Probably the canonical way they want you to solve it, Thevenin/Norton equivalents:
- Recognize the sources are VSRC + R Thevenin circuits.
- Transform them to Norton equivalents.
- 20V + 30R --> 2/3 A || 30R
- 50V + 30R --> 5/3 A || 30R
- Now the entire circuit is parallel. Three resistors in parallel, and two current sources in parallel.
- The unknown resistor is drawing 3/2 A (given). The node is supplied with 2/3 + 5/3 = 7/3 A. That leaves 5/6 A total flowing into the two parallel 30R resistors.
- 5/6A * (30R || 30R) = 5/6A * (15R) = 12.5V
- Rx = 12.5V / 1.5A = 8 1/3 R.
(Using fractions just because it's better than writing out decimal thirds.)
Or you could do nodal analysis, or superposition, or whatever.
However, the special-case observation notes that:
- Two Thevenin sources are in parallel, and have equal resistances.
- The equivalent of these is the average of the voltages, and half the resistance. (More generally, any number of parallel Thevenin sources has an equivalent where the voltage is the weighted average, and the resistance is all resistors in parallel. It is up to the student to prove this...)
- Halfway between 20 and 50V is 35V. Half of 30R is 15R.
- 1.5A into 15R is 22.5V, subtract that from 35V (= 12.5V).
- 12.5V / 1.5A = 8 1/3 ohm.
Tim