How can I find the ratio $\frac{V_2}{V_1}$ here? 
Find the $\frac{V_2}{V_1}$ ratio


The answer is $\frac{V_2}{V_1}=\frac{8}{23}$ but I'm not very familiar with circuits and I can't understand how book solves. I (maybe naively) tried simply to use Kirchhoff laws with $V_1$ and $V_2$ unknown (are, in principle, all solvable circuit problems, solvable with Kirchhoff laws?). Using these directions for currents and for unknown $V$.

My attempt
I wrote
\begin{equation}
\begin{cases}
-i_1+2i_2+V_1 = 0 \\ 
-3 i_1 - 3 i_2 + 4 i_3 - 2i_2 = 0\\
-V_2 - 4 i_3 = 0
\end{cases}
\end{equation}
This doesn't look so useful, because $V_1$ and $V_2$ unknown. Kirchhoff laws allows to write 3 equations, but we have too many unknowns here! Exploiting the system I can for example say that
\begin{equation}
\frac{V_1}{V_2} = \frac{1}{3}-\frac{11 }{12 } \cdot \frac{ i_2}{ i_3}
\end{equation}
but this doesn't answer the question if I don't know the ratio $\frac{i_2}{i_3}$.

 A: 
Kirchhoff laws allows to write 3 equations, but we have too many unknowns here!

The problem in your approach is that you assume voltage sources connected to both $V_1$ and $V_2$. With that, any ratio $V_2/V_1$ is possible. Try solving the problem by connecting the voltage source on $V_1$ only, and think of $V_2$ just as voltage measurement (nothing is connected there). Or vise-versa, voltage source on $V_2$, and $V_1$ is voltage measurement.

Since your target are voltages, the best is to use Millman method (see circuit below for reference)
$$\varphi_2 \bigl( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \bigr) = \varphi_1 \frac{1}{R_1} + \varphi_0 \frac{1}{R_2} + \varphi_3 \frac{1}{R_3}$$
$$\varphi_3 \bigl( \frac{1}{R_3} + \frac{1}{R_4} \bigr) = \varphi_2 \frac{1}{R_3} + \frac{1}{R_4} \varphi_0$$
$$\varphi_1 - \varphi_0 = V_1$$
This gives 3 equations and 4 unknowns ($\varphi_0$, $\varphi_1$, $\varphi_2$, $\varphi_3$). Since potential difference is what it matters, you can set one of the potentials to any value you want, e.g. $\varphi_0 = 0$. With this you can express all remaining potentials as a function of $V_1$. Note that $V_2 = \varphi_3 - \varphi_0$, and with $\varphi_0 = 0$ it becomes $V_2 = \varphi_3$. From this you can find the ratio $V_2/V_1$.

If you still insist on solving this problem via currents, then the equations are
$$i_1 = i_2 + i_3$$
$$V_1 - i_1 R_1 - i_2 R_2 = 0$$
$$-i_2 R_2 + i_3 R_3 + i_3 R_4 = 0$$
From the above equations you can find expressions for all three currents ($i_1$, $i_2$, $i_3$) as a function of voltage $V_1$. By noting that $V_2 = i_3 R_4$ it is trivial to find the ratio $V_2/V_1$.

A: You have done most of the work right. There is a small part you are missing.
Assume that some device is supplying $V_{1}$. Another device is supplying $V_{2}$.
You will immediately get $i_{1} = i_{4}$.
You should get the result with this...
