Is my Kirchoff's rules solution with mixed current and voltage source correct?

Question

Problem: Find \$I_{1}, I_{2}, I_{4}, I_{6}\$ using Kirchoff's rule

^{simulate this circuit – Schematic created using CircuitLab}

\$E\$ is the voltage source of 8 Volts and \$I_{general}\$ is the current source of 3 Amps.

My steps:

This is what I got:

I think there are 5 branches and 3 nodes and also 3 independent loops. So the number of needed equations = \$3-1 = 2\$ Using this I got equations:

For node 1: \$-I_{6}+I_{1}+I_{general}\$

For node 3: \$-I_{4}-I_{1}+I_{2}+I_{6}\$

Also I think there are 2 loops needed for equations: the bottom rectangle and the left one:

Loop1: \$1-3-1\$

Loop2: \$3-2-3\$

So we heave 2 more equations:

1) \$I_{1}R_{1}+I_{6}R_{6}=0\$

2) \$I_{2}R_{2}+I_{4}R_{4}+I_{2}R_{3}=E\$

So we got system of equations to solve: \begin{cases} -I_{6}+I_{1}+I_{general}=0 \\ -I_{4}-I_{1}+I_{2}+I_{6}=0 \\ I_{1}R_{1}+I_{6}R_{6}=0 \\ I_{2}R_{2}+I_{4}R_{4}+I_{2}R_{3}=E \end{cases}

Solving which, gives this result:

$$I_{1}=-2.571\\ I_{2}=-1.244\\ I_{4}=1.755\\ I_{6}=0.42$$

Question: Is there anything that I did right? I can get 0 in product of sum of I's but for that I need to change some signs a bit. Did I solve this correctly or there is something wrong in it or completely wrong?

Answer 1

Assuming your bottom node is assigned the value of \$0\:\textrm{V}\$ and assigning \$V_1\$ to your "1" node and \$V_2\$ to your "2" node, I get:

$$\begin{align*} \frac{V_1}{R_1} + \frac{V_1}{R_6} + 3 = 0\;\;\;\therefore V_1&=-3\cdot\left(R_1\vert\vert R_6\right)\\&= -25\frac{5}{7}\:\textrm{V}\\\\ \frac{V_2}{R_2+R_3} + \frac{V_2}{R_4} = 3 + \frac{8\:\textrm{V}}{R_2+R_3}\;\;\;\therefore V_2&=\left(3 + \frac{8\:\textrm{V}}{R_2+R_3}\right)\cdot\left(R_4\vert\vert \left[R_2+R_3\right]\right)\\&= 70.\overline{2}\:\textrm{V} \end{align*}$$

So I get:

$$\begin{align*} I_4=\frac{V_2}{R_4}&= 1.7\overline{5}\:\textrm{A}\\\\ I_1=\frac{V_1}{R_1}&= -2\frac{4}{7}\:\textrm{A}\\\\ I_2=\frac{8\:\textrm{V}-V_2}{R_2+R_3}&=-1.2\overline{4}\:\textrm{A}\\\\ I_6=\frac{-V_1}{R_6}&=\frac{3}{7}\:\textrm{A} \end{align*}$$

In short, I think you did just fine.