Current Source in a circuit
There's a nice schematic editor on this site. You should use it:







simulate this circuit &ndash; Schematic created using CircuitLab

I've labeled the nodes above and made a choice about which one I decided to call \$0\:\textrm{V}\$. You really only need to know the values for \$V_X\$, \$V_Z\$, and \$R_4\$ to figure out the current, \$I\$, using the obvious equation for the given arrow direction, \$I=\frac{V_Z-V_X}{R_4}\$. \$V_Y\$ isn't needed (though you could include it.) So you only need to solve the following:

$$\begin{align*}
\frac{V_X}{R_2}+\frac{V_X}{R_4} &amp;= \frac{V}{R_2}+\frac{V_Z}{R_4} + I_1\\\\ \frac{V_Z}{R_3}+\frac{V_Z}{R_4}+I_1 &amp;= \frac{V_X}{R_4}
\end{align*}$$

In matrix form, it's:

$$\begin{align*}
V_X\cdot\left(\frac{1}{R_2}+\frac{1}{R_4}\right)+V_Z\cdot\frac{-1}{R_4}&amp;=\frac{V}{R_2}+I_1\\\\V_X\cdot\frac{-1}{R_4}+V_Z\cdot\left(\frac{1}{R_3}+\frac{1}{R_4}\right)&amp;=-I_1
\end{align*}$$

And you can either solve that symbolically or numerically, if you have values handy. With the values in hand, you can then figure out \$I\$.



A symbolic solution could be done in the following way. Multiply equation 1 by \$R_2\vert\vert R_4\$:

$$\begin{align*}
V_X\cdot 1+V_Z\cdot\frac{-\left(R_2\vert\vert R_4\right)}{R_4}&amp;=\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right)
\\\\
V_X\cdot\frac{-1}{R_4}+V_Z\cdot\left(\frac{1}{R_3}+\frac{1}{R_4}\right)&amp;=-I_1
\end{align*}$$

Now, zero out \$V_X\$ in the second equation by adding to it the first equation multiplied by \$\frac{1}{R_4}\$:

$$\begin{align*}
V_X\cdot 1+V_Z\cdot\frac{-\left(R_2\vert\vert R_4\right)}{R_4}&amp;=\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right)
\\\\
V_X\cdot 0+V_Z\cdot\left(\frac{1}{R_3}+\frac{1}{R_4}-\frac{\left(R_2\vert\vert R_4\right)}{R_4^2}\right)&amp;=-I_1+\frac{R_2\vert\vert R_4}{R_4}\cdot\left(\frac{V}{R_2}+I_1\right)
\end{align*}$$

Divide the second equation so that \$V_Z\$ is multiplied by 1:

$$\begin{align*}
V_X\cdot 1+V_Z\cdot\frac{-\left(R_2\vert\vert R_4\right)}{R_4}&amp;=\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right)
\\\\
V_X\cdot 0+V_Z\cdot 1&amp;=\frac{-I_1+\frac{R_2\vert\vert R_4}{R_4}\cdot\left(\frac{V}{R_2}+I_1\right)}{\frac{1}{R_3}+\frac{1}{R_4}-\frac{\left(R_2\vert\vert R_4\right)}{R_4^2}}
\end{align*}$$

Now, multiply the second equation by \$\frac{R_2\vert\vert R_4}{R_4}\$ and add that into the first equation to zero out \$V_Z\$:

$$\begin{align*}
V_X&amp;=\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right)+\frac{R_2\vert\vert R_4}{R_4}\cdot V_Z
\\\\
&amp;=\left(R_2\vert\vert R_4\right)\cdot\left[\frac{V}{R_2}+\frac{V_Z}{R_4}+I_1\right]
\\\\
V_Z&amp;=\frac{-I_1\cdot R_4+\left(R_2\vert\vert R_4\right)\cdot\left(\frac{V}{R_2}+I_1\right)}{1+\frac{R_4}{R_3}-\frac{\left(R_2\vert\vert R_4\right)}{R_4}}
\end{align*}$$

Remember that \$I=\frac{V_Z-V_X}{R_4}\$, so if you are really wanting to have fun with algebra, plug in the values and simplify further.