0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

At node 1, we have:$$0.001+i_1-i_4-i_5=0$$. At node 2, we have: $$i_4+i_5+i_2=0$$.

$$i_1=\frac{0-V_1}{1000}=\frac{-V_x}{1000}$$(Vx is the voltage across R2) $$i_2=\frac{0-V_2}{6000}$$ $$i_3=0.001A$$ $$i_4=i_1+i_3=-i_2-i_5$$ $$i_5=\frac{V_1-V_2}{1000}=\frac{V_x}{1000}$$

Use the equation at node 1, to find an expression for i4 and plug that into the equation at node 2: $$i_1+0.001+i_2=0$$ $$\frac{-V_x}{1000}+0.001-\frac{V_2}{6000}=0$$

But there are V2 and Vx to solve in one equation.

\$\endgroup\$
2
  • \$\begingroup\$ Your first equation(at node 1) should also include \$i_5\$ (going out of the node). Also, your expression for \$i_1\$ should be \$-\frac{V_x}{1000}\$ look at the polarity of \$V_x\$ with respect to \$i_1\$ \$\endgroup\$
    – Big6
    Commented Sep 24, 2017 at 23:41
  • \$\begingroup\$ It still not solvable. there is one equation at the end, but Vx and V2 \$\endgroup\$ Commented Sep 25, 2017 at 0:01

1 Answer 1

0
\$\begingroup\$

You have too much variables and arrows (not wrong, but unnecessary). I redrawed the circuit:

enter image description here

You obviously missed that the voltage over R4 can be expressed in terms of Vx. This is Kirchoffs voltage law.

I have written the current equations for nodes 1 and 2. There are 2 unknowns and 2 equations.

Hopefully clear now.

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.