Invisible Transmission Line

We will calculate the equivalent impedance of the circuit, as seen from the source side. We will make it equal to the load resistance to ensure that the current is the same. We will solve the equation to the load resistance. .

First, we have the load resistor and the capacitor in parallel

$Z_1= \frac{-jX_C R}{R-jX_C}$

This is in series with the inductance:

$Z_2= jX_L+\frac{-jX_C R}{R-jX_C}$

Finally this is parallel to the other capacitor:

$Z_3= \frac{-jX_C Z_2}{Z_2-jX_C}= \frac{-jX_C \left( jX_L+\frac{-jX_C R}{R-jX_C}\right)}{ jX_L+\frac{-jX_C R}{R-jX_C} -jX_C}$

For the current to be the same this has to be equal to $R$ , $Z_3=R$ , or

$-jX_C \left( jX_L+\frac{-jX_C R}{R-jX_C}\right)= R \left(jX_L+\frac{-jX_C R}{R-jX_C} -jX_C\right)$

or

$X_LX_C - \frac{X_C^2R}{R-jX_C}=jX_L R -jX_C R- \frac{jX_C R^2}{R-jX_C}$

or

$X_LX_C({R-jX_C}) - {X_C^2R}=(jX_L R -jX_C R)({R-jX_C})- {jX_C R^2}$

or

$X_LX_C R-jX_LX_C^2 - X_C^2R=jX_L R^2 -jX_C R^2 +X_L X_CR - X_C^2R - jX_C R^2$

There are several terms that cancel each other and we are left with:

$-jX_LX_C^2 =j(X_L - 2X_C)R^2$

Therefore the resistance is

$R = \sqrt{\frac{ X_LX_C^2}{2X_C-X_L }}=\sqrt{\frac{ X_LX_C}{2-X_L/X_C }}$

Here is a sketch of the solution. The actual work is fairly tedious, since it has to be kept symbolic the entire way.

1) We know that for a particular value of the resistance, the source draws current as if that resistance is the only thing connected between its terminals.

$I_S = \frac{V_S}{R}$

2) Calculate the inductor current:

$I_L = I_S - \frac{V_S}{-j X_C}$

3) Calculate the voltage across the resistor:

$V_R = V_S - I_L (j X_L)$

4) Calculate the current through the resistor:

$I_R = I_L - \frac{V_R}{-j X_C}$

5) Determine an alternate expression for the voltage across the resistor:

$V_R = I_R \, R$

6) Equate the two expressions for the voltage across the resistor:

$I_R \, R = V_S - I_L (j X_L)$

7) Solving the last expression for the resistance yields:

$\large{R = \sqrt{\frac{X_L X_C}{2 - \frac{X_L}{X_C}}}}$

A numerical parameter sweep of the resistance confirms the legitimacy of this result.