66in 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
Akash Anand Good workUpvote·0· Reply ·2013-06-20 00:25:40
Soumyadeep Basu Why will current flow through the 6 ohm resistor?- Akash Anand Try to make a single point of the 2 end points of ammeter, you yourself get the answer.
- Soumyadeep Basu Oh, i get it.
Aditya Agarwal sir but it necessarily does not mean that any shorted wire( in this case the ammeter) will behave as a point.- Akash Anand @Aditya..yes it does
