Let's solve **Step 1: Find the Thevenin equivalent to the left of terminals a, b (using the test source method)**. --- ### **Step 1: Thevenin Resistance (\(R_{th}\)) using Test Source Method** #### **A. Remove all independent sources** - The 4A current source is independent. **Replace it with an open circuit**. #### **B. Apply a test source (\(V_{test}\))** - Apply \(V_{test}\) at terminals \(a, b\) (from \(a\) to \(b\)), and calculate the resulting current \(I_{test}\) entering at \(a\). - \(R_{th} = \frac{V_{test}}{I_{test}}\) #### **C. Redraw the Circuit** - All resistors remain. - The dependent source remains: \(160i_x\), where \(i_x\) is the current through \(40\,\Omega\). #### **D. Find relationships** Let \(i_x =\) current through **40Ω** (from a to b). Let \(I_{test}\) be the current injected into \(a\). ##### **Nodal Analysis at node a:** Let the voltage at \(a\) be \(V_{test}\) (since \(b\) is at 0V). - The current through \(80\,\Omega\) from \(a\) to left junction: \(\frac{V_{test} - V_1}{80}\) - The current through \(40\,\Omega\) from \(a\) to \(b\): \(\frac{V_{test}}{40} = i_x\) - The current from the dependent source (from left junction to \(a\)): \(160 i_x\) (enters \(a\)) ##### **At the left junction:** - Current through \(20\,\Omega\) (from left to \(a\)): \(\frac{V_1 - V_{test}}{20}\) - Current through \(60\,\Omega\) from left junction to ground (\(b\)): \(\frac{V_1}{60}\) **KCL at node \(a\):** \[ I_{test} = \frac{V_{test} - V_1}{80} + \frac{V_{test}}{40} + 160 i_x \] But \(i_x = \frac{V_{test}}{40}\), so: \[ I_{test} = \frac{V_{test} - V_1}{80} + \frac{V_{test}}{40} + 160\left(\frac{V_{test}}{40}\right) \] \[ I_{test} = \frac{V_{test} - V_1}{80} + \frac{V_{test}}{40} + 4V_{test} \] **KCL at left junction:** \[ \text{Current in} = \text{current out} \] \[ \frac{V_1 - V_{test}}{20} = \frac{V_1}{60} + \frac{V_1 - V_{test}}{80} + 160 i_x \] But notice that \(i_x\) flows only through the 40Ω. The dependent source is between left junction and \(a\), so it appears in both KCLs. But let's focus on the simpler way: #### **Combine resistors:** - The 80Ω and 40Ω are in series between \(a\) and \(b\): total \(= 120\,\Omega\). - The 20Ω and 60Ω are not directly in series or parallel due to the dependent source. Let’s apply a test voltage \(V_{test}\) across \(ab\): - The current through 40Ω: \(i_x = \frac{V_{test}}{40}\). - The current through 80Ω: \(i_{80} = \frac{V_{test}}{80}\). - The dependent **current source** \(160i_x\) is in parallel with the 80Ω resistor, from the junction of (20Ω, 80Ω, and 60Ω) to node a. Let's call the voltage at the left junction \(V_1\): - Current through 20Ω: \(\frac{V_1 - V_{test}}{20}\) - Current through 60Ω: \(\frac{V_1}{60}\) - Current through 80Ω: \(\frac{V_1 - V_{test}}{80}\) - Current through dependent source: \(160i_x = 4V_{test}\) (since \(i_x = V_{test}/40\)), from left junction to \(a\). **KCL at left junction:** \[ \frac{V_1 - V_{test}}{20} + \frac{V_1 - V_{test}}{80} = \frac{V_1}{60} + 4V_{test} \] \[ \left( \frac{1}{20} + \frac{1}{80} \right)(V_1 - V_{test}) = \frac{V_1}{60} + 4V_{test} \] \[ \frac{5}{80}(V_1 - V_{test}) = \frac{V_1}{60} + 4V_{test} \] \[ \frac{5V_1 - 5V_{test}}{80} = \frac{V_1}{60} + 4V_{test} \] Multiply both sides by 240 (LCM of 80 and 60) to clear denominators: \[ 3(5V_1 - 5V_{test}) = 4V_1 + 960V_{test} \] \[ 15V_1 - 15V_{test} = 4V_1 + 960V_{test} \] \[ 15V_1 - 4V_1 = 960V_{test} + 15V_{test} \] \[ 11V_1 = 975V_{test} \] \[ V_1 = \frac{975}{11} V_{test} \] Now, **find \(I_{test}\):** \[ I_{test} = \frac{V_{test}}{40} + \frac{V_{test}}{80} + 4V_{test} \] \[ = \frac{2V_{test}}{80} + \frac{V_{test}}{80} + 