
Sourish Ghosh Is it b?
Akshay Ginodia is the answer (a)?
Dwijaraj Paul Chowdhury No....it's d
Please post the method and explanations as well :)
in amm r-> 0
so I flows through it avoiding the upper 2 resistance (2,2)
so now there's 2&6 ohm in parallel & 3&3ohms in series.
therefore request R equivalent = 3+3+ (1/((1/2)+(1/6)))=6+1.49=7.49 ohms
now V=IR
I=V/R=9/7.49=1.201=1.2A
