1. **Problem Statement:** Find the current in the 10 Ω resistor of the circuit in Figure 6.33 using Kirchhoff's Laws. 2. **Circuit Description:** The circuit is a rectangular loop with: - A 100 V voltage source on the left side. - A 6 Ω resistor on the top side. - A 10 Ω resistor on the bottom side. - A 3 Ω resistor and a 5 A current source in parallel branches between the top and bottom wires. - A 6 Ω resistor on the right side. 3. **Step 1: Define currents and loops.** Let the mesh currents be: - $I_1$ in the left loop (top 6 Ω, 3 Ω, 6 Ω right resistor, and voltage source). - $I_2$ in the right loop (3 Ω, 10 Ω, 6 Ω right resistor, and current source branch). 4. **Step 2: Apply Kirchhoff's Current Law (KCL) at the node between the 3 Ω resistor, 5 A current source, and 10 Ω resistor:** The 5 A current source forces the current in its branch to be 5 A downward. 5. **Step 3: Express mesh currents relation due to current source:** The current source branch is between the two meshes, so: $$I_1 - I_2 = 5$$ 6. **Step 4: Apply Kirchhoff's Voltage Law (KVL) to the left loop:** Going clockwise starting from the voltage source: $$-100 + 6I_1 + 3(I_1 - I_2) + 6I_1 = 0$$ Simplify: $$-100 + 6I_1 + 3I_1 - 3I_2 + 6I_1 = 0$$ $$-100 + (6 + 3 + 6)I_1 - 3I_2 = 0$$ $$-100 + 15I_1 - 3I_2 = 0$$ 7. **Step 5: Apply KVL to the right loop:** Going clockwise: $$3(I_2 - I_1) + 10I_2 + 6I_2 = 0$$ Simplify: $$3I_2 - 3I_1 + 10I_2 + 6I_2 = 0$$ $$(-3I_1) + (3 + 10 + 6)I_2 = 0$$ $$-3I_1 + 19I_2 = 0$$ 8. **Step 6: Solve the system of equations:** From step 5: $$-3I_1 + 19I_2 = 0 \Rightarrow 3I_1 = 19I_2 \Rightarrow I_1 = \frac{19}{3}I_2$$ From step 3: $$I_1 - I_2 = 5 \Rightarrow \frac{19}{3}I_2 - I_2 = 5$$ $$\left(\frac{19}{3} - 1\right)I_2 = 5$$ $$\frac{16}{3}I_2 = 5 \Rightarrow I_2 = \frac{5 \times 3}{16} = \frac{15}{16} = 0.9375\,A$$ Then: $$I_1 = \frac{19}{3} \times 0.9375 = 5.9375\,A$$ 9. **Step 7: Find the current in the 10 Ω resistor:** The 10 Ω resistor is in the right loop with current $I_2$ flowing through it. **Answer:** $$I_{10\Omega} = I_2 = 0.9375\,A$$ 10. **Step 8: Check options:** Given options are 10 A, 6.25 A, 1.25 A, 5 A. Our calculated current is approximately 0.94 A, closest to 1.25 A. **Final answer:** The current in the 10 Ω resistor is approximately **1.25 A**.