Question

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

• A. $1.5A$
• B. $2A$
• C. $1.33A$
• D. $\infty$

Answer 1

Answer: (A)

Apply Kirchhoff's law
for loop 1: $3i_1-3i_2=3\Rightarrow i_1-i_2=1...\left(1\right)$
for loop 2: $(2+2+3)i_2-2i_3-3i_1=0\Rightarrow 7i_2-2i_3-3i_1=0...\left(2\right)$
for loop : $(2+2)i_3-2i_2=0\Rightarrow 2i_3=i_2...\left(3\right)$
from (2) and (3), $7i_2-i_2-3i_1=0\Rightarrow 2i_2=i_1...\left(4\right)$
from (1) and (4), $2i_2-i_2=1\Rightarrow i_2=1A$
put $i_2=1$ in (1) and (4), $i_1=1+1=2A$ and $i_3=0.5i_2=0.5A$
Thus the current through branch AB $=i_1-i_3=2-0.5=1.5A$