# Voltage calculation with the nodal analysis

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?

• 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 May 15, 2015 at 18:12
• $$e_1=v_e+\frac{e_1-v_e}{R1}$$ in this line you are adding a current and a voltage which isn't correct May 15, 2015 at 18:14
• @Sada93 You're right. I didn't see that. So the entire rest of the calculations is totally wrong. May 15, 2015 at 18:16
• 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. May 15, 2015 at 18:18

## 1 Answer

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.