Sure! Here is the rewritten content in a clear, typed format: --- ### Problem 1 **Find \(V_A\) using Thevenin's theorem** (Circuit diagram shown) #### Steps: 1. **Calculate \(R_{th}\) and \(V_{th}\):** - \(R_{th}\): Combine resistances: \[ R_{th} = \frac{1k\Omega + 2k\Omega}{2k\Omega + 2k\Omega} = 2k\Omega \] \[ R_{th} = \frac{3k\Omega}{2k\Omega + 2k\Omega} = 2k\Omega \] 2. **Find \(V_{th}\):** \[ V_{th} = 2V \] 3. **Find \(V_0\):** \[ V_0 = V_A = 2V \] \[ V_0 = \frac{1}{2} \cdot 2V = 1V \] --- ### Problem 2 **Find \(I_0\) using source transformation** (Circuit diagram shown) #### Steps: 1. **Source transformation:** - Voltage source 6V with 2kΩ in series transforms to a current source: \[ I = \frac{6V}{2k\Omega} = 3.33mA \] - New circuit: 3.33mA current source in parallel with 2kΩ, followed by 3kΩ resistor and two 2kΩ resistors in parallel. 2. **Current divider rule:** \[ I_0 = \frac{3.33 \cdot 3k}{3k + 2k} \] \[ I_0 = 2mA \] --- **Summary of Results:** - **Problem 1:** \(V_A = 1V\) - **Problem 2:** \(I_0 = 2mA\) --- Let me know if you want this in a different format or style!