1. **State the problem:** We need to find the branch currents, especially $I_3$, in a circuit with three loops and given resistors and voltage/current sources using loop analysis. 2. **Identify knowns and unknowns:** - Voltage sources: 75 V (left loop) and 13 V (right loop) - Current source: 13 A in the middle loop (pointing upward) - Resistors: 4 Ω, 6 Ω, and 8 Ω - Loop currents: $I_1$, $I_2$, $I_3$ 3. **Set up the equations:** Because there is a current source of 13 A in the middle loop, we know: $$I_2 = 13\ \text{A}$$ 4. **Write KVL equations for the other two loops:** - **Loop 1 (left loop):** Going clockwise: $$75 - 4(I_1 - I_3) - 6 (I_1 - I_2) = 0$$ - **Loop 3 (right loop):** Going clockwise: $$13 - 6 (I_3 - I_1) - 8 I_3 = 0$$ 5. **Simplify the equations:** - Loop 1: $$75 - 4I_1 + 4I_3 - 6I_1 + 6I_2 = 0$$ $$75 + 4I_3 + 6(13) - 10I_1 = 0$$ $$75 + 4I_3 + 78 - 10I_1 = 0$$ $$153 + 4I_3 - 10I_1 = 0$$ Rearranged: $$10I_1 - 4I_3 = 153$$ - Loop 3: $$13 - 6I_3 + 6I_1 - 8I_3 = 0$$ $$13 + 6I_1 - 14I_3 = 0$$ Rearranged: $$6I_1 - 14I_3 = -13$$ 6. **Solve simultaneously:** From equation 1: $$10I_1 - 4I_3 = 153$$ From equation 2: $$6I_1 - 14I_3 = -13$$ Multiply equation 2 by $5$ and equation 1 by $3$ to eliminate $I_1$: $$30I_1 - 70I_3 = -65$$ $$30I_1 - 12I_3 = 459$$ Subtract second from first: $$(-70I_3) - (-12I_3) = -65 - 459$$ $$-70I_3 + 12I_3 = -524$$ $$-58I_3 = -524$$ Solve for $I_3$: $$I_3 = \frac{524}{58} = 9.0345 \approx 9.03 \, A$$ 7. **Find $I_1$: Substitute $I_3$ back into equation 1:** $$10I_1 - 4(9.0345) = 153$$ $$10I_1 - 36.138 = 153$$ $$10I_1 = 189.138$$ $$I_1 = 18.9138 \, A$$ 8. **Confirm $I_2$: Given as 13 A.** **Final answer:** $$I_3 \approx 9.03 \, A$$