Question

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 1

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.