1. **State the problem:** We need to find the current across the 150 Ω resistor using the superposition theorem in the given circuit with two voltage sources: 30 V on the left and 54 V on the right. 2. **Superposition theorem:** Analyze the circuit by considering one voltage source at a time, turning the other into a short circuit (0 V). --- 3. **Case 1: Only 30 V source active (54 V source replaced by short):** - The circuit now has the 30 V source connected in series with 10 Ω and 30 Ω resistors in series. - The 150 Ω resistor is connected from the node between 10 Ω and 30 Ω resistors to ground. Calculate node voltage at the junction (voltage across 150 Ω) using voltage division and parallel relations. Total resistance between source and ground ignoring 150 Ω branch for now is $10 + 30 = 40\ \Omega$. Let the current through the 10 Ω and 30 Ω be $I_1$. Since 150 Ω resistor is connected from the node to ground, the voltage at the node divides current. Using node voltage method: Let node voltage be $V_1$. At node: $$\frac{V_1}{150} = \frac{30 - V_1}{30} + \frac{30 - V_1}{10}$$ Rewrite right side: $$\frac{30 - V_1}{30} + \frac{30 - V_1}{10} = (30 - V_1)\left(\frac{1}{30} + \frac{1}{10}\right) = (30 - V_1)\frac{4}{30} = \frac{2}{15}(30 - V_1)$$ So, $$\frac{V_1}{150} = \frac{2}{15}(30 - V_1)$$ Multiply both sides by 150: $$V_1 = 150 \times \frac{2}{15} (30 - V_1) = 20 (30 - V_1)$$ Distribute: $$V_1 = 600 - 20 V_1$$ Bring like terms together: $$V_1 + 20 V_1 = 600$$ $$21 V_1 = 600$$ Solve for $V_1$: $$V_1 = \frac{600}{21} = \frac{200}{7} \approx 28.57\ \text{V}$$ Current through 150 Ω resistor: $$I_{150,1} = \frac{V_1}{150} = \frac{28.57}{150} \approx 0.1905\ \text{A}$$ Direction: From node to ground. --- 4. **Case 2: Only 54 V source active (30 V source replaced by short):** - Now the 54 V source is on the right, with 10 Ω and 30 Ω resistors plus 150 Ω resistor same as before. Label node voltage at the junction as $V_2$. Node voltage equation: Voltage on the right side is 54 V; left side is shorted to 0 V. At node $V_2$: $$\frac{V_2}{150} = \frac{V_2}{10} + \frac{54 - V_2}{30}$$ Rewrite right side: $$\frac{V_2}{10} + \frac{54 - V_2}{30} = \frac{3 V_2}{30} + \frac{54 - V_2}{30} = \frac{3 V_2 + 54 - V_2}{30} = \frac{2 V_2 + 54}{30}$$ So, $$\frac{V_2}{150} = \frac{2 V_2 + 54}{30}$$ Multiply both sides by 150: $$V_2 = 5(2 V_2 + 54) = 10 V_2 + 270$$ Bring terms together: $$V_2 - 10 V_2 = 270$$ $$-9 V_2 = 270$$ Solve for $V_2$: $$V_2 = -\frac{270}{9} = -30\ \text{V}$$ Current through 150 Ω resistor: $$I_{150,2} = \frac{V_2}{150} = \frac{-30}{150} = -0.2\ \text{A}$$ Negative sign means direction opposite to assumed. --- 5. **Total current (by superposition):** $$I_{150} = I_{150,1} + I_{150,2} = 0.1905 - 0.2 = -0.0095\ \text{A}$$ Magnitude approx 9.5 mA; direction opposite to first assumed direction. **Final answer:** The current across the 150 Ω resistor is approximately $$\boxed{-0.0095\ \text{A}}$$ (9.5 mA flowing opposite to the initial assumed direction).