A) \[4\,A\]
B) \[6\,A\]
C) \[3\,A\]
D) \[5\,A\]
Correct Answer: A
Solution :
From Kirchhoff?s first law, in an electric circuit the algebraic sum of the currents meeting at any junction is zero, ie, \[\Sigma i=0\]
\[\therefore \] Taking inward direction of current as positive and outward as negative, we have \[1A-3A-2A+I=0\] \[\Rightarrow \] \[I=4A\]
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