question_answer

The current I in B circuit is           

A)  0.4 A           

B)  \[-0.4\text{ }A\]        

C)  0.8 A                          

D)  \[-0.08\text{ }A\]                   

Correct Answer: B

Solution :

 At junction A, \[{{I}_{3}}={{I}_{1}}+{{I}_{2}}\] ?.(i) In loop ABCDA, from the Kirchhoffs Voltage Law \[-30{{I}_{1}}-40{{I}_{3}}+40=0\] \[-30{{I}_{1}}-40({{I}_{1}}+{{I}_{2}})+40=0\] \[7{{I}_{1}}+4{{I}_{2}}=4\]   ...(ii) In loop ADEFA, from the Kirchhoffs Voltage Law \[-40{{I}_{2}}-40({{I}_{1}}-{{I}_{2}})=-120\] \[{{I}_{1}}+{{I}_{2}}=3\]        ...(iii) From Eqs. (i) and (ii), we get \[{{I}_{1}}=-0.4A\]