1. **Problem Statement:** Find the voltage $v_o$ across the 2 $\Omega$ resistor using the Superposition Theorem. 2. **Superposition Theorem:** The voltage across an element in a linear circuit with multiple independent sources is the algebraic sum of the voltages caused by each independent source acting alone, with all other independent sources turned off (voltage sources replaced by short circuits and current sources replaced by open circuits). 3. **Step 1: Consider the 40 V voltage source alone** (turn off the 8 A current source by opening it). - The 8 A current source is open. - The circuit reduces to the 2 $\Omega$ resistor in series with the 3 $\Omega$ and 5 $\Omega$ resistors, and the 40 V source in parallel with the 5 $\Omega$ resistor. 4. **Calculate equivalent resistance seen by the 40 V source:** - The 3 $\Omega$ resistor is in series with the 2 $\Omega$ resistor. - The 5 $\Omega$ resistor is in parallel with the 40 V source, so voltage across 5 $\Omega$ resistor is fixed at 40 V. 5. **Voltage across 2 $\Omega$ resistor due to 40 V source:** - Since the 40 V source is in parallel with the 5 $\Omega$ resistor, the voltage across the 5 $\Omega$ resistor is 40 V. - The current through the 3 $\Omega$ resistor is $I = \frac{40 V}{5 \Omega} = 8 A$. - Voltage across 3 $\Omega$ resistor is $V = IR = 8 A \times 3 \Omega = 24 V$. - Total voltage across 2 $\Omega$ and 3 $\Omega$ resistors is $V_{2+3} = 40 V$ (since 40 V source is across 5 $\Omega$ resistor and the rest). - Voltage across 2 $\Omega$ resistor is proportional to its resistance: $$ v_{o1} = 40 V \times \frac{2 \Omega}{2 \Omega + 3 \Omega} = 40 V \times \frac{2}{5} = 16 V $$ 6. **Step 2: Consider the 8 A current source alone** (turn off the 40 V voltage source by shorting it). - The 40 V source is replaced by a short circuit. - The 8 A current source is in parallel with the 3 $\Omega$ resistor. 7. **Calculate voltage across 2 $\Omega$ resistor due to 8 A current source:** - The 3 $\Omega$ and 5 $\Omega$ resistors are in series, total resistance $R_{35} = 3 + 5 = 8 \Omega$. - The 8 A current source is in parallel with the 3 $\Omega$ resistor, so the current splits. - The 2 $\Omega$ resistor is in series with the parallel combination of 3 $\Omega$ resistor and 8 A current source, and the 5 $\Omega$ resistor. 8. **Find voltage across 2 $\Omega$ resistor:** - The 8 A current source forces 8 A through the parallel branch. - The voltage across the 3 $\Omega$ resistor is $V = IR = 8 A \times 3 \Omega = 24 V$. - This voltage appears across the 2 $\Omega$ resistor and 5 $\Omega$ resistor in series. - Total resistance in series with 2 $\Omega$ resistor is 5 $\Omega$. - Current through 2 $\Omega$ and 5 $\Omega$ resistors is $I = \frac{24 V}{2 \Omega + 5 \Omega} = \frac{24 V}{7 \Omega} \approx 3.43 A$. - Voltage across 2 $\Omega$ resistor is: $$ v_{o2} = I \times 2 \Omega = 3.43 A \times 2 \Omega = 6.86 V $$ 9. **Step 3: Total voltage $v_o$ across 2 $\Omega$ resistor:** $$ v_o = v_{o1} + v_{o2} = 16 V + 6.86 V = 22.86 V $$ **Final answer:** $$ v_o \approx 22.86 V $$