Voltage calculation with the nodal analysis

Question

I'm having problems with the calulation of \$v\$ in this circuit.

^{simulate this circuit – Schematic created using CircuitLab}

With the nodal analysis (K.C.L. at the node 1) I got:

$$\frac{e_1-v_e}{R_1}+\frac{e_1}{R_2}+\frac{e_1-v}{R_3}=0$$

Now: $$e_1=v_e+\frac{e_1-v_e}{R_1}\implies e_1\left(1-\frac{1}{R_1}\right)=v_e\left(1-\frac{1}{R_1}\right)\implies e_1=v_e$$

Then the previous equation becomes:

$$\frac{v_e-v_e}{R_1}+\frac{v_e}{R_2}+\frac{v_e-v}{R_3}=0\implies\frac{v_e}{R_2}+\frac{v_e}{R_3}=\frac{v}{R_3}=>v=v_e\frac{R_3}{R_2}+v_e$$

Now, since the (given) quantities are: \$v_e=30\text{V}\$,\$R_2=2\text{k}\Omega\$ and \$R_3=0.8\text{k}\Omega\$ I get \$v=30\frac{0.8}{2}+30=30\frac{8}{10}\frac{1}{2}+30=12+30=42\$V, which is wrong because the solution reports it as \$v=10\$V.

Could anybody kindly tell me where am I wrong?

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

Node 0 is your reference.

$$\frac{e_1-v_e}{R_1}+\frac{e_1}{R_2}+\frac{e_1-e_3}{R_3}=0$$

Node 2, there is no need for KCL because of the voltage source.

$$e_2 = v_e$$

Node 3.

$$\frac{e_3-e_1}{R_3} + \frac{e_3 - v_e}{R_4}=0$$

Two equations, two unknowns.

Assume unknown currents leave node.