1. **State the problem:** We need to find the current through the 10 Ω resistor in a circuit with two voltage sources (6 V and 4 V) and three resistors (2 Ω, 3 Ω, and 10 Ω) using the principle of superposition. 2. **Apply superposition principle:** Consider one voltage source at a time while replacing the other voltage source with a short circuit. 3. **Case 1: Only 6 V source active (4 V source replaced by short):** - The 4 V source branch becomes a 3 Ω resistor connected directly across the 10 Ω resistor. - The 6 V source is in series with 2 Ω resistor. - The 3 Ω and 10 Ω resistors are in parallel. Calculate equivalent resistance of 3 Ω and 10 Ω in parallel: $$ R_{eq1} = \frac{3 \times 10}{3 + 10} = \frac{30}{13} \approx 2.31\ \Omega $$ Total resistance in this case: $$ R_{total1} = 2 + 2.31 = 4.31\ \Omega $$ Current from 6 V source: $$ I_1 = \frac{6}{4.31} \approx 1.39\ A $$ Voltage across parallel combination (3 Ω and 10 Ω): $$ V_{parallel1} = I_1 \times 2.31 \approx 1.39 \times 2.31 = 3.21\ V $$ Current through 10 Ω resistor: $$ I_{10,1} = \frac{V_{parallel1}}{10} = \frac{3.21}{10} = 0.321\ A $$ 4. **Case 2: Only 4 V source active (6 V source replaced by short):** - The 6 V source branch becomes a 2 Ω resistor connected directly across the 10 Ω resistor. - The 4 V source is in series with 3 Ω resistor. - The 2 Ω and 10 Ω resistors are in parallel. Calculate equivalent resistance of 2 Ω and 10 Ω in parallel: $$ R_{eq2} = \frac{2 \times 10}{2 + 10} = \frac{20}{12} = 1.67\ \Omega $$ Total resistance in this case: $$ R_{total2} = 3 + 1.67 = 4.67\ \Omega $$ Current from 4 V source: $$ I_2 = \frac{4}{4.67} \approx 0.856\ A $$ Voltage across parallel combination (2 Ω and 10 Ω): $$ V_{parallel2} = I_2 \times 1.67 \approx 0.856 \times 1.67 = 1.43\ V $$ Current through 10 Ω resistor: $$ I_{10,2} = \frac{V_{parallel2}}{10} = \frac{1.43}{10} = 0.143\ A $$ 5. **Combine currents from both cases:** - Both currents through the 10 Ω resistor flow in the same direction (towards the node connecting the two branches), so total current is the algebraic sum: $$ I_{10} = I_{10,1} + I_{10,2} = 0.321 + 0.143 = 0.464\ A $$ **Final answer:** The current through the 10 Ω resistor is approximately **0.464 A**.