question_answer

What is the current flowing in arm\[AB\]?

A) \[\frac{35}{4}A\]                             

B) \[\frac{13}{7}A\]

C) \[\frac{5}{7}A\]                                

D) \[\frac{7}{5}A\]

Correct Answer: B

Solution :

Key Idea: As per Kirchhoff's law\[\Sigma iR=V\] Let \[{{i}_{1}},\,\,{{i}_{2}}\] be current in the two loops of the given circuit then Applying Kirchhoff?s law to loop\[1\], we get                 \[(\Sigma iR=V)\]                 \[8{{i}_{1}}+2({{i}_{1}}+{{i}_{2}})=10\] \[\Rightarrow \]                      \[10{{i}_{1}}+2{{i}_{2}}=10\] \[\Rightarrow \]                         \[5{{i}_{2}}+{{i}_{2}}=5\]                     ? (i) Applying Kirchhoff?s law to loop\[2\], we get             \[4{{i}_{2}}+2({{i}_{1}}+{{i}_{2}})=8\] \[\Rightarrow \]                      \[6{{i}_{2}}+2{{i}_{1}}=8\] \[\Rightarrow \]                        \[{{i}_{1}}+3{{i}_{2}}=4\]                                      ? (ii) From Eq. (i) and (ii), we get                 \[{{i}_{1}}=\frac{11}{14}A,\,\,{{i}_{2}}=\frac{15}{14}A\] Hence current flowing is arm \[AB\] is                 \[{{i}_{1}}+{{i}_{2}}=\frac{11}{14}+\frac{15}{14}=\frac{26}{14}=\frac{13}{7}A\]