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 \[{{i}_{1}}+{{i}_{2}}=\frac{11}{14}+\frac{15}{14}=\frac{26}{14}=\frac{13}{7}A\]You need to login to perform this action.
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