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Please post the method and explanations as well :)

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2 Answers

Soumyadeep Chatterjee ·Jun 19 '13 at 10:13

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

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Akash Anand Good work
Upvote·0· Reply ·Jun 20 '13 at 0:25
Soumyadeep Basu Why will current flow through the 6 ohm resistor?
Upvote·0· Reply ·Jun 26 '13 at 23:52
Akash Anand Try to make a single point of the 2 end points of ammeter, you yourself get the answer.
Upvote·0· Reply ·Jun 27 '13 at 1:08
Soumyadeep Basu Oh, i get it.
Upvote·0· Reply ·Jun 27 '13 at 1:41
Aditya Agarwal sir but it necessarily does not mean that any shorted wire( in this case the ammeter) will behave as a point.
Upvote·0· Reply ·Jul 9 '13 at 7:58
Akash Anand @Aditya..yes it does
Upvote·0· Reply ·Jul 9 '13 at 22:42
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