Electronic – Total resistance of circuit

First of all, the circuit drawn is a high-pass filter.

At very high frequencies, the inductor has very high impedance and thus, the output \$V_{ab}\$ should essentially equal the input.

At very low frequencies, the inductor has very low impedance and thus, the output should be essentially zero.

The frequency where the output voltage is \$\frac{1}{\sqrt{2}}\$ times the input voltage is given by

$$f_0 = \frac{R}{2\pi L} $$

So, either the circuit drawn does not reflect the circuit measured or something went terribly wrong with your measurement apparatus (or operator of said apparatus).

For completeness, the transfer function is, by inspection:

$$H(j\omega) = \frac{j\omega L/R}{1 + j\omega L/R}\quad,\quad\omega = 2\pi f$$

$$|H(f)| = \frac{2\pi fL/R}{\sqrt{1 + (2\pi fL/R)^2}} $$

In this case whenever you are calculating the circuit resistance you want the total resistance seen by your only voltage source (18 V battery)

you better start from the node C, Such that i can see a 1 KΩ series with another 1 KΩ

The equivalent resistance of this combination is a 2 KΩ resistance so your circuit now looks like this

i can see now a 2.2 KΩ (Node B) parallel to a 2 KΩ (Node C)

We can now simply estimate the equivalent resistance of this combination which should be equals to

Now we place this resistance at node (B) and remove the node (C) branch (Open circuit) or vice verse

Such that your new circuit will look like this

Now i can see a 22/21 KΩ Series with a 1 KΩ resistance

Such that they are combined into a new resistance of value 43/21 KΩ

Your circuit will look something like this

Now this circuit can be easily solved you have a 43/21 KΩ parallel with a 2.2 KΩ
and this whole combination equivalent resistance is series with the 5.6 KΩ resistance