1. **State the problem:** Calculate the total inductance $L_T$ of the given circuit with inductors $L_1$ to $L_7$ having values 1mH, 3mH, 2mH, 4mH, 5mH, 6mH, and 7mH respectively. 2. **Identify series and parallel connections:** - $L_1$ (1mH) and $L_3$ (3mH) are in series. - $L_2$ (2mH) branches downward from the node between $L_1$ and $L_3$. - $L_6$ (6mH) is below $L_2$, connecting to $L_4$ (4mH). - $L_4$ (4mH) and $L_5$ (5mH) are in series. - $L_7$ (7mH) is below $L_6$, linking the bottom circuit. 3. **Calculate series combinations:** - Series inductors add directly: $$L_{13} = L_1 + L_3 = 1 + 3 = 4\text{mH}$$ - Series $L_4$ and $L_5$: $$L_{45} = L_4 + L_5 = 4 + 5 = 9\text{mH}$$ 4. **Analyze parallel branches:** - $L_2$ (2mH) and $L_6$ (6mH) are connected in parallel between nodes. - Calculate their parallel equivalent inductance: $$\frac{1}{L_{26}} = \frac{1}{L_2} + \frac{1}{L_6} = \frac{1}{2} + \frac{1}{6} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$$ - So, $$L_{26} = \frac{3}{2} = 1.5\text{mH}$$ 5. **Combine $L_{26}$ with $L_7$ in series:** - $$L_{267} = L_{26} + L_7 = 1.5 + 7 = 8.5\text{mH}$$ 6. **Combine $L_{267}$ with $L_{45}$ in parallel:** - $$\frac{1}{L_{total_parallel}} = \frac{1}{L_{267}} + \frac{1}{L_{45}} = \frac{1}{8.5} + \frac{1}{9}$$ - Calculate common denominator and sum: $$\frac{1}{8.5} \approx 0.1176, \quad \frac{1}{9} = 0.1111$$ $$\Rightarrow \frac{1}{L_{total_parallel}} = 0.1176 + 0.1111 = 0.2287$$ - So, $$L_{total_parallel} = \frac{1}{0.2287} \approx 4.37\text{mH}$$ 7. **Finally, add $L_{13}$ in series with $L_{total_parallel}$:** - $$L_T = L_{13} + L_{total_parallel} = 4 + 4.37 = 8.37\text{mH}$$ **Final answer:** $$L_T \approx 8.37\text{mH}$$