
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. \[X = - 29\,{\text{V}},Y = - 26\,{\text{V}}\]

Answer
506.7k+ views
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.
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.
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