Solve the circuit by nodal analysis and find  V_a .
Circuit with Four Nodes

Solution
a) Choose a reference node, label node voltages:
Choosing a reference node and labelling node voltages

b) Apply KCL to each node:

Node 1:
 -Is_2 +\frac{V_1 - V_2}{R_3}+\frac{V_1 - V_3}{R_1}=0 \to 6V_1 - V_2 - 5V_3=5 (1)

Node 2:
 Is_3 +\frac{V_2 - V_1}{ R_3} + \frac{V_2}{R_2}=0 \rightarrow - 4V_1+9V_2=-40 (2)

Node 3:
 -Is_1 - Is_3 +\frac{V_3 - V_1}{ R_1}=0 \rightarrow V_3 - V_1=4 (3)
(1), (2) and (3) imply that  V_1=37v, \, V_2= 12v and  V_3=41 .

c) Find the required quantities:
Applying KVL
If we apply KVL in the loop shown above:

 -V_1 - V_a +V_2=0 \rightarrow V_a= -25 v .

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