1. **Problem Statement:** Find the branch currents $I_1$, $I_2$, and $I_3$ in the circuit shown in Fig. 1a using mesh analysis. 2. **Mesh Analysis Formula:** Mesh analysis involves writing Kirchhoff's Voltage Law (KVL) equations for each mesh (loop) in the circuit. The sum of voltage drops equals the sum of voltage rises in each loop: $$\sum V = 0$$ 3. **Step 1: Define Mesh Currents** - Assign mesh currents $I_1$, $I_2$, and $I_3$ to the three loops as indicated. 4. **Step 2: Write KVL Equations for Each Mesh** - For Mesh 1 (with resistors 5Ω and 6Ω and voltage source 15V): $$5I_1 + 6(I_1 - I_2) = 15$$ - For Mesh 2 (with resistors 6Ω, 10Ω, 4Ω and voltage source 10V): $$6(I_2 - I_1) + 10I_2 + 4(I_2 - I_3) = -10$$ - For Mesh 3 (with resistor 4Ω): $$4(I_3 - I_2) = 0$$ 5. **Step 3: Simplify Equations** - Mesh 1: $$5I_1 + 6I_1 - 6I_2 = 15 \Rightarrow 11I_1 - 6I_2 = 15$$ - Mesh 2: $$6I_2 - 6I_1 + 10I_2 + 4I_2 - 4I_3 = -10 \Rightarrow -6I_1 + 20I_2 - 4I_3 = -10$$ - Mesh 3: $$4I_3 - 4I_2 = 0 \Rightarrow -4I_2 + 4I_3 = 0$$ 6. **Step 4: Write System of Equations** $$\begin{cases} 11I_1 - 6I_2 = 15 \\ -6I_1 + 20I_2 - 4I_3 = -10 \\ -4I_2 + 4I_3 = 0 \end{cases}$$ 7. **Step 5: Solve the System** - From third equation: $$4I_3 = 4I_2 \Rightarrow I_3 = I_2$$ - Substitute $I_3 = I_2$ into second equation: $$-6I_1 + 20I_2 - 4I_2 = -10 \Rightarrow -6I_1 + 16I_2 = -10$$ - Now system reduces to: $$\begin{cases} 11I_1 - 6I_2 = 15 \\ -6I_1 + 16I_2 = -10 \end{cases}$$ - Multiply first equation by 6 and second by 11 to eliminate $I_1$: $$66I_1 - 36I_2 = 90$$ $$-66I_1 + 176I_2 = -110$$ - Add equations: $$140I_2 = -20 \Rightarrow I_2 = -\frac{20}{140} = -\frac{1}{7} \approx -0.1429$$ - Substitute $I_2$ into first equation: $$11I_1 - 6(-0.1429) = 15 \Rightarrow 11I_1 + 0.8574 = 15 \Rightarrow 11I_1 = 14.1426 \Rightarrow I_1 = \frac{14.1426}{11} \approx 1.2857$$ - Recall $I_3 = I_2 = -0.1429$ 8. **Final Answer:** $$I_1 \approx 1.29\,A, \quad I_2 \approx -0.14\,A, \quad I_3 \approx -0.14\,A$$ The negative sign indicates the actual current direction is opposite to the assumed direction for $I_2$ and $I_3$.