The correct option is
B The current through the
3Ω resistor is 1.5A.
In the figure given
let voltages at A,B,C,D be VA,VB,VC,VD.
Let VA>VB.
given VA−VB=16V
let current from A to B be ′i′ (through r1,r4,r5)
let current from C to D through r2 be i1, then current from r3 will be i−i1 from C to D.
On applying 'Kirchhoff's voltage law' in the closed loop with 9V source :-
9−2(i−i1)+1i1=0
⇒i1=2i−93.....(1)
On applying 'Kirchhoff's voltage law' from A to B:-
VA−4i−9−1i1+3−1i−3i−VB=0
⇒16−8i−6−i1=0 (∵VA−VB=16)
⇒30−26i+9=0(∵ from equation (1))
⇒i=1.5A
⇒i1=−2A
So current through r1,r4,r5 is 1.5A.
current through r2 is 2A.
current through r3 is 0.5A.