For the circuit shown in the figure, the source frequency is 5000rad/sec. The mutual inductance between the magnetically coupled inductors is 5mH with their self inductances being 125mH and 1mH. The Thevenin's impedance, Z_th, between the terminals P and Q in Ω is ____________ (rounded off to 2 decimal places).

The correct answer is 5.32 (or approximately 5.33).

1. Identification of Parameters from the Circuit Diagram:
From the given circuit diagram, the values of the components are as follows:

• Source angular frequency: $\omega =5000m\mathrm{rad/s}$
• Self-inductance of primary coil: $L_1=125\mathrm{mH}=0.125H$
• Self-inductance of secondary coil: $L_2=1\mathrm{mH}=0.001H$
• Mutual inductance: $M=5\mathrm{mH}=0.005H$
• Capacitance in the secondary loop: $C=50\mathrm{\mu F}=50\times {10}^{-6}F$

2. Calculate the Impedances of each component:
Let us calculate the individual inductive and capacitive reactances:
$Z_{L1}=j\omega L_1=j\left(5000\right)\left(0.125\right)=j625\Omega$
$Z_{L2}=j\omega L_2=j\left(5000\right)\left(0.001\right)=j5\Omega$

$Z_C=\frac{1}{j\omega C}=-j\frac{1}{5000\times 50\times {10}^{-6}}=-j4\Omega$

3. Equivalent Impedance of the Coupled Inductor:
The total impedance of the secondary loop is:
$Z_s=Z_{L2}+Z_C=j5-j4=j1\Omega$
Using the reflected impedance formula, the total input impedance looking into the primary side of the magnetically coupled inductor is:
$Z_{\mathrm{in}}=Z_{L1}+\frac{(\omega M{)}^2}{Z_s}=j625+\frac{{25}^2}{j1}$
Since $\frac{1}{j}=-j$:
$Z_{\mathrm{in}}=j625-j625=0\Omega$

4. Calculate the Thevenin Impedance, $Z_{\mathrm{th}}$:
Now we simplify the network looking into terminals P and Q:

• The coupled inductor acts as a short circuit ($0\Omega$).
• The series branch containing the $2\Omega$ resistor and this coupled inductor has a total impedance of:
  $Z_{\mathrm{branch}}=2+0=2\Omega$
• This branch is in parallel with the vertical $4\Omega$ resistor:
  $Z_{\mathrm{parallel}}=4\parallel 2=\frac{4\times 2}{4+2}=\frac{8}{6}=1.33\Omega$
• Finally, this parallel combination is in series with the $4\Omega$ input resistor connected to terminal P:
  $Z_{\mathrm{th}}=4+1.33=5.33\Omega$