Why Is Current in AB Finite in a Circuit with Zero Resistance?

[sbhit2001]

PROBLEM-
http://ijso2013.hbcse.tifr.res.in/files/qna/theory-2008.pdf
questions 1-10 to 1-12.
ATTEMPT AT SOLUTION
In the first question, I am unable to understand how the current in AB is equal to 0.2A when the resistance of AB is 0(galvanometer is ideal). As current is equal to V/R, it would become infinite by this formula as r=0. I also tries using Kirchoff's laws and took the two loops with AB as one side and the sides with the resistances as the other sides but then also I got 2 equations and three unknowns . I fail to understand why current in AB is finite whereas resistance is zero. Please provide some sort of clue...

 

[gneill]

If the galvanometer truly has a resistance of zero then the potential drops will occur elsewhere. Fortunately there are resistances between the galvanometer and the voltage source that serve this purpose.

An ideal galvanometer behaves just like a piece of wire with no resistance, or in the real world, like a wire with negligible resistance compared to other resistances in the circuit.

 

[sbhit2001]

But shouldn't there be a resistance necessarily between A and B due to current flowing from A to B?

 

[gneill]

But shouldn't there be a resistance necessarily between A and B due to current flowing from A to B?

A wire carries current just fine without worrying about resistance.

If you wish, assign a resistance to it (say, r_g), solve the problem then take the limit as r_g → 0. But this is an unnecessary complication.

You can use KVL across the galvanometer just fine by setting its potential drop to zero.