A) \[{{\varepsilon }_{1}}-\left( {{i}_{1}}+{{i}_{2}} \right)R-{{i}_{1}}{{r}_{1}}=0\]
B) \[{{\varepsilon }_{2}}-{{i}_{1}}{{i}_{2}}R-{{\varepsilon }_{1}}-{{i}_{1}}{{r}_{1}}=0\]
C) \[-{{\varepsilon }_{2}}-\left( {{i}_{1}}+{{i}_{2}} \right)R+{{i}_{2}}{{r}_{2}}=0\]
D) \[{{\varepsilon }_{1}}-\left( {{i}_{1}}+{{i}_{2}} \right)R+{{i}_{1}}{{r}_{1}}=0\]
Correct Answer: A
Solution :
Apply Kirchhoffs second law, also called loop rule. The algebraic sum of the changes in potential in complete transversal of a mesh (closed loop) is zero. ie, \[\sum{V=0}\] Here \[{{\varepsilon }_{1}}-\left( {{i}_{1}}+{{i}_{2}} \right)R-{{i}_{1}}{{r}_{1}}=0\]
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