Electrical – Applying Cramer’s rule to find out the transfer function of the circuit

Your expression for \$H(s)\$ is correct:

$$H(s)=\frac{RCs}{1+RCs}$$

where \$\tau=RC\$ is the time constant of the high pass filter. The most important features of this transfer function are the location of the pole (\$s_{\infty}=-1/RC\$) and the location of the zero (\$s_0=0\$). Note that the location of the pole and the zero is of course the same for the transfer function in your lecture notes. However, the gain factor in your notes is wrong. The consequence of this is that for \$s\rightarrow\infty\$ (i.e. for very high frequencies) your transfer function converges to unity gain, whereas the one in your notes converges to \$1/RC\$.

Mutual inductance (M) and coupling factor (k): -

So if you have two inductors of 10uH each coupled at 70% then M = 7uH.

This means that the 10 uH coupled inductor becomes a 100% coupled transformer with leakage inductance of 3 uH in each limb.

Because the transformer is 1:1 (in my example) you can simply electrically connect the secondary components (including the extra series 3 uH leakage inductance) across the "now" 7 uH primary. Here's what it should now look like: -

^{simulate this circuit – Schematic created using CircuitLab}

If the turns ratio wasn't 1:1 (i.e. both coupled inductors were not identical in value) then it becomes a little trickier because you have to impedance transpose the components on the secondary (including the new secondary inductor due to leakage) before you can connect them across the primary inductor.

This wiki page shows how a mutually coupled arrangement of inductors is equivalent to this: -

So if you start with L1=10 uH as per my example and have 7 uH mutual coupling you end up with a common inductor of 7 uH and two teed off inductances of 3 uH as an equivalent model.