In a portion of some large electrical network, current in certain branches are known. The values of \[{V_A}{V_B}\] and \[{V_C}{V_D}\] are and X and Y respectively, where X and Y are:

A. \[X = 29\,{\text{V}},Y = 26\,{\text{V}}\]
B. \[X = 58\,{\text{V}},\,Y = 52\,{\text{V}}\]
C. \[X = - 58\,{\text{V}},\,Y = - 52\,{\text{V}}\]
D. \[X = - 29\,{\text{V}},Y = - 26\,{\text{V}}\]

Hint: Determine the value of the electric current flowing from the branch at point C using Kirchhoff’s current law. Then apply Kirchhoff’s voltage law to the electrical network between the points A and B and C and D to determine the potential difference between points A and B and the points C and D.

Complete step by step answer:
We have given a large electrical network in which there are various resistors, currents flowing and batteries. We have asked to determine the potential difference between the points A and B and the potential difference between the points C and D.

Let us redraw the given electrical network with all the electric current values shown in the network.

Let us first determine the value of the electric current \[{I_x}\] in the above electrical network. Let us apply Kirchhoff’s current law to the given electrical network.
\[7\,{\text{A}} = 2\,{\text{A}} + 3\,{\text{A}} + {I_x}\]
\[ \Rightarrow 7\,{\text{A}} = 5\,{\text{A}} + {I_x}\]
\[ \Rightarrow {I_x} = 7\,{\text{A}} - 5\,{\text{A}}\]
\[ \Rightarrow {I_x} = 2\,{\text{A}}\]
Hence, the current entering the branch of the given network is \[2\,{\text{A}}\].
To determine the potential difference between the points A and B of the given electrical network, we should apply Kirchhoff voltage law between the points A and B.
\[{V_A} - \left( {7\,{\text{A}}} \right)\left( {2\,\Omega } \right) - \left( {3\,{\text{V}}} \right) - \left( {5\,{\text{V}}} \right) - \left( {3\,{\text{A}} + 2\,{\text{A}}} \right)\left( {4\,\Omega } \right) - \left( {4\,{\text{V}}} \right) - \left( {2\,{\text{A}}} \right)\left( {6\,\Omega } \right) - {V_B} = 0\]
Here, \[{V_A}\] is the potential at point A and \[{V_B}\] is the potential at point B.
\[ \Rightarrow {V_A} - 14 - 3 - 5 - 20 - 4 - 12 - {V_B} = 0\]
\[ \Rightarrow {V_A} - 58 - {V_B} = 0\]
\[ \Rightarrow {V_A} - {V_B} = 58\,{\text{V}}\]
Therefore, the potential difference between the points A and B is \[58\,{\text{V}}\]. Hence, the determined value of X is \[58\,{\text{V}}\].
\[X = 58\,{\text{V}}\]
To determine the potential difference between the points A and B of the given electrical network, we should apply Kirchhoff voltage law between the points A and B.
\[{V_C} - \left( {9\,{\text{V}}} \right) + \left( {2\,{\text{A}}} \right)\left( {8\,\Omega } \right) - \left( {5\,{\text{V}}} \right) - \left( {3\,{\text{A}} + 2\,{\text{A}}} \right)\left( {4\,\Omega } \right) - \left( {4\,{\text{V}}} \right) - \left( {3\,{\text{A}}} \right)\left( {10\,\Omega } \right) - {V_D} = 0\]
Here, \[{V_C}\] is the potential at point B and \[{V_D}\] is the potential at point D.
\[ \Rightarrow {V_C} - 9 + 16 - 5 - 20 - 4 - 30 - {V_D} = 0\]
\[ \Rightarrow {V_C} - 52 - {V_D} = 0\]
\[ \Rightarrow {V_C} - {V_D} = 52\,{\text{V}}\]
Therefore, the potential difference between the points C and D is \[52\,{\text{V}}\]. Hence, the determined value of Y is \[52\,{\text{V}}\].
\[\therefore Y = 52\,{\text{V}}\]

Hence, the correct option is B.

Note: The students should not forget to determine the electric current originating from branch at point C in the given electrical network. If this current value is not used while applying Kirchhoff’ voltage and current law, then the final values of the potential differences will be incorrect.