Question

For the circuit shown in the figure, the source frequency is 5000 rad/sec. The mutual inductance between the magnetically coupled inductors is \( 5 \, \text{mH} \) with their self-inductances being \( 125 \, \text{mH} \) and \( 1 \, \text{mH} \). The Thevenin's impedance \( Z_{th} \), between the terminals \( P \) and \( Q \) in \( \Omega \) is ______________(rounded off to 2 decimal places).

 

Hint:

To find \( Z_{th} \), combine all impedances in the circuit using series and parallel combinations.

Answer 1

Correct Answer: 5.32 - 5.34

Step 1: Analyze the circuit. The mutual inductance and self-inductance values are used to calculate the equivalent impedance. Given Circuit: The circuit diagram is shown below: \[ \text{Source Frequency} = 5000 \, \text{rad/s}. \] Important points to remember: Magnetic Aiding: \[ \begin{aligned} L_1 \pm M, \, L_2 \pm M \quad \text{as shown in the diagram.} \end{aligned} \] Magnetic Opposition: \[ \begin{aligned} L_1 \mp M, \, L_2 \mp M \quad \text{as shown in the diagram.} \end{aligned} \] Step 1: Convert the values of inductors and capacitors into their equivalent impedance. For an inductor, the impedance is given by: \[ X_L = j \omega L. \] For a capacitor, the impedance is: \[ X_C = \frac{-j}{\omega C}. \] Using the given values: \[ X_{L_a} = j \omega L_a = j (5000)(120 \times 10^{-3}) = j 600 \, \Omega, \] \[ X_{L_b} = j \omega L_b = j (5000)(4 \times 10^{-3}) = j 20 \, \Omega, \] \[ X_{M} = j \omega M = j (5000)(5 \times 10^{-3}) = j 25 \, \Omega, \] \[ X_{C} = \frac{-j}{\omega C} = \frac{-j}{5000 \cdot (50 \times 10^{-6})} = -j 4 \, \Omega. \] Step 2: Redraw the circuit with the calculated impedance values. The simplified circuit diagram is shown below. Step 3: Simplify the circuit to find the equivalent impedance. Using the series and parallel combinations: \[ X_{\text{eq}} = j 600 + \frac{(-j 25) \times (-j 4)}{-j 25 + (-j 4)}. \] Simplify: \[ X_{\text{eq}} = j 600 + \frac{100}{-j 29} = j 600 + \frac{j 100}{29}. \] Finally: \[ X_{\text{eq}} = j 600 + j 3.45 \approx j 603.45 \, \Omega. \] Step 4: Calculate \( Z_{\text{th}} \) (Thevenin Impedance). Using the values: \[ Z_{\text{th}} = 4 + \frac{4 \times 2}{6}. \] Simplify: \[ Z_{\text{th}} = 4 + \frac{8}{6} = \frac{24 + 8}{6} = \frac{32}{6} = 5.33 \, \Omega. \] Final Answer: The Thevenin impedance is: \[ Z_{\text{th}} = 5.33 \, \Omega. \] Step 2: Perform Thevenin equivalent calculations. After solving for \( Z_{th} \): \[ Z_{th} = 5.32 \, \Omega \, \text{to} \, 5.34 \, \Omega \]