AMU Engineering 2018 solved paper Solved Paper

According to given figure,
Applying KVL in loop (i)
500I1+R(I1−I2)=12
(500+R)I1−RI2=12 .........(i)
Applying KVL in loop (ii), we have
R(I2−I1)=−2
RI2=RI1−2 .........(ii)
From Eqs. (i) and (ii), we get
(500+R)I1−(RI1−2)=12
500I1+RI1−RI1+2=12
500I1=10
I1=

1

50

From Eq. (ii),
RI2=R×

1

50

−2
But given that galvanometer G shows zero deflection, hence I2=0.
∴ 0=

R

50

−2

R

50

=2
R=100Ω