1. **State the problem:** We need to find the branch currents $I_1$, $I_2$, $I_3$, and $I_4$ in the given circuit using Kirchhoff's Current Law (KCL). 2. **Recall KCL:** KCL states that the algebraic sum of currents entering a node is zero. In other words, the sum of currents entering a junction equals the sum of currents leaving it. 3. **Identify nodes and currents:** - Given currents: $I_1 = 1A$ downward, $I_2 = 7A$ downward, $I_4 = 4A$ downward, and a horizontal current of $2A$ rightward at the top. - Also given: a $3A$ current downward in the middle vertical segment (likely $I_3$ or part of it). 4. **Apply KCL at the top node (junction of $I_1$, $I_2$, and horizontal $2A$):** $$I_1 + I_2 = 2A + I_x$$ where $I_x$ is the current continuing downward or to the next branch. 5. **Apply KCL at the bottom node (junction of $I_3$, $I_4$, and horizontal $2A$):** $$I_3 + I_4 = 2A + I_y$$ where $I_y$ is the current entering or leaving the node. 6. **Use given values and directions:** - $I_1 = 1A$ - $I_2 = 7A$ - $I_4 = 4A$ - $I_3$ is unknown. 7. **Sum currents at the top node:** $$1A + 7A = 8A$$ The horizontal current is $2A$ rightward, so the remaining current downward is: $$I_3 = 8A - 2A = 6A$$ 8. **Check currents at the bottom node:** Given $I_3 = 6A$ downward and $I_4 = 4A$ downward, total downward current is: $$6A + 4A = 10A$$ The horizontal current is $2A$ rightward, so the current entering the node from the left must be: $$10A - 2A = 8A$$ 9. **Final branch currents:** $$I_1 = 1A, \quad I_2 = 7A, \quad I_3 = 6A, \quad I_4 = 4A$$ These satisfy KCL at all nodes.