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

A) +3V
B) +2V
C) -2V
D) None of these

Hint: Kirchhoff’s proposed two laws that help in solving complex electronics circuits one is KCL (Kirchhoff’s current Law) and another one is KVL (Kirchhoff’s voltage law). These two laws are used for finding the equivalent resistance.

Complete step by step solution:
We will be using KVL, Kirchhoff has defined the voltage loop law as: In any loop network that is closed the net total voltage in the entire loop is equivalent to the sum of all the voltage drops which is within the loop and is also equal to zero.
Here, consider two loops in the clockwise direction for the two areas. Here the current “I” flowing in the 4V and current ${I_1}$ is flowing in 2V and current $\left( {I - {I_1}} \right)$ is flowing in 1V. Apply KVL in the first loop:
$2 - 2{I_1} + 4 - {I_1} = 0$ ;
$ \Rightarrow 6 - 3{I_1} = 0$;
Do the needed calculation:
$ \Rightarrow {I_1} = \dfrac{6}{3}$;
$ \Rightarrow {I_1} = 2A$;
Now,
${V_{PQ}} = {V_p} - {V_Q}$;
Now, we know that:
V = IR;
Put the above relation in the above equation:
$\Rightarrow {V_{PQ}} = 2 - {I_1}R$;
$\Rightarrow {V_{PQ}} = 2 - \left( 2 \right)\left( 2 \right)$;
The value of ${V_{PQ}}$is:
$ \Rightarrow {V_{PQ}} = + 2V$;

Option (B) is the correct option. Therefore, the potential difference ${V_{PQ}}$ is +2V.

Note: Here, there is no electronic device between the midpoint in the circuit and the point Q, so there is no need to choose a long path in calculating the potential difference, instead choose the shortest path between the point P and Q but before that calculate the current in the circuit by using Kirchhoff’s voltage loop law.