4V_{test} \] \[ = \frac{3V_{test}}{80} + 4V_{test} \] \[ = \frac{3V_{test} + 320V_{test}}{80} \] \[ = \frac{323V_{test}}{80} \] But this misses the 20Ω and 60Ω branches, as current can also flow through those. So, actually: \[ I_{test} = (\text{current through 40Ω}) + (\text{current through 80Ω}) + (\text{current through dependent source}) \] But the current entering \(a\) is: - From \(a\) to \(b\) through 40Ω: \(i_x = V_{test}/40\) - From \(a\) to left through 80Ω: \((V_{test} - V_1)/80\) - From left to \(a\) through dependent source: \(160i_x = 4V_{test}\) So, the total current into \(a\) is: \[ I_{test} = \frac{V_{test}}{40} + \frac{V_{test} - V_1}{80} + 4V_{test} \] Plug in \(V_1 = \frac{975}{11} V_{test}\): \[ I_{test} = \frac{V_{test}}{40} + \frac{V_{test} - \frac{975}{11}V_{test}}{80} + 4V_{test} \] \[ = \frac{V_{test}}{40} + \frac{V_{test} (1 - \frac{975}{11})}{80} + 4V_{test} \] \[ = \frac{V_{test}}{40} + \frac{V_{test} \left(\frac{11 - 975}{11}\right)}{80} + 4V_{test} \] \[ = \frac{V_{test}}{40} + \frac{V_{test} \left(\frac{-964}{11}\right)}{80} + 4V_{test} \] \[ = \frac{V_{test}}{40} - \frac{964 V_{test}}{880} + 4V_{test} \] \[ \frac{964}{880} = \frac{241}{220} \] So: \[ I_{test} = \frac{V_{test}}{40} - \frac{241 V_{test}}{220} + 4V_{test} \] Now, get all terms over a common denominator (let's use 220): \[ \frac{V_{test}}{40} = \frac{11 V_{test}}{440} \] \[ 4 V_{test} = \frac{4 \times 220 V_{test}}{220} = \frac{880 V_{test}}{220} \] So: \[ I_{test} = \frac{11 V_{test}}{440} - \frac{241 V_{test}}{220} + \frac{880 V_{test}}{220} \] \[ = \frac{11 V_{test}}{440} + \frac{880 V_{test} - 241 V_{test}}{220} \] \[ = \frac{11 V_{test}}{440} + \frac{639 V_{test}}{220} \] \[ = \frac{11 V_{test}}{440} + \frac{2 \times 639 V_{test}}{2 \times 220} \] \[ = \frac{11 V_{test}}{440} + \frac{639 V_{test}}{220} \] \[ = \frac{11 V_{test}}{440} + \frac{319.5 V_{test}}{110} \] But let's just convert \(\frac{639}{220}\) to the same denominator as \(\frac{11}{440}\): \[ \frac{639}{220} = \frac{2 \times 639}{2 \times 220} = \frac{1278}{440} \] So: \[ I_{test} = \frac{11 V_{test}}{440} + \frac{1278 V_{test}}{440} = \frac{(11+1278)V_{test}}{440} = \frac{1289 V_{test}}{440} \] Therefore, \[ R_{th} = \frac{V_{test}}{I_{test}} = \frac{V_{test}}{\frac{1289 V_{test}}{440}} = \frac{440}{1289} \approx 0.341 \Omega \] --- ### **Step 2: Thevenin Voltage (\(V_{th}\))** Now **find the open-circuit voltage** at \(a\) (with 4A source acting, \(i_x\) as before). Set \(i_{ab}=0\). Find voltage at \(a\) relative to \(b\): Let \(i_x\) be the current through the 40Ω resistor. - The 4A current splits between 60Ω, 20Ω, and rest of the circuit. - The dependent source and resistors complicate a simple mesh/nodal analysis, but let's apply nodal analysis at \(a\): Let \(V_a = V_{th}\), \(V_1\) is node between 60Ω, 20Ω, 80Ω. - KCL at node \(V_1\): \[ 4 = \frac{V_1}{60} + \frac{V_1 - V_a}{20} + \frac{V_1 - V_a}{80} + 160 i_x \] But with open circuit at \(a\), \(i_x = 0\) (since no current flows in open circuit), so dependent source is zero. \[ 4 = \frac{V_1}{60} + \frac{V_1 - V_a}{20} + \frac{V_1 - V_a}{80} \] Combine the two terms: \[ \frac{V_1 - V_a}{20} + \frac{V_1 - V_a}{80} = (V_1 - V_a)\left(\frac{1}{20} + \frac{1}{80}\right) = (V_1 - V_a)\left(\frac{5}{80}\right) \] So: \[ 4 = \frac{V_1}{60} + \frac{5(V_1 - V_a)}{80} \] \[ 4 = \frac{V_1}{60} + \frac{5V_1 - 5V_a}{80} \] Multiply both sides by 240 (LCM): \[ 960 = 4V_1 + 15V_1 - 15V_a \] \[ 960 = 19V_1 - 15V_a \] Now, KCL at node \(a\) (open circuit, so only path to ground is via 80Ω and 40Ω): \[ 0 = \frac{V_a - V_1}{80} + \frac{V