1. **Problem Statement:** Calculate the voltage drop across the 2 Ω resistor in the given circuit using the superposition theorem. 2. **Superposition Theorem:** This theorem states that in a linear circuit with multiple independent sources, the voltage/current for any element is the algebraic sum of the voltages/currents caused by each independent source acting alone, with all other independent sources turned off (replaced by their internal resistances). 3. **Step 1: Consider the 24 A current source active, turn off the 6 A source.** - Turning off the 6 A current source means replacing it with an open circuit. - The right branch with 6 Ω resistor is open, so no current flows there. - The 24 A current source and 4 Ω resistor are in series with the 2 Ω resistor. - Total resistance in this path: $$R_{total1} = 4 + 2 = 6\ \Omega$$ - Current through 2 Ω resistor is 24 A (same as current source). - Voltage drop across 2 Ω resistor: $$V_1 = I \times R = 24 \times 2 = 48\ V$$ 4. **Step 2: Consider the 6 A current source active, turn off the 24 A source.** - Turning off the 24 A current source means replacing it with an open circuit. - The left branch with 4 Ω resistor is open, so no current flows there. - The 6 A current source and 6 Ω resistor are in series with the 2 Ω resistor. - Total resistance in this path: $$R_{total2} = 6 + 2 = 8\ \Omega$$ - Current through 2 Ω resistor is 6 A (same as current source). - Voltage drop across 2 Ω resistor: $$V_2 = I \times R = 6 \times 2 = 12\ V$$ 5. **Step 3: Combine the effects of both sources.** - The currents from each source cause voltage drops in the same direction across the 2 Ω resistor. - Total voltage drop: $$V = V_1 + V_2 = 48 + 12 = 60\ V$$ **Final answer:** The voltage drop across the 2 Ω resistor is $$60\ V$$.