Answer
$V_Q=-10V$
Work Step by Step
Let us suppose that the current direction is counter clockwise. Starting from point Q, according to loop rule
$\epsilon_1-iR_2-\epsilon_2-iR_1=0$
$\epsilon_1-\epsilon_2=iR_1+iR_2$
$\epsilon_1-\epsilon_2=i(R_1+R_2)$
$i=\frac{\epsilon_1-\epsilon_2}{R_1+R_2}$
We plug in the known values to obtain:
$i=\frac{150-50}{3+2}=\frac{100}{5}=20A$
As $i=20>0$, we know our assumption is correct. Now, starting from point Q in counter clockwise direction:
$V_Q+\epsilon_1-iR_2=V_P$
$V_Q=V_P-\epsilon_1+iR_2$
We plug in the known values to obtain:
$V_Q=100-150+20(2)$
$V_Q=-10V$
