1. **Problem Statement:** Find the total current through resistor R3 in the given circuit with two voltage sources VS1 = 24 V, VS2 = 18 V, and resistors R1 = 1.2 kΩ, R2 = 1.2 kΩ, and R3 = 3.3 kΩ. 2. **Circuit Analysis Approach:** We will use the node voltage method or mesh current method to find the current through R3. Here, we consider the node between R1, R2, and R3 as node A. 3. **Assign currents:** Let the current through R3 be $I_{R3}$. The currents through R1 and R2 are given as 5.5 mA and 1.8 mA respectively, and the total current entering the node is the sum of currents through R1 and R2 minus the current from VS2. 4. **Calculate voltage at node A:** Voltage drop across R1 is $V_{R1} = I_{R1} \times R1 = 5.5 \times 10^{-3} \times 1200 = 6.6$ V. Voltage at node A relative to ground (bottom node) is $V_A = VS1 - V_{R1} = 24 - 6.6 = 17.4$ V. 5. **Calculate voltage at node B (between R2 and VS2):** Voltage drop across R2 is $V_{R2} = I_{R2} \times R2 = 1.8 \times 10^{-3} \times 1200 = 2.16$ V. Voltage at node B relative to ground is $V_B = VS2 + V_{R2} = 18 + 2.16 = 20.16$ V. 6. **Calculate voltage difference across R3:** $V_{R3} = V_A - 0 = 17.4$ V (since the other end of R3 is grounded). 7. **Calculate current through R3:** Using Ohm's law, $I_{R3} = \frac{V_{R3}}{R3} = \frac{17.4}{3300} = 0.00527$ A or 5.27 mA. 8. **Check options:** Given options are 5.5 mA, 1.8 mA, 12.8 mA, 7.3 mA. The closest current to our calculation is 5.5 mA. **Final answer:** The total current through R3 is approximately **5.5 mA**.