Dear...
Question is very simple
..
See specification of ammerter is 9 ohm and 10 mA
That means it can have maximum current 10 mA with it
So from I which enters through A ...10 mA will goes through Ammeter and then 0.9 ohm resistance and finallt leave through terminal B
Also the net current (I-10) will pass through resistance 0.1 ohm and leave from B
SO final current entering A is I and leaving b is (I-10)+10
So The potential accross AB will be equal to potential accros terminal of Ammeter and accross resistance 0.9 ohm
So aply equation
9(10mA) + 0.9(10mA) = (I-10)mA ( 0.1)
99 = (0.1* I - 1 )
so 0.1*I = 100
or I = 1000 mA
or I = 1 A
Note : for furthure doutb ask !
Good wishes
