Solution: This problem is based on the inverse derivation of the node voltage method, that is, the node voltage equation is already available and the original circuit diagram is needed.

Node A: The equation can be changed to (VA-VB)/2+VA= 1. Therefore, the 1A current source connected to the node points to the node a; There is a 2 Ω resistor between nodes A and B; The resistance between nodes A and D (ground) is 1ω.

Node C: (VC-VB)/ 1+VC/2 = 0. The resistance between nodes B and C is 1ω, and the resistance between nodes C and D (ground) is 2ω.

The general structure of the circuit thus obtained is shown in the left figure:

List the voltage equation of node B: (VB-VA)/2+VB/R+(VB-VC)/1= 0, and expand it to get:

-VA+(3+2/R)VB-2VC=0. So: 3+2/R=5, we get: r = 1 (ω).

Change it to B grounding, as shown on the right.

Node A: (VD-VA)/1+1= VA/2, simplified as: 3VA-2VD=2.

Node D: VD/1+(VD-VA)/1+1+(VD-VC)/2 = 0. Simplification: 2VA+VC-5VD=2.

Node C: VC/ 1+(VC-VD)/2 = 0. Simplified: 3VC-VD=0.