A) 1.33 A
B) zero
C) 2.00 A
D) 1.00 A
Correct Answer: A
Solution :
The circuit diagram can be redrawn as
For the loop ABCDA, \[+2-2I+2-2I=0\] ?(i) For the loop ABFEA, \[2-2I+2-2(I-I)=0\] \[4-2I-2I+2I=0\] \[4-4I+2I=0\] \[2=2I-I\] ?(ii) From Eqs. (i) and (ii), we get \[2=2I-I,2=I+I\] \[4=3I,\,I=4/3=1.33\,A\]
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