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**Hint:**In this question we have been asked to calculate the potential difference between the two points of the given part of circuit. To solve this question, we shall calculate the current flowing through each arm and later using Ohm's law, we shall calculate the potential across the given points. While solving we must remember that the potential difference in the direction of current is taken as negative.

**Complete answer:**

In the given diagram, the current in the arm PO is not given. Therefore, we shall calculate the current in the arm PO.

Now, in the given figure, we can see that the current in arm OM is 6 Amperes. This current is then split between arm PO and OQ. The current flowing in arm OQ is 3 Amperes. Therefore, the remaining current will flow in the arm OP. The magnitude of the current shall be 3 Amperes and the direction as shown in the figure below.

Now, using Kirchhoff’s rule

We can say that,

\[{{V}_{P}}-10V+3\times 1-3\times 1-3\times 3={{V}_{Q}}\]

Therefore, on solving

We get,

\[{{V}_{P}}-{{V}_{Q}}=10+3-3+9\]

Therefore,

\[{{V}_{P}}-{{V}_{Q}}=19V\]

Therefore, the potential across points PQ is 19 Volts.

**Note:**

The Ohm’s Law states that the current in a circuit is directly proportional to the voltage applied and inversely proportional to the total resistance in the circuit. Potential difference or voltage can be defined as the amount of work needed per unit charge to move that charge between two given points. The unit of potential difference was named after Alessandro Volta.

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