CGPET 2018 Solved Paper

By Kirchhoff's loop rule;
In loop (1),
−2i1−i3+4i2=0
or 4i2=2i1+i3 .......(i)
In loop(2),
−2(i1−i3)+1(i3)+4(i2+i3)+i3=0
⇒−2i1+2i3+i3+4i2+4i3+i3=0
⇒4i2+8i3=2i1 ........(ii)
In loop (4),
−4i2−4(i2+i3)−4i2+12=0
or 12i2+4i3=12 ........(iii)
From Eqs. (i) and (ii), we get
4i2=4i2+8i3+i3
⇒ 9i3=0 or i3=0 .......(iv)
From Eq. (iv), option (a) is incorrect.
From Eqs. (iii) and (iv), we get
12i2+4×0=12 or i2=1A
Now, from Eq. (i), we get
4×1=2i1+0
⇒i1=2A
Hence, I1=i1+i2=2+1=3A ........(v)
and I2=i1−i3=2−0=2A ........(iv)
From Eqs. (v) and (vi), option (c) and (d) are incorrect.
Now,
VP−VQ=0×1=0⇒Vp=VQ
andVP−VS=2×2=4
∴VP>Vs
orVQ>Vs ........(vii)
From Eq. (vii), option (b) is correct.