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OK. I think a key is understanding that your voltage source is an ideal
voltage source plus its internal resistance, which with test equipment
would generally be a well defined 50 ohms. I had to bang this into my head
when I started measuring crystal parameters.

Now, if you've accepted that you don't even need to draw the L and C in
your circuit (because their reactances cancel), you can just leave them out
and draw a series circuit of: V source = 1 v, Rg = 50 and Rloss which is
unknown. (Rg means R of generator.)

Write the equation for current using Ohm's law. I = E/R where I is 1 V and
R is 50 + Rloss.

So I = 1/(50 + Rloss)

You already know that voltage across Rloss is 0.26 volts and it's also
equal to Rloss times the current. So we can say,

0.260 = Rloss * 1/(50 + Rloss)

Manipulate that equation with a little algebra. My steps would look like
this:

Rloss = 0.26*(50+Rloss), then

0.74*Rloss = 13

Rloss = 13/0.74 = 17.57 ohms

73-

Nick, WA5BDU


On Sun, Sep 8, 2019 at 10:50 AM Steve Ratzlaff <ratzlaffsteve@...>
wrote:

Hi Nick,

I still don't understand how to solve the problem. Please show this
math-challenged dummy all the steps needed to solve the problem.

Thanks.

73,

Steve AA7U