Let's suppose we want to find $I_x$ e $V_y$ here
$I_x=2\,$A is trivial, but book says $V_y=-4\,$V, Why? Is $V_y$ fixed if I don't know anything else about the elements of the circuit, except what is given by this graph? After all every elements has its own resistance and (except the pure resistance) a emf. Let's consider this
Resistances are not given by the graph of the problem so I suppose $V_y$ is independent from resistances: I suppose that components, no matter the resistance, play with their emfs to adjust current and satisfy the purpose to which they were projected (a current 3 times the other current, etc). So I suppose for simplicity that all the resistances have the same value $R_a=R_b=R_c=R$ where $R$ is the upper resistance (the resistance alone on the top): this is the probably the simpler choice. Now I impose (the equation are dimensionally correct, suppose a dimensional constant fix this, for example $emf=2R$ means "if $R=1\,\Omega$ then $emf=2\,$V" etc)
\begin{equation}
emf_a=emf_c=6R \qquad emf_b=8R
\end{equation}
So the circuit is solved by (I use 2nd Kirchhoff law by starting from up-left in each loop)
\begin{cases}
8R – i_1 R + 6 R – (i_1+i_2) R = 0 \\
– i_2 R + 6 R – i_2 R + i_1 R – 8 R = 0
\end{cases}
If $R\neq 0$ this system gives $i_1=6$ and $i_2=2$. All conditions are satisfied (no matter $R$):
-
on the left a up current of $8\,$ A
-
in the center a down current that is $3$ times the current going right through the alone resistance
-
$V_G – V_F$ is $2$ times $V_B – V_A$
But $V_y$ (i.e. $V_B-V_A$) is $- 2R$ (in the sense "if $R=1\, \Omega$ then $V_y=-2\,$V"). Can so $V_y$ have any negative value depending on resistances of elements?
Edit
Thank you for the answer and to explaim me this customary convention that I didn't know. But still I can't understand how to solve this problem. Assuming that the 3 components (let's call them $a$, $b$ and $c$) have a emf but a negligible resistance, the solving system is (I use $E$ for emf, and I suppose all emf toward down; furthermore I use current as in my 2nd scheme)
\begin{equation}
\begin{cases}
E_b – E_a = 0 \\
-i_2 R + E_c – E_b = 0 \\
i_1 + i_2 = 8 \\
i_1 = 3 i_2 \\
E_c = 2 E_a
\end{cases}
\end{equation}
This give currents $i_1=6$ and $i_2=2$, but also
\begin{equation}
E_a = 2 R \qquad E_b = 2 R \qquad E_c = 4 R
\end{equation}
So $V_y = -2 R$, that is $-4\,$V only if $R=2\,\Omega$.
Question: isn't in the problem missing the data $R=2\,\Omega$? Is it possible to solve the problem if $R$ is not given, if all is given is the first graph I put in the question?
Best Answer
It is customary to assume that the independent and dependent source resistances are all zero unless otherwise stated.
Considering the resistor one can write $V_{\rm y} - 2 V_{\rm y}= I_{\rm x}R \Rightarrow V_{\rm y} = - I_{\rm x}R$ where $R$ is the resistance of the resistor which, given the book solution, must be $2\,\Omega$.