In the circuit shown, what is the potential difference $V_{P_Q}$?

 

$V_{MN}=4$ Volt on the basis of this currents in different resistors can be obtained.

Now $\begin{array}{l}{V}_P-1X\frac{2}{3}+4=V_Q\\ V_{P-}{V}_Q=\frac{2}{3}-4=-\frac{10}{3}V\end{array}$