To the Ineffable All,
Some of you got the right answer and some of you did not. Nodal and mesh methods are powerful, but it is like shooting a fly with a howitzer gun. Why get involved with multiple simultaneous equations or matrices when it is unnecessary? I will use the combination Thevenin/Norton theorem, and you all judge whether is is simpler and easier.
Calculate the voltage across Rx when its value is changed to infinity (35 volts). Calculate the current present in Rx when its value is change to zero (7/3 amps). The Thevenin resistance is thereby 35/(7/3) = 15 ohms. Therefore, we are looking at a Thevenin circuit of 35 volts in series with 15 ohms. Now, simply calculate Rx to give a series current of 1.5 amps (35/(15+Rx) = 1.5). Rx is easily found to be 8.33... .
Ratch
The OP talked about using the mesh method and since the problem is clearly homework, using a different procedure might not count.
Simon posted a problem a while back that required complex numbers and seemed rather difficult to do, with or without shortcuts. But methodically writing the mesh or node equations results in an answer:
https://www.eevblog.com/forum/beginners/mesh-analysis/On page 6 (at the end), I reduced this problem to 2 equations for nodal analysis and on page 7 there is a mesh analysis with a bunch more equations.
The fact that the components include capacitors and inductors matters not at all. Even the phase shift in the sources doesn't matter as long as the computer is doing the work. I would not like to solve this by hand. Been there, done that - with a sliderule.
That was a fun thread! It took awhile to get back into the groove since I graduated in '73 and never used the stuff again.
I solved it by writing one equation:
(20 - 1.5 Rx) / 30 + (50 - 1.5 Rx) / 30 = 1.5
This rapidly gives
70 - 3 Rx = 45
From which
Rx = (70 - 45) / 3 = 25/3 = 8.33
For me, that was simple and direct and was the work of a few seconds.
Part of the art of solving problems is to be able to look at them and see which path will lead to the quickest answer with least effort. This comes from practice in solving many problems, but sometimes brute force can be simplest.
SORRY GUYS.
I miss read the results table
it is 8.333 somthing
I think i was looking at and changing I(V2): instead of I(R2):
--- Operating Point ---
V(n003): 12.4997 voltage
V(n001): 20 voltage
V(n002): 50 voltage
I(R3): 1.25001 device_current
I(R2): -1.50002 device_current
I(R1): -0.250011 device_current
I(V2): -1.25001 device_current
I(V1): -0.250011 device_current
My apologies.
BILL.
it is 8.333 somthing
Correct, where the "somthing" is just more 33333333333333333333...