{oc}\) at a-b.** Let’s assign node voltages: Let \( V_a \) = \( V_{th} \) (since b is ground). - The 4A current splits between 20Ω+dep. source and 80Ω. Let \( i \) be the current through 80Ω branch (\( i = i_x \)), then \( 4 - i \) through 20Ω branch. KVL around left loop: - Through 60Ω: \( 4A \) flows, so voltage at top left node is \( V_1 = 60 \times 4 = 240V \). - Through 20Ω: \( (4-i) \) flows, so voltage drop = \( 20(4-i) \). - Dependent current source: in series, but for KVL, treat as a black box. At node a: - Current through 80Ω = \( i \) - \( V_a = V_{th} = 80i \) At top left node: \( V_1 \) At node a: \( V_{th} \) KVL from top left to a: \[ V_1 - 20(4-i) = V_{th} \] \[ 240 - 80 + 20i = V_{th} \] \[ 160 + 20i = V_{th} \] But also, the dependent current source supplies \( 160i \) from left node to node a. At node a, KCL: \[ \text{Current into node a from 20Ω+dep}: (4-i) + 160i \] \[ \text{Current into node a from 80Ω}: -i \] \[ \text{Sum} = 0 \] \[ (4-i) + 160i - i = 0 \] \[ 4 + 158i = 0 \] \[ i = -\frac{4}{158} = -\frac{2}{79} \, \text{A} \] Now plug into \( V_{th} = 160 + 20i \): \[ V_{th} = 160 + 20 \left(-\frac{2}{79}\right) \] \[ V_{th} = 160 - \frac{40}{79} \] \[ V_{th} \approx 159.49 \text{ V} \] --- ## **Step 1.4: Thevenin Equivalent** \[ \boxed{ \begin{aligned} V_{th} &\approx 159.49~\text{V} \\ R_{th} &= -\frac{80}{159}~\Omega \approx -0.503~\Omega \end{aligned} } \] --- ## **Step 2: SPICE Simulation** You would use SPICE to build the circuit and measure the open-circuit voltage and input resistance by applying a test source. If any discrepancy, it may be due to sign conventions, interpretation of dependent sources, or a mistake in equations. --- ### **Summary** - **Thevenin Voltage \( V_{th} \):** ≈ 159.49 V - **Thevenin Resistance \( R_{th} \):** ≈ -0.503 Ω (negative, due to dependent source active behavior) If you need the **Norton equivalent**, just use \( I_{N} = \frac{V_{th}}{R_{th}} \). If you need a SPICE netlist, let me know!