
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
554.4k+ 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.
Recently Updated Pages
Which of the following is not a national political class 12 social science CBSE

State tangent law in magnetism class 12 physics CBSE

Ethylidene chloride and ethylene dichloride are A Chain class 12 chemistry CBSE

Explain how the laws of segregation and the law of class 12 biology CBSE

Which of the following graphs shows the variation of class 12 physics CBSE

For the same angle of incidence in media A B and C class 12 physics CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

When was the first election held in India a 194748 class 12 sst CBSE

