To solve this circuit using source transformation, follow these steps: ### Step 1: Source Transformation of Voltage Source 1. **Transform the 32V Voltage Source and4kΩ Resistor**: - The voltage source (32V) in series with a 4kΩ resistor can be transformed into a current source. - Current \( I = \frac{V}{R} = \frac{32V}{4kΩ} = 8 \, \text{mA} \). - The new current source is 8 mA in parallel with the 4kΩ resistor. ### Step 2: Simplify the Circuit 2. **Combine Parallel Current Sources**: - The 8 mA current source (from the transformation) and the 12 mA current source are in parallel. They can be combined into a single current source: - Total current = 8 mA + 12 mA = 20 mA. 3. **Combine Parallel Resistors**: - Combine the 4kΩ (from the transformation), 3kΩ, and 6kΩ resistors in parallel: - \( \frac{1}{R_{\text{eq}}} = \frac{1}{4kΩ} + \frac{1}{3kΩ} + \frac{1}{6kΩ} \). - \( R_{\text{eq}} = 1.5kΩ \). ### Step 3: Analyze the Simplified Circuit 3. **Circuit Analysis**: - With the equivalent 20 mA current source and 1.5kΩ resistor, calculate the voltage across the equivalent resistor: - \( V_{\text{eq}} = I \times R_{\text{eq}} = 20 \, \text{mA} \times 1.5kΩ = 30 \, \text{V} \). ### Step 4: Finding \( v_ \) 4. **Calculate \( v_ \)**: - \( v_ \) is the voltage across the 4kΩ resistor in parallel with the equivalent 1.5kΩ resistor. - Since the voltage across resistors in parallel is the same, \( v_ = V_{\text{eq}} = 30 \, \text{V} \). Thus, the voltage \( v_ \) is 30V.