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:


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:


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?

  • \$\begingroup\$ You've lost a term divided by R1 somewhere. Clearly R1 has to be an influence on the answer. There should also be some influence by R4 because if R4 was zero ohms v = V_e \$\endgroup\$
    – Andy aka
    May 15, 2015 at 18:12
  • 5
    \$\begingroup\$ $$e_1=v_e+\frac{e_1-v_e}{R1}$$ in this line you are adding a current and a voltage which isn't correct \$\endgroup\$
    – Sada93
    May 15, 2015 at 18:14
  • \$\begingroup\$ @Sada93 You're right. I didn't see that. So the entire rest of the calculations is totally wrong. \$\endgroup\$
    – sl34x
    May 15, 2015 at 18:16
  • \$\begingroup\$ Yeah. Whenever you see a bold statement such as $ e_1 = v_e $ it cant be correct as there has to be a voltage drop across those resistors. \$\endgroup\$
    – Sada93
    May 15, 2015 at 18:18

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Node 0 is your reference.


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.


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