1. **Problem Statement:** Determine the current $I$ in the given circuit using Kirchhoff's Current Law (KCL). 2. **Understanding the Circuit:** - A 12V source is connected to a 3\Omega resistor. - After the 3\Omega resistor, the current $I$ flows upward to a node. - From this node, the circuit splits into three parallel branches: 1) 2\Omega resistor then 11V source, 2) 1\Omega resistor to ground, 3) 2\Omega resistor then 9V source. 3. **Apply KCL at the node after the 3\Omega resistor:** The current $I$ entering the node equals the sum of currents leaving through the three branches: $$I = I_1 + I_2 + I_3$$ 4. **Find voltages at the node:** Let the node voltage be $V$ (with respect to ground). 5. **Express each branch current in terms of $V$:** - Branch 1 (2\Omega and 11V source): Voltage at the far end is 11V, so current flows from node to 11V source through 2\Omega resistor: $$I_1 = \frac{V - 11}{2}$$ - Branch 2 (1\Omega resistor to ground): $$I_2 = \frac{V - 0}{1} = V$$ - Branch 3 (2\Omega and 9V source): Voltage at far end is 9V, current: $$I_3 = \frac{V - 9}{2}$$ 6. **Express $I$ in terms of $V$:** $$I = \frac{V - 11}{2} + V + \frac{V - 9}{2} = \frac{V - 11 + 2V + V - 9}{2} = \frac{4V - 20}{2} = 2V - 10$$ 7. **Find $I$ using the 3\Omega resistor and 12V source:** The current $I$ flows through the 3\Omega resistor from 12V source to node at voltage $V$: $$I = \frac{12 - V}{3}$$ 8. **Set the two expressions for $I$ equal:** $$2V - 10 = \frac{12 - V}{3}$$ 9. **Solve for $V$:** Multiply both sides by 3: $$3(2V - 10) = 12 - V$$ $$6V - 30 = 12 - V$$ Add $V$ to both sides: $$6V + V - 30 = 12$$ $$7V - 30 = 12$$ Add 30 to both sides: $$7V = 42$$ Divide both sides by 7: $$V = 6$$ 10. **Find $I$ using $V=6$:** $$I = 2V - 10 = 2*6 - 10 = 12 - 10 = 2$$ **Final answer:** $$\boxed{I = 2\text{ A}}$$