question_answer

The potential difference between points \[\text{A}\]    and \[\text{B}\] is:                                                       [BHU M-2003]

A)  \[\frac{20}{7}V\]                                            

B)  \[\frac{40}{7}V\]

C)  \[\frac{10}{7}V\]                                            

D)   Zero

Correct Answer: D

Solution :

                 Key Idea: When resistors are connected in series same current flows through them. The given circuit can be redrawn as follows: In this circuit, resistances of \[8\,\Omega \] and \[6\,\Omega \] are connected in series and resistances of \[4\,\Omega \] and\[3\,\Omega \] are also connected in series. Hence, \[R'=8\,\Omega +6\,\Omega =14\,\Omega \] \[R''=4\,\Omega +3\,\Omega =7\,\Omega \] The given circuit now reduces to as shown : \[{{i}_{1}}=\frac{10}{7}A\] \[{{i}_{2}}=\frac{10}{14}A\] From Kirchhoff?s law                                 \[{{V}_{B}}-{{V}_{A}}=8{{i}_{2}}-4{{i}_{1}}\]                                 \[{{V}_{B}}-{{V}_{A}}=8\times \frac{10}{14}-4\times \frac{10}{7}=0\] Hence, potential difference between points A and B is zero.