CGPET 2016 Solved Paper

We have 3 loops
In loop I,

9−3i1−6(i1−i2)=0
9=9i1−6i2 ......(i)
In loop II,

−4i2−(i2−i3)x2-i2+6(i1−i2)=0
6i1−13i2+2i3=0 .....(ii)
In loop III,

−2i3−4+2(i2−i3)=0
⇒ −2i3+2i2−2i3=4
⇒ 4i3−2i2=4 ....(iii)
Substituting for i_1 and i_3 in EQ,(ii) from Eq. (i) and (iii), we get
i1=9+6

i2

9

i3=(−4)+2i+

2

4

6(1+

2

3

i2)−13i2+2(1+

1

2

i2)=0
⇒ 6+4i2−13i2+2+i2=0
⇒ 8−8i2=0
∴ i2=1A
Thus, the potental drop across 4Ω resistor,
=i2×R=1×4=4V