1. **State the problem:** Given the currents $I_1=10$ A, $I_2=12$ A, and $I_5=8$ A, find the magnitude and direction of currents $I_3$, $I_4$, $I_6$, and $I_7$ in the diamond-shaped network using Kirchhoff’s Current Law (KCL). 2. **Label nodes and incoming/outgoing currents:** - Node $a$: currents $I_1$ entering - Node $b$: currents $I_2$ entering - Node $c$: currents $I_3$, $I_6$ (unknown directions) - Node $d$: currents $I_4$, $I_5$, $I_7$ (unknowns involve $I_4$, $I_7$) 3. **Apply KCL at node $a$:** Current entering = current leaving $$I_1 = I_3 + I_4$$ So, $$I_4 = I_1 - I_3 = 10 - I_3$$ 4. **Apply KCL at node $b$:** Similar relation gives $$I_2 = I_4 + I_7$$ So, $$I_7 = I_2 - I_4 = 12 - I_4$$ 5. **Apply KCL at node $c$:** Currents entering = currents leaving $$I_3 + I_6 = I_1$$ But we need direction info; instead, more direct is to write KCL at $c$: $$I_3 = I_6 + I_5$$ Given $I_5=8$ A, $$I_3 = I_6 + 8$$ 6. **Apply KCL at node $d$:** Currents entering = currents leaving $$I_6 + I_4 = I_5 + I_7$$ Rearranged, $$I_7 = I_6 + I_4 - I_5$$ 7. **Substitute $I_4$ from step 3:** $$I_7 = I_6 + (10 - I_3) - 8 = I_6 + 2 - I_3$$ 8. **From step 4, $I_7 = 12 - I_4 = 12 - (10 - I_3) = 2 + I_3$** 9. **Equate the two expressions for $I_7$:** $$2 + I_3 = I_6 + 2 - I_3$$ Simplify: $$I_3 = I_6 - I_3$$ $$2I_3 = I_6$$ So, $$I_6 = 2I_3$$ 10. **From step 5, $I_3 = I_6 + 8$, substitute $I_6$:** $$I_3 = 2I_3 + 8$$ $$-I_3 = 8$$ $$I_3 = -8$$ A (negative means opposite direction) 11. **Calculate other currents:** $$I_6 = 2 imes (-8) = -16$$ A (opposite direction) $$I_4 = 10 - I_3 = 10 - (-8) = 18$$ A $$I_7 = 2 + I_3 = 2 - 8 = -6$$ A (opposite direction) **Final answers:** - $I_3 = 8$ A opposite assumed direction - $I_4 = 18$ A - $I_6 = 16$ A opposite assumed direction - $I_7 = 6$ A opposite assumed direction These signs indicate their actual directions are opposite to the initially assumed ones in the figure.