1. **State the problem:** Use the Superposition theorem to calculate the voltage drop across the 3 \(\Omega\) resistor in the given circuit. 2. **Identify sources:** The circuit has two current sources (15 A each, opposite directions) and one voltage source (20 V). 3. **Superposition theorem:** Analyze the circuit by considering one independent source at a time while replacing others with their internal resistances (current sources open-circuited, voltage sources short-circuited). 4. **Step 1: Consider only the 20 V voltage source.** - Replace both 15 A current sources with open circuits. - The circuit reduces to a series-parallel resistor network with the 20 V source. - Calculate equivalent resistance and current through the 3 \(\Omega\) resistor. 5. **Calculate equivalent resistance:** - The 1 \(\Omega\) and 2 \(\Omega\) resistors are in series: \(R_{12} = 1 + 2 = 3\ \Omega\). - The 6 \(\Omega\) resistor is in parallel with \(R_{12}\): $$R_{p} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2\ \Omega$$ - The total resistance in series with the 3 \(\Omega\) resistor is: $$R_{total} = 2 + 3 = 5\ \Omega$$ 6. **Calculate current from 20 V source:** $$I = \frac{V}{R_{total}} = \frac{20}{5} = 4\ A$$ 7. **Voltage drop across 3 \(\Omega\) resistor (due to 20 V source):** $$V_{3,1} = I \times 3 = 4 \times 3 = 12\ V$$ 8. **Step 2: Consider only the 15 A current sources.** - Replace the 20 V voltage source with a short circuit. - The two 15 A current sources are in opposite directions; their net effect is zero current in the middle branch. - Therefore, no current flows through the 3 \(\Omega\) resistor from the current sources alone. 9. **Voltage drop across 3 \(\Omega\) resistor (due to current sources):** $$V_{3,2} = 0\ V$$ 10. **Step 3: Combine results using superposition:** $$V_3 = V_{3,1} + V_{3,2} = 12 + 0 = 12\ V$$ **Final answer:** The voltage drop across the 3 \(\Omega\) resistor is \(\boxed{12\ V}\).