1. **State the problem:** We need to find the currents $i_1$, $i_2$, and $i_3$ in the circuit using Kirchhoff's Current Law (KCL), which states that the algebraic sum of currents entering and leaving a node is zero. 2. **Apply KCL at node A:** Currents entering node A: 10 A (from left), $i_1$ (upward) Currents leaving node A: 8 A (top left), 12 A (to node B), 4 A (bottom left), 2 A (upward inside node A) Write the equation: $$10 + i_1 = 8 + 12 + 4 + 2$$ Simplify the right side: $$10 + i_1 = 26$$ Solve for $i_1$: $$i_1 = 26 - 10 = 16\,A$$ 3. **Apply KCL at node B:** Currents entering node B: 12 A (from A), $i_2$ (upward) Currents leaving node B: 14 A (downward), 4 A (back to A) Write the equation: $$12 + i_2 = 14 + 4$$ Simplify the right side: $$12 + i_2 = 18$$ Solve for $i_2$: $$i_2 = 18 - 12 = 6\,A$$ 4. **Apply KCL at node C:** Currents entering node C: 14 A, 4 A (both going toward node C as given) Current leaving node C: $i_3$ (upward) Write the equation: $$14 + 4 = i_3$$ Simplify: $$i_3 = 18\,A$$ **Final answers:** $$i_1 = 16\,A, \quad i_2 = 6\,A, \quad i_3 = 18\,A$$