three inductors are connected in parallel named A, B, C. La=25H, Lb = 30H, Lc=35H, Mutual Inductance indicated by M is Mab = 10H, Mbc=15H, Mca=10H. Also A and B are aiding each other which is in opposition to C. Calculate the equiv impedance of the system

To calculate the equivalent impedance of the parallel inductors A, B, and C, we need to consider both the self-inductances of each inductor and their mutual inductances. Given: - \( L_a = 25 \, \text{H} \) - \( L_b = 30 \, \text{H} \) - \( L_c = 35 \, \text{H} \) - \( M_{ab} = 10 \, \text{H} \) - \( M_{bc} = 15 \, \text{H} \) - \( M_{ca} = 10 \, \text{H} \) Inductors A and B aiding means their mutual inductance contributes positively, while C opposes, which means its effect is negative concerning the aiding inductors. ### Step 1: Determine the equivalent inductance for the combination The formula for the equivalent inductance \( L_{\text{eq}} \) of inductors in parallel with mutual coupling can be expressed as: \[ \frac{1}{L_{\text{eq}}} = \frac{1}{L_a} + \frac{1}{L_b} + \frac{1}{L_c} - \frac{2}{M_{ab}} - \frac{2}{M_{bc}} - \frac{2}{M_{ca}} \] Substituting the values, we have: 1. Self-inductances: \[ \frac{1}{L_a} = \frac{1}{25} \] \[ \frac{1}{L_b} = \frac{1}{30} \] \[ \frac{1}{L_c} = \frac{1}{35} \] 2. Mutual inductances: \[ \frac{2}{M_{ab}} = \frac{2}{10} = 0.2 \] \[ \frac{2}{M_{bc}} = \frac{2}{15} \approx 0.1333 \] \[ \frac{2}{M_{ca}} = \frac{2}{10} = 0.2 \] ### Step 2: Calculate individual terms - Sum of the self-inductances: \[ \frac{1}{L_a} + \frac{1}{L_b} + \frac{1}{L_c} = \frac{1}{25} + \frac{1}{30} + \frac{1}{35} = 0.04 + 0.0333 + 0.0286 \approx 0.1019 \] - Sum of the mutual inductances: \[ - \left( \frac{2}{M_{ab}} + \frac{2}{M_{bc}} + \frac{2}{M_{ca}} \right) = - (0.2 + 0.1333 + 0.2) = -0.5333 \] ### Step 3: Combine all terms Combining these results: \[ \frac{1}{L_{\text{eq}}} = 0.1019 - 0.5333 \approx -0.4314 \] Thus, we have: \[ L_{\text{eq}} = \frac{1}{-0.4314} \approx -2.32 \, \text{H} \] ### Step 4: Calculate the equivalent impedance The impedance \( Z \) of an inductor is given by: \[ Z = j \omega L \] Thus, for \( L_{\text{eq}} \), since the value is negative, it is clear that the system is unstable or improperly coupled, indicating that the equivalent impedance would not be physically realizable. If we assume a stable operation mode, we must consider configurations to avoid such an indication, as negative equivalent inductance typically does not make physical sense in standard scenarios. ### Summary Given this scenario indicates wrong configurations or calculations needing verification, to ensure the physical realizable system, we need proper characterization. The calculations show an unstable equivalent inductance value leading to the conclusion that either further analysis of circuits or stable component arrangements are required.