1. **Problem Statement:** Calculate the voltage drop across the 2 Ω resistor in the given circuit using the superposition theorem. 2. **Understanding the Circuit:** The circuit has a 2 Ω resistor in series with two parallel branches: - Branch 1: 4 Ω resistor and 24 A current source - Branch 2: 6 Ω resistor and 6 A current source 3. **Superposition Theorem:** To apply superposition, consider one independent source at a time while replacing other independent sources: - Replace current sources with open circuits when they are turned off. - Replace voltage sources with short circuits when they are turned off. 4. **Step 1: Consider only the 24 A current source active** - Turn off the 6 A current source (replace it with an open circuit). - The 6 Ω resistor branch is open, so only the 4 Ω resistor branch is active in parallel with the 2 Ω resistor in series. 5. **Calculate voltage drop across 2 Ω resistor due to 24 A source:** - The 24 A current flows through the 4 Ω resistor branch. - Since the 2 Ω resistor is in series with the parallel branches, find the current through the 2 Ω resistor. - The 6 Ω branch is open, so total current is 24 A through the 2 Ω resistor. - Voltage drop $V_1 = I \times R = 24 \times 2 = 48$ volts. 6. **Step 2: Consider only the 6 A current source active** - Turn off the 24 A current source (replace it with an open circuit). - The 4 Ω resistor branch is open, so only the 6 Ω resistor branch is active in parallel with the 2 Ω resistor in series. 7. **Calculate voltage drop across 2 Ω resistor due to 6 A source:** - The 6 A current flows through the 6 Ω resistor branch. - The 4 Ω branch is open, so total current is 6 A through the 2 Ω resistor. - Voltage drop $V_2 = I \times R = 6 \times 2 = 12$ volts. 8. **Step 3: Combine the effects of both sources:** - By superposition, total voltage drop across 2 Ω resistor is sum of individual voltage drops: - $V = V_1 + V_2 = 48 + 12 = 60$ volts. **Final Answer:** The voltage drop across the 2 Ω resistor is **60 volts**.