1. **Problem Statement:** Find the voltage $v_0$ across the 2-ohm resistor using the Superposition Theorem in the given circuit. 2. **Superposition Theorem:** This theorem states that in a linear circuit with multiple independent sources, the voltage across (or current through) an element is the algebraic sum of the voltages (or currents) caused by each independent source acting alone, with all other independent sources turned off (voltage sources replaced by short circuits and current sources by open circuits). 3. **Step 1: Consider the 40V voltage source alone** - Turn off the 8A current source (open circuit). - The 5Ω resistor is in series with the 40V source. - The 3Ω and 2Ω resistors are in series on the left side. - Calculate the voltage $v_0^{(1)}$ across the 2Ω resistor. Calculate total resistance seen by the 40V source: $$R_{total} = 3 + 2 + 5 = 10\ \Omega$$ Current from 40V source: $$I = \frac{40}{10} = 4\ A$$ Voltage across 2Ω resistor: $$v_0^{(1)} = I \times 2 = 4 \times 2 = 8\ V$$ 4. **Step 2: Consider the 8A current source alone** - Turn off the 40V voltage source (replace with short circuit). - The 5Ω resistor is now in parallel with the 3Ω resistor and 2Ω resistor series combination. Calculate equivalent resistance of 3Ω and 2Ω in series: $$R_{series} = 3 + 2 = 5\ \Omega$$ Now, 5Ω resistor in parallel with this 5Ω series: $$R_{parallel} = \frac{5 \times 5}{5 + 5} = \frac{25}{10} = 2.5\ \Omega$$ Current source of 8A splits equally because resistors are equal: Current through 2Ω resistor branch: $$I_{2\Omega} = 8 \times \frac{5}{5 + 5} = 4\ A$$ Voltage across 2Ω resistor: $$v_0^{(2)} = I_{2\Omega} \times 2 = 4 \times 2 = 8\ V$$ 5. **Step 3: Sum the contributions** $$v_0 = v_0^{(1)} + v_0^{(2)} = 8 + 8 = 16\ V$$ **Final answer:** $$\boxed{v_0 = 16\ V}$$