An air-core radio-frequency transformer as shown has a primary winding and a secondary winding. The mutual inductance M between the windings of the transformer is _______ μH. (Round off to 2 decimal places)

The correct answer is 51.12.

Based on the circuit diagram shown in the image:

We are given the following parameters:
- Frequency of the AC source, $f=100\text{kHz}={10}^5\text{Hz}$
- Resistance in the primary circuit, $R=22\text{Ω}$
- Peak-to-peak voltage across the resistor, $V_{R,\text{p-p}}=5.0\text{V}$
- Open-circuit peak-to-peak voltage across the secondary winding, $V_{2,\text{p-p}}=7.3\text{V}$

Step 1: Calculate the primary current
Since the primary winding is in series with the resistor, the peak-to-peak current in the primary circuit ($I_{1,\text{p-p}}$) is determined by the voltage drop across the resistor $R$:

$I_{1,\text{p-p}}=\frac{V_{R,\text{p-p}}}{R}=\frac{5.0}{22}\text{A}$

Step 2: Relation for induced voltage in the secondary winding
The voltage induced in the secondary winding under open-circuit conditions is due to the mutual inductance $M$ and the rate of change of the primary current:

$V_{2,\text{p-p}}=\omega M{I}_{1,\text{p-p}}$

where the angular frequency $\omega$ is:

$\omega =2\pi f=2\pi \times {10}^5\text{rad/s}$

Step 3: Solve for the mutual inductance $M$
Substituting the known values into the voltage equation:

$7.3=(2\pi \times {10}^5)\times M\times \left(\frac{5.0}{22}\right)$

Rearranging the formula to solve for $M$:

$M=\frac{7.3\times 22}{2\pi \times {10}^5\times 5.0}$

$M=\frac{160.6}{10\pi \times {10}^5}=\frac{160.6}{\pi \times {10}^6}\text{H}$

$M\approx 5.11205\times {10}^{-5}\text{H}=51.12\text{μH}$

Rounding off to two decimal places, the mutual inductance between the windings is 51.12 μH.