What is the current in branch AB of the circuit shown?

1.5 A

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2 A

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1.33 A

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Solution

The correct option is A $1.5 A$
Apply Kirchhoff's law
for loop 1: $3 i_{1} - 3 i_{2} = 3 \Rightarrow i_{1} - i_{2} = 1 . . . (1)$
for loop 2: $(2 + 2 + 3) i_{2} - 2 i_{3} - 3 i_{1} = 0 \Rightarrow 7 i_{2} - 2 i_{3} - 3 i_{1} = 0 . . . (2)$
for loop : $(2 + 2) i_{3} - 2 i_{2} = 0 \Rightarrow 2 i_{3} = i_{2} . . . (3)$
from (2) and (3), $7 i_{2} - i_{2} - 3 i_{1} = 0 \Rightarrow 2 i_{2} = i_{1} . . . (4)$
from (1) and (4), $2 i_{2} - i_{2} = 1 \Rightarrow i_{2} = 1 A$
put $i_{2} = 1$ in (1) and (4), $i_{1} = 1 + 1 = 2 A$ and $i_{3} = 0.5 i_{2} = 0.5 A$
Thus the current through branch AB $= i_{1} - i_{3} = 2 - 0.5 = 1.5 A$

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