1. **State the problem:** We need to find the current through the 10 Ω resistor in the given circuit using the superposition principle. The circuit has two voltage sources (50 V and 25 V) and resistors of 25 Ω, 15 Ω, and 10 Ω. 2. **Understand the superposition principle:** This principle states that the current due to multiple independent sources is the algebraic sum of the currents caused by each source acting alone, with all other independent sources replaced by their internal resistances (voltage sources replaced by short circuits). 3. **Analyze the circuit with only the 50 V source active:** - Replace the 25 V source with a short circuit. - The circuit now has the 50 V source, 25 Ω and 15 Ω resistors on the top branch, and the 10 Ω resistor on the right vertical branch. 4. **Calculate equivalent resistance for the 50 V source case:** - The 25 Ω and 15 Ω resistors are in series: $$R_{top} = 25 + 15 = 40\ \Omega$$ - The 10 Ω resistor is in parallel with the 40 Ω branch: $$R_{eq} = \frac{40 \times 10}{40 + 10} = \frac{400}{50} = 8\ \Omega$$ 5. **Calculate total current from the 50 V source:** $$I_{total} = \frac{50}{8} = 6.25\ A$$ 6. **Find current through the 10 Ω resistor (50 V source only):** Using current division: $$I_{10,50V} = I_{total} \times \frac{R_{top}}{R_{top} + 10} = 6.25 \times \frac{40}{50} = 6.25 \times 0.8 = 5\ A$$ 7. **Analyze the circuit with only the 25 V source active:** - Replace the 50 V source with a short circuit. - The circuit now has the 25 V source, 25 Ω resistor on the top left, 15 Ω resistor on the top right, and 10 Ω resistor on the right vertical branch. 8. **Calculate equivalent resistance for the 25 V source case:** - The 25 Ω and 15 Ω resistors are in series: $$R_{top} = 25 + 15 = 40\ \Omega$$ - The 10 Ω resistor is in parallel with the 40 Ω branch: $$R_{eq} = \frac{40 \times 10}{40 + 10} = 8\ \Omega$$ 9. **Calculate total current from the 25 V source:** $$I_{total} = \frac{25}{8} = 3.125\ A$$ 10. **Find current through the 10 Ω resistor (25 V source only):** Using current division: $$I_{10,25V} = I_{total} \times \frac{R_{top}}{R_{top} + 10} = 3.125 \times \frac{40}{50} = 3.125 \times 0.8 = 2.5\ A$$ 11. **Combine currents from both sources:** Assuming currents flow in the same direction through the 10 Ω resistor, $$I_{10,total} = I_{10,50V} + I_{10,25V} = 5 + 2.5 = 7.5\ A$$ **Final answer:** The current through the 10 Ω resistor is $$\boxed{7.5\ A}$$.