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.You need to login to perform this action.
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