question_answer

Current \[{{I}_{1}}\] in the following circuit is

A) \[0.4\,A\]

B)                    \[-\,0.4\,A\]

C) \[0.8\,A\]          

D)         \[-\,0.8\,A\]

Correct Answer: B

Solution :

[b] Applying \[{{1}^{st}}\] law of Kirchoffs, \[{{I}_{3}}={{I}_{1}}+{{I}_{2}}\]                              ...(i) Applying 2nd law of Kirchhoff in Mesh ABCFA \[-30{{I}_{1}}+40-40{{I}_{3}}=0\]                  ...(ii) in Mesh FCDEF \[40{{I}_{3}}-40-80+40{{I}_{2}}=0\]   ...(iii) putting \[{{I}_{3}}\] from (i) in (ii) and (iii) we get \[-30{{I}_{1}}+40-40{{I}_{1}}-40{{I}_{2}}=0\] or, \[-70{{I}_{1}}-40{{I}_{2}}=-40\] or,   \[7{{I}_{1}}+4{{I}_{2}}=4\]                     ...(iv) and \[40{{I}_{1}}+40{{I}_{2}}-40-80+40{{I}_{2}}=0\] or, \[40{{I}_{1}}+80{{I}_{2}}=120\] or,  \[4{{I}_{1}}+8{{I}_{2}}=12\] or,    \[2{{I}_{1}}+4{{I}_{2}}=6\]                    ...(v) (iv)\[-\](v)   gives \[5{{I}_{1}}=-2\] \[{{I}_{1}}=-0.4\,\,A\]